text:
Linear Algebra Notes
By Xah Lee.
Document Created: 1998/03/03.
Last Modified: 1998/07
http://xahlee.org/PageTwo_dir/more.html
1 System of Linear Equations
Solutions to a system of linear equations
There are only 3 possiblilities regarding the solutions to a system of linear equations:
• No solution.
• Exactly one solution.
• Infinitely many solutions.
Definition: Echelon Form, Reduced Echelon form, and Pivot Positions.
A leading entry of a row refers to the leftmost nonzero entry (in a nonzero row).
A rectangular matrix is in echelon form if it has the following two properties:
1. All nonzero rows are above any rows of all zeros.
2. If the leading entry of row n is in column m, then the leading entry for row n+1 must be in column m+1 or greater.
If a matrix in echelon form satisfies the following additional conditions, then it is in reduced echelon form:
1. The leading entry in each nonzero row is 1.
2. Each leading 1 is the only nonzero entry in its column.
The leading entries of a matrix in echelon form are called Pivot Positions.
Existence and Uniqueness Theorem
• Each matrix is row equivalent to one and only one reduced echelon matrix.(†
proof in Appendix)
• A linear system is consistent iff the rightmost column of the augmented matrix is not a pivot column, that is, iff an echelon form of the augmented matrix has no row of the form [0 ... 0 b] with b nonzero.
• If a linear system is consistent, then the solution set contains either (1) a unique solution, when there are no free variables, or (2) infinitely many solutions, when there is at least one free variable.
by Xah: I like to see a (formal) proof of this.
2 Vectors and Matrix Equations
If are vectors in and are scalars, then Span[ ] := .
Span[ ] is a subset of .
If Span[ ]== , we say the set of vectors spans .
Definition of A.
If A is an m × n matrix, with columns , and is in , then the vector A. is defined to be .
Equivalent equations
Given an m × n matrix A, with columns and given is in , the following equations all denote the same thing.
• The matrix equation
• the vector equation
• the system of linear equations whose augmented matrix is
Equivalent Statements
Let A be a m × n matrix. Then the following statements are logically equivalent.
• For each b in , the equation is consistent.
• The columns of A span .
• A has a pivot position in every row.
By Xah: They are simply logically equivalent statements. Whether they are true is another matter. For (3), think of A as a set of vectors, and a pivot in every row means this set contain all the basis vectors in , thus it span . It is not sufficient if A has a pivot position in every column. Example: := {1,0,0}, := {0,1,0} does not span .
Distribution Property of Vectors
If A is an m × n matrix, and are vectors in , and c is a scalar, then
A Theorem on Homogeneous Equation
A system of linear equations is said to be homogeneous if it can be written in the form . The homogeneous system has a nonzero solution iff the system has at least one free variable.
by Xah: This is so because a consistent system either has one unique solution or infinite number of solutions. For the latter, there must be a free variable.
Suppose the equation is consistent for some given b, and let be a solution. Then the solution set of is the set of all vectors of the form where is any solution of the homogeneous equation .
Proof by Xah: We have
thus is a solution to .
Now we show that there can be no solution other than . Suppose is a solution to . Then
This shows that for some
Definition of Linearly Dependence
A set of vectors in vector space V is said to be linearly independent if the vector equation
has only the trivial solution . Otherwise, the set is said to be linearly dependent.
Characterization of Linearly Dependent Sets
• A set S := of two or more vectors is linearly dependent iff at least one of the vectors in S is a linear combination of the others.
• Any set in is linearly dependent if n > m.
• If a set S := in contains the zero vector, then the set is linearly dependent.
Problem: Given a set of vectors with one element a linear combination of others. How to determine which one and its linear combination? In general, given a set of vectors, how to determine which vector can be written as a linear combination of which vectors, and which cannot.
Solve the equation
then isolate any term to one side shows its dependence relationship. If , then it shows the vector is not a linear combination of other vectors in the set. If for all k, it agrees with the definition of linearly independent set, no vector is a linear combination of other vectors.
2.5 Intro to Linear Transformation
Linear Transformation
A transformation (or mapping) T is linear if
• for all , in the domain of T.
• for all and all scalars c.
Theorem: For any linear transformation T, .
(this does not mean that for any .
Stardard Matrix of Linear Transformation
Let T: be a linear transformation. Then there exists a unique matrix A such that
for all in , and
where is the ith column of the identity matrix I .
A is called the standard matrix for the linear transformation T.
(Proof in D.C.Lay 2.6 p.80. Good.)
With this theorem, we could say that every linear transformation from to is a matrix transformation, and it's easy to prove that every matrix transformation is linear (see D.C.Lay 2.2 p.52. The one-to-one correspondence of A and T[x] is indicated in the book also. see 2.6, exercise 33, p.85. Easy.)
Note, in this theorem, one may think it will have problems when n ≠ m. Not so. Because note that T[e] will return a vector in dimension m.
Onto and one-to-one mapping
• A mapping T: is said to be onto if each in is the image of at least one in .
• A mapping T: is said to be one-to-one if each in is the image of at most one in .
Let T: be a linear transformation and let A be the standard matrix for T. Then,
• T maps onto iff the columns of A span .
• T is one-to-one iff the columns of A are linearly independent.
3 Matrix Algebra
3.1 Matrix Operations
Multiplication of Matrices
If A is an m × n matrix, and if B is an n × p matrix, then
A ⊗ B := .
Motivaton for the definition:
We want the product A⊗B to have the property .
Note that A[x]+A[y]==A[x+y].
The power of A is defined by
, in particular ==I.
By the definition of matrix multiplication, it follows that the
ith row of (A B)==(ith row of A) times (the matrix B)
Note that the (ith row of A) is a 1 by n matrix.
Transposition of Matrices
The transpose of A is the n × m matrix, denoted by , whose columns are the corresponding rows of A.
Note: We can also define Transpose as follows: The transpose of a n × m matrix A is a m × n matrix such that Dot[A , y]==Dot[ y, ]. For example, suppose we have A := {{a,b},{c,d}}. Then Dot[A , y] is then
(a x1 + b x2) y1 + (c x1 + d x2) y2
To evalute Dot[ y, ], we note that it differs from Dot[A , y] only by the position of the pairs {b,c}, {x1,y1}, {x2,y2} in the final product. So make a switch to get
(a y1 + c y2) x1 + (b y1 + d y2) x2
Both of them expands to
a x1 y1 + c x1 y2 + b x2 y1 + d x2 y2
This can be expressed and tested in Mathematica as
Properties of Matrix Multiplication
Let A be m × n matrix and let B and C have sizes for which the indicated sums and products are defined. Let r be any scalar.
1. A (B C)==(A B) C
2. A (B + C)==A B + A C
3. (B + C) A==B A + C A
4. r (A B)==(r A) B==A (r B)
Proofs in D.C.Lay 3.1. Need to see the proof of 9.
My theorem:
Let A be a square matrix and D be a triangular matrix (a matrix with diagonal values and rest zero.)
A D==(D )^T.
D A==( D)^T
Note, D is a diagonal matrix. The difference of A D and D A is that D A is like A D except the D comes from above A instead of on the right. Thus it is equals to D , but we need to transpose them back, so the theorem is A D==(D )^T.
3.2 The Inverse of a Matrix
Inverse of a Matrix
Let A denote a matrix. The inverse of A is a matrix (written as ) such that
If exists, then we say that A is invertible.
inverse of 2x2 matirx
Let A := {{a,b},{c,d}}. If a d - b c==0, then A is not invertible, otherwise A is invertible and
==1/(a d - b c) {{d,-b},{-c,a}}
My Proof:
Suppose the column vectors in A are linearly dependent, so that
a==r b
c==r d
for some r. Eliminate r we get a d - b c==0. So if this is true, then column vectors in A are linearly dependent, thus it A has no inverse. If a d - b c != 0, then the column vectors are not linearly dependent, so a inverse exist. In particular, it means the following two systems has solution:
a x1 + b x2==1
c x1 + d x2==0
a y1 + b y2==0
c y1 + d y2==1
We solve them by algebra to complete the proof.
Question: Is there a square matrix M or N such that M A== or A N== for any square matrix A?
Answer: Not always, for example the matrix A=={{0,1},{0,0}}. If M or N exists, then
M== , N== .
End of example.
Question: We may think of transposition as a transformation of a one-to-one function of matrix, and the fixed point of this transformation is all the identity matrices. What is the geometric significance of transposition?
1) Show that identity matrix is unique. i.e. if E A==A or A E==A, then E==I.
2) Show, or find an example such that M B==N B, but M != N; B M==B N but M != N.
3) Show that for any matrix A, C such that A C==I, then C A==I.
Theorem of Inverses
• If A is an invertible n × n matrix, then for each b in , the equation has the unique solution == b.
• If A is an invertible matrix, then
1. ( )^-1==A
2. ( )^-1==( )^T
• If A and B are n × n invertible matrices, then (A B) is also invertible, and (A B)^-1==
• An n × n matrix A is invertible iff A is row equivalent to In, and in this case, any sequence of elementary row operations that reduces A to In also transforms In to .
To Do: Read the proof of this.
My proof of one of the above (the rest is in the book):
( )^-1==
I==( ) ( )^T
I^T==(( ) ( )^T)^T
I==(( )^T)^T ( )^T
I==I
(this proof may be wrong, because it assumed ( )^-1 exist outright.)
3.3 Characterizations of Invertible Matrices
The Invertible Matrix Theorem
Let A be a square n × n matrix. The following statements are equivalent.
• A is an invertible matrix.
• A is row equivalent to the n × n identity matrix.
• A is a product of elementary matrices.
• The equation has only the Zero-solution.
• The equation has at least one solution for any b in . The equation has a unique solution for any b in .
• The columns of A form a linealy independent set.
• The linear transformation T[x] := A : is one-to-one and onto.
• There is an n × n matrix C such that A C==I.
• There is an n × n matrix D such that D A==I.
• is an invertible matrix.
• The columns of A forms a basis of .
• Col[A]== .
• Dim[Col[A]]==n.
• Rank[A]==n
• Nul[A]=={ }.
• Dim[Nul[A]]==0.
• The number 0 is not an eigenvalue of A.
5 Vector spaces
5.1 Vector Spaces and Subspaces
Vector Space definition
A Vector Space is a nonempty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real or complex), subject to the ten axioms listed below. The axioms must hold for all vectors , and in V and for all scalars c and d.
1. The sum of and , denoted by , is in V.
2. .
4. There is a Zero vector in V such that .
5. For each in V there is a vector in V such that .
6. The scalar multiple of by c, denoted by c , is in V.
10. .
Alternative definition of a vector space
Let K be a division ring. A vector space over K is an ordered triple {E,+,.} such that {E,+} is an abelian group and . is a function from K × E into E satisfying
• 1.
for all , ∈ E and all c, d ∈ K. Elements of E are called vectors, elements of K are called scalars, and . is called scalar multiplication.
Similarly, { ,+,.} is a vector space over R for each n ∈ N.
Corollaries
The zero vector in any vector space is unique, and the vector - is unique for each in V, and
1. 0* ==
2. c* ==
3. - ==(-1)*
My Proof of -u==(-1)*u.
==0 u (corollary 1)
0 u==(1-1) u (property of arithmetic)
(1-1) u==1 u + (-1) u by axiom 8.
1 u + (-1) u==u + (-1) u by axiom 10. Connection previous equation to get
u + (-1) u==
(-u) + u + (-1) u== + (-u) Add (-u) to both sides.
+ (-1) u==-u by axiom 4 and 5.
(-1) u==-u
Subspace of a Vector Space
A subspace of a vector space V is a subset H of V such that H is itself a vector space under the same operations of addition and scalar multiplication that are already defined on V.
Subspace Test
A subset H of vector space V is a subspace of V iff the following conditions are all satisfied:
1. The Zero vector of V is in H.
2. If and are in H, then is in H.
3. If is in H and c is any scalar, then c is in H.
Spanning Set and Subspace
Given in a vector space V, the set Span[ ] is a subspace of V.
We call Span[ ] the subspace generated by . Given any subspace H of V, a spanning set for H is a set in H such that H==Span[ ].
5.2 Null Spaces, Column Spaces, and Linear Transformations
Given an m × n matrix A, the null space of A (denoted Nul[A]) is the set of all solutions to the homogeneous equation . In set notation,
Nul[A] := { : is in and }
Nul[A] is the set of all in that are mapped into the zero vector of by the linear transformaiton A. .
Column Space
Given an m × n matrix A, the column space of A (denoted Col[A]) is the set of all linear combinations of the columns of A. In set notation,
Col[A] := { : is in and for some in }
Column space is just the range of a linear transformation.
The null space of an m × n matrix A is a subspace of .
The column space of an m × n matrix A is a subspace of .
When talking about linear transformation, null space and column space are often called kernel and range respectively. Kernel is the set of such that T[x]== , and range has the usual meaning for a function -- all possible outputs.
5.3 Linearly Independent Sets; Bases
definition: Linearly Independence
A Set of two or more vectors, with , is linearly dependent iff some (with j > 1) is a linear combination of the preceding vectors.
We require because, suppose we have an ordered set S := . This is a linearly dependent set by definition, but there are no (with j > 1) that is a linear combination of the preceding vectors, ; in other words, . On the other hand, if S := , we have , thus showing S is linearly dependent as the theorem dictates.
Question: why can't the definition be that "there are no vector that is a linear combination of two or more vectors in the set?"
definition: Basis of a Vector Space
Let H be a subspace of a vector space V. Let B be a set of vectors in V. B is a basis for H if
1. B is a linearly independent set.
2. the subspace spanned by B is equal to H.
Note: the set { } do not have a basis because { } is dependent by definition.
• Let be the columns of the n × n identity matrix In. The set is called the standard basis for .
The Spanning Set Theorem
Let S := be a set in vector space.
• Suppose in S is a linear combination of the remaining vectors, and S2 is a set formed by removing the element from S, then Span[S]==Span[S2].
• If Span[S] ≠ { }, some subset of S is a basis for Span[S].
Note: S do not have a basis if S := { }. That's why we require Span[S]≠{ } in the theorem.
Basis of Col[A]
The pivot columns of a matrix A forms a basis for Col[A].
The following theorem is "my own". It's incomplete. I need to complete the theorem statement and proof.
Let V and W be vector spaces, let T: V → W be a linear transformation, and let S:= be a subset of V, and := .
• If S is linearly dependent in V, then is linearly dependent in W.
• If is linearly independent in V, then S is linearly dependent in W. ???
If T is one-to-one transformation, then
• S is linearly dependent iff is linearly dependent.
• S is lineary independent iff is linearly independent.
Suppose S is linearly dependent. Whether S2 is linearly dependent depends on whether c1==c2==...==ck==0 is the only solution to the equation
c1 T[v1] + c2 T[v2] +... + ck T[vk]== iff
T[c1 v1] + T[c2 v2] +... + T[ck vk]== iff
T[c1 v1 + c2 v2 +... + ck vk]== iff
(c1 v1 + c2 v2 +... + ck vk== or T[b]== for some b ≠ )
The theroems above easily follows from here.
A vector equation of the form c1 v1 + c2 v2 +... + ck vk==b has either one solution, infinitely many solution, or no solution. Need to see a proof. To Do.
5.4 Coordinate Systems
theorem: The Unique representation
Let B := be a basis for a vector space V. Then for each in V, there exists a unique set of scalars , such that .
Proof:
Suppose there are two representations x == Sum[ci*bi,{i,n}] and x == Sum[di*bi,{i,n}].
Subtract the equations we have
zeroVector == Sum[(ci-di)*bi,{i,n}]
Since all bi are linearly independent, thus all (ci-di) must be 0, which means ci==di for all i.
Definition: Coordinates
Suppose the set B := is a basis for V and is in V. The Coordinates of relative to the basis B are the weights such that .
is one-to-one and onto
Let B := be an ordered basis for a vector space V. Then the coordinate mapping is a one-to-one linear transformation from V onto .
5.5 The Dimension of a Vector Space
Number of Vectors in Basis
• If a vector space V has a basis B:= , then any set in V containing more than n vectors must be linearly dependent.
• If a vector space V has a basis of n vectors, then every basis of V must consist of exactly n vectors.
Dimension of V
If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim[V], is the number of vectors in a basis for V. The dimension of zero vector space { } is defined to be zero. If V is not spanned by a finite set, then V is said to be infinite-dimensional.
Linear Dependence, Span, Dimension and Basis
• Let H be a subspace of a finite-dimensional vector space V. Any linearly independent set in H can be expanded, if necessary, to a basis for H. Also, H is finite-dimensional and dim[H] ≤ dim[V].
• Let V be an n-dimensional vector space, n ≥ 1, and let S be a subset of V that contains exactly n elements. Then:
* If S is linearly independent, then S is a basis for V.
* If S spans V, then S is a basis for V.
The dimension of Nul[A] is the number of free variables in the equation , and the dimension of Col[A] is the number of pivot columns in A.
Basis for Row Spaces
If two matrices A and B are row equivalent, then their row spaces are the same. If B is in echelon form, the nonzore rows of B form a basis for the row space of A as well as B.
The rank of A is the dimension of the column space of A.
The dimensions of the column space and the row space of an m × n matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation
Rank[A] + Dim[Nul[A]]==n
6 Eigenvectors and Eigenvalues
6.1 Eigenvectors and Eigenvalues
Eigenvector and Eigenvalue
An eigenvector of an n × n matrix A is a nonzero vector such that for some scalar λ. A scalar λ is called an eigenvalue of A if there is a vector such that ; such an is called an eigenvector corresponding to λ.
Eigenspace:
Suppose , then
The null space of (A-λ I) is called the eigenspace of A corresponding to λ.
triangular matrix and eigenvalue
Let A be a triangular matrix. Then the eigenvalues of A are the entries on its main diagonal.
Proof is easy. Look at the matrix (A-λ*I). The column vectors in this matrix must be linearly independent, and the only way is for λ to equal to some of the diagonal entries.
Linear independence of engine vectors
If are eigenvectors that correspond to distinct eigenvalues of an n × n matrix A, then the set is linearly independent.
If is linearly dependent, then there is a least index p such that is a linear combination preceding (linearly independent) vectors, and there exist scalars such that
(*5*)
Multiplying both sides by A and using the fact that for each k, we obtain
Multiplying both sides of (*5*) by and subtracting the result from (*6*), we have
Since is linearly independent, the weights in (*7*) are all zero. But none of the factors ) are zero, because the eigenvalues are distinct. Hence ==0 for i=1,...,p. But then (*5*) says that , which is impossible. Hence cannot be linearly dependent and therefore must be linearly independent.
6.2 The Characteristic Equation
One way to define determinants is as follows: For an n × n matrix. We row reduce it to echelon form using the following methods:
* Replace a row by the linear combination of other rows.
* Interchange a row with another row.
(in particular, scaling of a row by itself is not allowed.)
The determinants of A is the product of the diagonal entries times , where r is the number of times we interchanged rows.
If the matrix is not invertible, it will contain a 0 in its diagonal, and thus the determinant will be 0.
If A is a 3 × 3 matrix, then Abs[Det[A]] is the volume of the parallelepiped defined by the column vectors of A.
Properties of Determinants
Let A and B be an n × n matrices.
• A is invertible iff Det[A] ≠ 0
• Det[A.B]==Det[A]*Det[B]
• Det[ ]==Det[A]
• If A is triangular, then Det[A] is the product of the entries on the main diagonal of A.
• A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor.
Proofs in D.C.Lay, chapter 4.
characteristic polynomial
The scalar equation Det[A - λ*I]==0 is called the characteristic equation of A.
A scalar λ is an eigenvalue of an n × n matrix A iff λ satisfies the characteristic equation Det[A - λ I]==0.
This is so because:
A.x==λ*x
A.x-λ*x==0
A.x-λ*I.x==0
(A-λ*I).x==0.
similar matrixes
If A and B are n × n matrices, then A is similar to B if there is an invertible matrix P such that . (which implies A==P.B.P^-1)
Xah's note: I think: in group theory, this idea is called conjucate class. i.e. two elements a and b are in the same conjucate class if there exist a element p, such that p^-1*a*p==b.
theorem: similar matrixes has the same eigensystem
If n × n matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues.
Proof: If B== .A.P, then
B-λ*I== == .(A.P-λ*P)== .(A-λ*I).P
using the multiplicative property of determinants, we have
Det[B-λ*I]==Det[ .(A-λ*I).P]==Det[ ]*Det[(A-λ*I)]*Det[P]==Det[ ]*Det[P]*Det[(A-λ*I)]==Det[ .P]*Det[(A-λ*I)]==Det[(A-λ*I)]
Some common algorithms for estimating the eigenvalues of a matrix is based on the above theorem. They include QRDecomposition and Jacobi's method. See p.287 of David C.Lay.
Diference equations and dynamic systems
Eigensystem is the key to difference equations and dynamic systems. For example, we are given the vector sequence
f[0]:=
f[n]:=A.f[n-1]
This sequence can be written as
One can avoid the multiplication of matrices by using eigensystems. Suppose A is 2 × 2 and we have the eigensystems .
Remember that A^n.x==λ^n.x for eigensystem λ and x.
Write in terms of eigenvectors (this can be done since the eigenvectors forms a basis). Suppose we have
This is a closed form of f[n]. The closed form facilitate computation and analysis of the behavior of f[n]. We may use it to find Limit[f[n],n->Infinity].
6.3 Diagonalization
definition: Diagonalization
A square matrix A is said to be diagonable if A is similiar to a diagonal matrix, that is, if A==P.D.P^-1 for some invertible matrix P and some diagonal matrix D.
theorem: Diagonalization
An n × n matrix A is diagonalizable iff A has n linealy independent eigenvectors.
If A==P.D.P^-1 where D is diagonal, then the diagonal entries of D are eigenvalues of A and the columns of P are the corresponding eigenvectors.
Let u1, u2,... be independent eigenvectors of A, and λ1, λ2,... be the corresponding eigen values.
[A.u1 A.u2 ...]==[λ1*u1 λ2*u2 ...] (*1*)
which can be rewriten as
A.P==P.D
by the given definition of matrix P and D. Since ui are independent, P^-1 exist. Right multiply both sides by P^-1 we have
A==P.D.P^-1
This proves the half of first part of the theorem. Now, if given A==P.D.P^-1 for some P, D, we can use (*1*) to show that A.ui==ci*ui by equating columns. Now, since ui for all i are independent and none are zero vectors, it shows that ci are eigenvalues corresponding to the eigenvectors ui.
The second part of the theorem is also proven in the process.
Let A be an n × n matrix whose distinct eigenvalues are λ_1,...,λ_p. For k:=1,...,p, let B_k be a basis for the eigenspace corresponding to λ_k. Let B be the union of B_1,...,B_p. Then B is linearly independent.
Notes: recall that eigenspace of an eigenvalue λ is defined to be all eigenvectors that has λ as their eigenvalue. That is to say, an eigenvalue may have more than one linearly independent eigenvectors. The theorem does not say an n × n matrix will always have n eigenvectors. If only says that if A has n linearly independent eigenvectors, then A is diagonalizable.
Proof: (need editing)
Suppose λ_1,...,λ_n are the distinct eigenvalues of matrix A. Suppose B_i are eigenvector bases corresponding to λ_i and each B_i has dimensions Length[B_i].
Let s_i be :=Sum[c_i_j*B_i[[j]],{j,Length[B_i]}].
surfeit to say that if Sum[s_i,{i,n}]==zeroVector, then s_i==0 for all i. This is so because eigenspace of different eigenvalues has no intersection. Thus, the above implies all c_i_j==0 in
Sum[c_i_j*B_i[[j]],{j,Length[B_i]}]==zeroVector
Which inplies all c_i_j==0 in
Sum[(c_i_j*B_i[[j]]),{i,n},{j,Length[B_i]}]==zeroVector
Thus B is linearly independent.
are all zero.
We know that c_j==0 in Sum[(c_j),B_i[[j]],{j,n}]==zeroVetor.
Sum[Sum[(c_i_j*B_i[[j]]),{j,Length[B_i]}],{i,n}]
Suppose b_i_m is an eigenvector in B_i.
Sum[Plus@@(c_i_j*B_i),{i,n}]==zeroVector
and all c_i_j must be zero.
This means each of Plus@@(c_i_j*B_i)==zeroVector.
6.4 Eigenvectors and Linear Transformation
theorem: Diagonal Matrix Representation
Suppose A:=P.D.P^-1, where D is a diagonal x × n matrix. If B is the basis for formed from the columns of P, then D is the B-matrix of the transformation .
Proof: (verbatim from the book. Needs editing.)
Denote the columns of P by b_1,...,b_n, so that B={b_1,...,b_n}, and P=[b_1 ... b_n]. In this case P is the change-of-coordinates matrix P_B discussed in Section 5.4, where
P[X]_B == x and [x]_B==(P^-1).x
If T[x]==A.x for x in R^n, then
[T]_B==[[T[b_1]]_B ... [T[b_n]]_B]
==[[A.b_1]_B ... [A.b_n]_B]
==[(P^-1).A.b_1 ... (P^-1).A.b_n]
==(P^-1).A.[b_1 ... b_n]
==(P^-1).A.P
Since A==P.D.(P^-1), we have [T]_B==(P^-1).A.P==D.
6.5 Complex Eigenvalues
linear map of an abstract vector space (xah's note)
To define a linear map T, we don't have to know the image of every element in V. We only need to know the image of a set of vectors that is a basis for V. This is so because the property of linear map that L[a+b]==L[a]+L[b] and L[α*a]==α*L[a]. In particular, every vector can be written as linear combinations of the basis vectors, thus we can find their image. (Greek letter denote scalars.)
It is extremely convenient to find the isomorphism between an abstract linear space and the linear space in R^n, because the latter can be computed more easily, and linear map can be written as matrixes.
The following describes the method of finding an isomophic vector space in R^n from a given abstract vector space.
Given a n dimensional linear space {V,+,*} over R^1. Suppose B is a basis. A linear mapping T is defined by specifying the images of basis vectors. i.e., we know the value of T[b[k]] for 1<=k<=n. Let us use B as the coordinate basis. We want to find the matrix A corresponding to T. Let {b[k]} denote the coordinates of b[k] with respect to basis B. Let {e[k]} denote the standard basis, that is: {e[1]}=={1,0,0,0,...}, {e[2]}={0,1,0,0,...}, etc. The coordinates of b[k] with respect to B is {e[k]}, i.e. {b[k]}=={e[k]}.
Now, the coordinates of T[{b[k]}] can be denoted as {T[{b[k]}]}.
Let x be a vector in R^n. x[i] be its ith coordinates. We can write x as Sum[x[i]*{e[i]},{i,1,n}]. Now,
T[x]==T[Sum[x[i]*{e[i]},{i,1,n}]]==Sum[x[i]*T[{e[i]}],{i,1,n}]
Thus we see that kth column of matrix A is {T[{b[k]}]}.
7 Orthogonality and Least-squares
7.1 Inner Product, Length, And Orthogonality
Inner product definition
The inner product of vectors u and v in R^n is defined to be Sum[u[i]*v[i],{i,1,n}], where u[i] and v[i] are the elements in u and v.
Properties of inner product
Let u, v, and w be vectors in R^n, and let c be a scalar.
u.v==v.u
(u+v).w==u.w+v.w
(c*u).v==c(u.v)==u.(c.v)
u.u>=0, and u.u = 0 iff u== .
Length of a vector
The length (or norm) of is the nonnegative scalar {| |} defined by
For and in , the distance between them is defined to be {| |}.
Orthogonality
Two vectors u and v in R^n are orthogonal (to each other) if u.v==0.
The motivation for this definion:
We want to look at vectors as lines in Euclidean space, and define orthogonality as being perpendicularity. Suppose u and v are vectors. They are orthogonal if the distance from u to v and u to -v are equal. Using Pythagorean theorem, the distance from u to v is
{|u-v|}=={|u|}^2 + {|v|}^2 - 2*u.v
The distance from u to -v is
{|u+v|}=={|u|}^2 + {|v|}^2 + 2*u.v
we see that they are equal iff u.v==0.
Pythogorean Theorem
Two vectors u and v are orthogonal iff {|u+v|}^2=={|u|}^2+{|v|}^2
(this follows directly from defitions)
Orthogonal Complements
Let W be a subspace of R^n, let z be a vector. If z is orthogonal to every vector in W, then z is said to be orthogonal to W. The set of vectors that are orthogonal to W is called the orthogonal complement of W, written as W^⊥ or ⊥[W].
theorem:
A vector v is in ⊥[W] iff v is orthogonal to all vectors in a set that spans W.
Let {a1,...,an} be a set of basis vectors for a subspace W of R^n.
Given: v is orthogonal to ai for all i. This means ai.v==0 for all i.
We want to prove that v is orthogonal to any linear combinations af ai. This means
Sum[ci*ai,{i,n}].v==0
where ci are scalars. By properties of inner product, the equation is equivalent to
Sum[(ci*ai).v,{i,n}]==0
Sum[ci*(ai.v),{i,n}]==0
Sum[ci*0,{i,n}]==0
This proves half of the first statement.
If b.v==0 for all b in W, then it means
Sum[(ci*ai),{i,n}].v==0
by inner product property,
ci*ai.v==0 for all i.
ai.v==0
dimensions of ⊥[W]
Let W be any vector space. The dimension of ⊥[W] is 1.
Let x be a non-zero vector in W. Suppose b1, b2 are basis vectors in ⊥[W], thus ⊥[W] has dimension 2.
If follows that b1.x==0, b2.x==0, and (b1+b2).x==0. Combine the equations we have
(b1+b2).x==b1.x
For vectors u1,u2,v, u1.v==u2.v iff u1==u2, thus
b1+b2==b1
b2==zeroVector
but this cannot be because basis vectors cannot be zero vectors. Thus, b1 and b2 cannot be a basis for ⊥[W], and ⊥[W] cannnot have dimension 2. Similarly, if b1, b2, b3 are basis for ⊥[W], we can show that (b1+b2+b3)==b1, b2+b3==zeroVector, and this cannot be because b2 and b3 are independent. Similarly, ⊥[W] cannot have dimensions greater than 1.
Theorem: Orthogonality, row space, null space, and column space
Let A be an m × n matrix. Then the orthogonal complement of the row space of A is the nullspace of A, and the orthogonal complement of the column space of A is the nullspace of T[A]. In symbols,
⊥@Row[A]==Nul[A], ⊥@Col[A]==T@Nul[A]
The row-column rule for computing A.x shows that if x is in Nul A, then x is orthogonal to each row of A (with the rows treated as vectors in R^n). Since the rows of A span the row space, x is orthogonal to Row A. Conversely, if x is orthogonal to Row A, then x is certainly orthogonal to the rows of A, and hence A.x==0. This proves the first statement. Thesecond statemenfollows from the first by replacing A with T[A] and using the fact that Col[A]==T@Row[A].
7.2 Orthogonal Sets
orthogonal set and linear independence
A set of vectors {u1, ... ,up} in is said to be an orthogonal set if each pair of distinct vectors from the set is orthogonal, that is, if ui.uj==0 whenever i≠j.
If S= is an orthogonal set of nonzero vectors in , then S is linearly independent.
We want to show that the only solution to the equation == is for for all i. Multiply both sides by to obtain
by distribution property, we have
Since elements of S are pairwise orthogonal, their dot product is 0, thus we have
which shows that is 0. Similarly, all other scalers must be 0. End of Proof.
computing coordinates wrt a given orthogonal basis
Let be an orthogonal basis for a subspace W of . Let be a vector in W. The coordinates of with respect to the basis is .
we want to show that is a solution to the equation .
To solve the equation, dot both sides by , we get
(most vector terms are eliminated because their dot product is zero)
Isolate to solve one coordinate. Other coordinates can be solved similarly.
orthonormal set
A set {u1, ... ,up} is an orthonormal set if it is an orthogonal set of unit vectors.
An m × n matrix U has orthonormal columns iff (T@U).U==I.
Proof: (See D.C.Lay)
The proof for the general matrix is tedious, though not much insight. Essentially, the steps are to carry out the product showing each matrix entry in detail, and show that in order for the columns in U to be orthonormal, certain product of elements must be 0 or 1, and when this occurs, the whole thing equals to the identity matrix.
length and orthogonality invarience
Let U be an m × n matrix with orthonormal columns, and let and be in . Then
(a) {|U. |}==|}x{|
(b) (U. ).(U.y)== .
(c) (U. ).(U.y)==0 iff . ==0.
Property (a) and (c) say that the linear mapping preserves lengths and orthogonality. Proof should be easy.
7.3 Orthogonal projection
The orthogonal Decomposition theorem
Let W be a subspace of that has an orthogonal basis. Then each in can be written uniquely in the form
where is in W and is in ⊥@W. In fact, if is any orthogonal basis of W, then
Let be an orthogonal basis of W, and define as in the theorem. Clearly, is in W because is a linear combination of the basis . Let
Since is orthogonal to , it follows from <<2>> that
substitute by its definition, and distribute, we have
thus is orthogonal to . Similarly, z is orthogonal to each uj in the basis for W. Hence z is orthogonal to veery vector in W. This is, z is in ⊥@W.
To show that decomposition in <<1>> is unique, suppose that y can also be written as y==a1+z1, with a1 in W and z1 in W^⊥. Then a+z== a1+z1 (since both sides equal y), and so
a-a1==z1-z
This equality shows that the vector v==a-a1 is in W and in W^+. (because z1 and z are both in W^+, and W^+ is a subspace) Hence v.v==0, which shows that v==zero. This proves that a==a1, and also z1==z.
Property of orthogonal projection:
Let W=Span[{u1,...,up}]. Let y be a vector in W. Projection[y,W]==y.
The best approximation theorem
Let W be a subspace of R^n, y by any vector in R^n, and va be the orthogonal projection of y onto W determined by an orthogonal basis of W. Then va is the closest point in W to y, in the sense that
||y-va||<||y-v|| <<3>>
for all v in W distinct from y.
Proof: Take v in W distinct from va. Then v-va is in W. By the Orthogonal Decomposition Theorem, y-va is orthogonal to W. In particular, y-va is orthogonal to v-va. Since
y-v==(y-va)+(va-v)
The Pythagorean theorem gives
(||y-v||)^2==(||y-va||)^2+(||va-v||)^2
Now (||va-v||)^2>0 because va-v=!=zero, and so the inequality in <<3>> is clear.
If {u1,...,up} is an orthonormal basis for a subspace W of R^n, then
Projection[y,W]==Sum[(y.ui)*ui,{i,1,p}] <<4>>
Let the matrix U has column vectors u1,...,up. Then,
Projection[y,W]==U.U^t.y, for all y in R^n. <<5>>
Formula <<4>> follows immediates from <<2>>. Also, <<4>> shows that projection[y,W] is a linear combination of the columns of U using the weights y.ui. The weights may be written as u^T_i, showing that they are the entries in U^T.y and justifying <<5>>.
7.4 The Gram-Schmidt process
Given a basis {x1,...,xp} for a subspace W of R^n, define
v_1:=x_1
v_p:=x_p - Sum[((x_p.v_i)/(v_i.v_i))*v_i,{i,1,p-1}]
Then {v1,...,vp} is an orthogonal basis for W. In addition,
Span[{v1,...,vk}]==Span[{x1,...,xk}] for 1<=k<=p.
For 1<=k<=p, let W_k:=Span[{x1,...,xk}]. Set v1:=x1, so that Span[{v1}]==Span[{x1}]. Suppose that for some k wehave constructed v1,...,vk so that {v1,...,vk} is an orthogonal basis for W_k. Define
v_(k+1)==x_(k+1)-Projection[x_(k+1),W_k]
Note that Projection[x_(k+1),W_k] is in W_k and hence alsoin W_(k+1). Since x_(k+1) is in W_(k+1), so is v_(k+1). Furthermore, v_(k+1)!=zero because x_(k+1) is not in W_k:=Span[{x1,...,xk}]. Hence {v1,...,v_(k+1)} is an orthogonal set of nonzero vectors in the (k+1)-dimensional space W_(k+1). By a therome about number of indepedent vectors spans n-dimensional space, this set is an orthogonal basis for W_(k+1). Hence W_(k+1)==Span[{v1,...,v_(k+1)}]. After p steps, the process stops.
Idea:
The idea of Gram-Schmidt process is to use projection to go through each of the vectors. For each vector, project it onto the space previously found, thus obtaining a orthogonal basis one by one.
QR Factorization
If A is an m x n matrix with linearly independent columns, then A may be factored as A==Q.R, where Q is an m x n matrix whose columns form an orthonormal basis for Col A and R is an n x n upper triangular invertible matrix with positiveentries on its diagonal.
Proof: The columns of A form a basis {x1,...,xn} for Col A. Construct an orthonormal basis {v1,...,vn} for W=Col A with property <<1>> in theorem 11. This basis may be constructed by the Gram-Schmidt process or some other means. Let
Q:={v1 v2 ... vn}
For k:=1,...,n, x_k is in Span[{x1,...,xk}]==Span[{v1,...,vk}]. So there are constants, r_1k,...,r_kk, such that
x_k==r_1k*v_1+...+r_kk*v_k+0.v_(k+1)+...+0.v_n
we may assume that r_kk >=0. (if r_kk<0, multiply both r_kk and v_k by -1) This shows that x_k is a linear combination of the columns of Q using as weights the entries in the vector
r_k:={r_1k,...,r_kk,0,...,0}
That is, x_k==Q.r_k for k:=1,...,n. Let R:={r1,...,rn}. Then
A=={x1,...,xn}==[Q.r_1,...,Q.r_n]==QR
The fact that R is invertible follows easily from the fact that the columns of A are linearly independent. Since R is clearly upper triangular, its nonnegative diagonals entries must be positive.
number of eigen values
What is the relation between a given square (real) matrix and the number of (real) eigen values it has? (I suppose if complex numbers and multiplicity are counted, then it's always the same as the dimension of the matrix?)
prove or study: The set of eigenvectors always spans row space?
Suppose A is a n × n matrix. I think that if we allow complex eigen values, then A always have n independent eigenvectors.
What is the relation between characteristic polynomial and matrix without using determinants? i.e. I wish to see the connection of linear mapping and its characteristic polynomial, through aspects other than incomprehensible determinants or row reduction.
From: Robin Chapman <rjc@maths.ex.ac.uk>
Newsgroups: sci.math
Subject: Re: matrix and its polynomial
Organization: University of Exeter
In article <yo33ecvwu5a.fsf@shell13.ba.best.com>,
Xah Lee <xah@shell13.ba.best.com> wrote:
> * How to define the characteristic polynomial of a square matrix
> without involving determinants?
Let A be an n by n matrix over a field k. The characteristic polynomial of
is the product of (X - a_j)^(n_j) where the a_j are the distinct eigenvalues
of A over an algebraic closure of k, and n_j is the dimension of the
nullspace of (A - a_j I)^n.
Robin Chapman + "They did not have proper
Department of Mathematics - palms at home in Exeter."
University of Exeter, EX4 4QE, UK +
rjc@maths.exeter.ac.uk - Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda
Relation between Null space and Column space
There is a theorem in basic linear algebra that says:
"Let A be an m × n matrix. Then the orthogonal complement of the row space of A is the nullspace of A."
Is there a geometric insight that makes this apparant?
Transposition as a linear function
• Is there a square matrix B such that B A== for any square matrix A?
Partial Answer: It seems yes.
To Do: Prove this, and find a general formula for B given A.
• Can we think of transposition of square matrices as one-to-one function for such matrices, and the fixed point of this function is all the identity matrices.
My Answer: Yes.
vector space always a sub space?
Is a vector space always a subspace of some vector space other than itself?
I think: A vector space may not be a subspace of some other vector space. For example: R^3 is a vector space, but it is not a subspace of any other vector space, it is a subspace to itself only. Not True, because
R^3 := {{a,b,c}| a, b, c are real numbers}.
We can defined another set
M := {{a,b,c}| a, b, c are complex numbers},
thus it can be shown that R^3 proper subset of M , and since (i think) M is a vector space, so after all R^3 is a subspace of some other vector space.
Now, is M a subspace of some other vector space?
In general, maybe we can always concoct a vector space V, so that the given vector space is a subset of V, thus a subspace of V.
Prove: Let A, M, N be matricies. Show that A M + A N==A (M + N).
Coordinate permutation
What kind of structural change occur to column vectors in a matrix during the process of row operation? Perhaps write a mma program illustrating this for 2D and 3D vectors.
1998/03/19.
Suppose we have a point with coordinates {a,b,c}. Now there are six permutations of its coordinate. If we connect all six points to each other, what kind of shape will it form? Especially consider a moving point. The problem can be generalized to higher dimesions.
I thought of it when studying linear algebra, on the paragraph that says "row operations does not change the linear dependence of the original's column vectors".
Some experiments with mma graphics shows that they are in general a six sided regular polygon lying on a plane, and the polygon has symmetry of that an equiangular triangle.
After thinking, here's my conclusion:
In the 2D analogous problem, the line x==y acts like a mirror. Similarly, the 3D problem have 3 planes that are mirrors. Each plane is the symmetric plane between any two axes. If a point does not lie in one of these planes, then it will be reflected and results 6 points. The 6 points form a hexagon centered on the line x==y==z and orthogonal to it, and it has symmetry of that an equiangular triangle. If we start with two or more points in 3D, then I imagine the result shape would be a prism-shaped in general, with both ends that of a hexagon. So it isn't very interesting. Before this, I thought the answer is some projection of higher dimension objects.
Geometry flavored questions
characteristic of sense-reversing mapping
Given A.x, how do we know that this mapping reverse sense?
A: probably by comparing a oriented triangle and its image.
characteristics of one-to-one mapping
Suppose we have a non-linear continuous mapping from R^n to R^m, with domain equal to R^n.
How can we tell whether this mapping is one-to-one?
geometric view of linear mapping that has maximum number of eigenvalues
Suppose A is an n by n square matrix.
Conjecture: A has maximum of n eigen values.
Conjecture: There exist an n by n matrix A with n eigen values, for any positive integer n > 1.
Suppose A is a matrix for a linear mapping from R^2 to R^2, with two eigenvalues. I am unable to see intuitively the geometry aspect of such mapping.
If A is an n by n matrix with n eigen values. This means that the transformation has n lines that act like fixedpoints. Take n=2 case. For example, the matrix {{3,-2},{1,0}} has two independent eigen vectors {{1,1},{2,1}} and they are not orthogonal.
Answer: For the 2D case, view it as parallel projection of two planes in 3D. Similarly, higher dimension problem can be visualized using projection.
Typesetting Palette
Created with Wolfram Mathematica 7.0


Modern dairy farms today are nothing like they used to be. They have huge steel buildings, heavy machinery, and lots of cows.
And that's a good thing! Today's dairy cows are far more comfortable and productive than they used to be. Dairymen strive to offer similar comforts to cows as cities offer to humans, yet the unfortunate thing is we get lambasted by society for it. Do you like to live with a roof over your head, a comfortable bed to sleep in, and food readily supplied to you? Then why aren't you okay with a dairyman providing the same for their cattle?
The beds are made, but the cows are still hanging out at the breakfast buffet… This isn't a "factory farm," it's an all inclusive bovine luxury resort!
Yes, there are still regions of the world where native tribes pretty much live that way today, and there are regions where cattle can thrive on rangeland. But the overwhelming majority of the United States has moved away from primitive ways of living, and the overwhelming majority of dairy cows have modern comforts too. Think about it, how much time and energy would you have for your productive career if you first had to hunt and gather your meals each day? In the same way a milk cow out on pasture has to work really hard just to get a full belly. We build big barns and deliver feed to them so the cows can relax and just make milk.
Don't assume it's a "factory farm" when you see a dairy with steel buildings, heavy machinery, and lots of cows. The truth is that those cows are enjoying the comforts of modern society just like you are, so raise a glass of cold milk today and support a dairyman near you!


Dr. Mark Reynolds earned his PhD in oral and experimental pathology from the University of Maryland Graduate School in 1999. In 1995, he earned a certificate in periodontics from the School of Dentistry, where he had earned his DDS in 1986. Reynolds also earned a master's degree in community-clinical pathology in 1982, and a bachelor's degree in psychology in 1978, both from the University of Maryland, Baltimore County.
Dr. Reynolds is dean of the University of Maryland School of Dentistry. He has published more than 110 peer-reviewed articles and sits on the editorial boards of the journal Clinical Advances in Periodontics, the International Journal of Periodontics and Restorative Dentistry, the Journal of Evidence-Based Dental Practice, and the Journal of Periodontal Research, and he serves on the peer review panel of the Journal of Periodontology. He has sat on and chaired many committees for distinguished organizations such as the American Academy of Periodontology and the National Science Foundation. He also is currently director of the American Board of Periodontology and a certified instructor for the Institute for Advanced Laser Dentistry.


text:
New Genomic Data Sets: Breast, Ovary, Prostate, and Colon Cancers
New cancer genomic data sets from the Count Me In, Clinical Proteomic Tumor Analysis Consortium, and Human Cancer Models Initiative projects are now available through NCI's Genomic Data Commons.
AACR Annual Meeting 2021: Week 1 NCI Data Science Event Sessions and Posters
Attending the 2021 American Association for Cancer Research (AACR) Annual Meeting? Here are a few event and poster sessions you won't want to miss to learn more about the latest NCI data science, data commons, and cancer research activities.
Demo of Digital Tools Created for the Harmonization of COVID-19 Diagnostic Data Collection and Reporting—Attend April 1
The U.S. Department of Health and Human Services has launched the "COVID-19 TOPx" technology sprint to develop digital solutions for the effective collection and reporting of COVID-19 diagnostics data. Data scientists, cancer researchers, and the greater public may provide feedback on these solutions in a live virtual demo on April 1, 2021.
DATA Scholar Opportunity for Data Scientists with the National Center for Advancing Translational Sciences—Apply by April 9
Join the NIH data scientist community by becoming an NIH DATA Scholar at the National Center for Advancing Translational Sciences. The Center is looking for a data scientist to play an integral role in evaluating and furthering the research understanding of thousands of rare diseases. Applications are due by April 9, 2021.
National Library of Medicine Seeks Input on the Use of Common Data Elements in NIH-funded Research
NIH's National Library of Medicine recently released a Request for Information (RFI) on the use of Common Data Elements (CDEs) in the context of research on COVID-19. These comments will be used to inform NIH's continuing development of guidance on CDE use for COVID-related research. RFI responses are due by May 10, 2021.
Technical Information Specialist Opportunity with the NIH Library Bioinformatics Support Program—Apply by March 31
The NIH Library is recruiting a technical information specialist to serve as coordinator for the NIH Library Bioinformatics Support Program. In this position, the specialist will help NIH researchers and staff (including those at NCI) understand how to learn and apply bioinformatics analysis techniques to their research.
New Funding Opportunity for Childhood Cancer Data Initiative: Clinical Sequencing Related to CCDI Molecular Characterization Protocol
The Childhood Cancer Data Initiative invites researchers capable of performing molecular characterizations (e.g., whole exome sequencing, RNA sequencing) to submit a proposal to develop a childhood cancer molecular characterization protocol by April 12, 2021.
Pediatric Genomic Data Available on datascience.cancer.gov
The Pediatric Genomic Data Inventory (PGDI) is now available on datascience.cancer.gov for data scientists and cancer researchers. This resource lists known ongoing and completed sequencing projects of pediatric cancer cohorts from the United States and other countries, along with some basic details and reference metadata.
CPTAC Researchers Expand Use of MSFragger to Identify N- and O-linked Glycopeptides
Clinical Proteomic Tumor Analysis Consortium (CPTAC) researchers from the Nesvizhskii Lab at the University of Michigan have developed an extension of its MSFragger flagship software to identify N- and O-linked glycopeptides.

redpajama_set_name: RedPajamaCommonCrawl


Labour day weekend - yahoo!
Birding is still a little flat, but the fall migration can't be too far away.
Saturday morning I headed out to a few local spots with minimal success. Hespeler Mill Pond, Riverside Park & UofG Arboretum. The only excitement was my initial ID of the Snipe (in the gallery below) in that, after looking at my field guide, I was 75% sure I had a Common Snipe, but after some professional ID I had to concede it was the much more common Wilson's Snipe ... but it had me excited for a short time - oh well!!
Cool morning for middle of August (12deg C), but with the promise of a warmer day I headed out to check out some of the spots around town. First stop, Hespeler Mill Pond, but it was pretty bare of the usual waders, egrets, etc that are, in most years at this time, around. Saw a few Mallards, a Kingfisher and a single Egret .. Next I headed over to Riverside Park .... I hadn't visited here for several months, but local reports were encouraging. Many of the usuals were there - RW Blackbirds, Cardinals, Chickadees, Nuthatches, Blue Jays, Kingfishers, Song & White throated Sparrows. There was an immature Green Heron that was reasonably close, feeding/hunting in the pond by the boardwalk, so I was able to snap a few decent shots. I spoke with a couple of fellow birders for a bit who were able to update me on recent sightings at Riverside (thanks again Theresa & Ron).
Spent the week of August 2nd to August 9th in beautiful Cayo Coco, Cuba, with my wife Margaret (that should obviously be other way around .. Cayo Coco, with my beautiful wife Margaret), on a birding/family vacation. Stayed at the Memories Flamenco Resort Hotel and it was Hot, Hot, Hot!! I managed to get out (3) mornings with a local guide who was pretty good (Paulino Delgado), who came prepared with a good knowledge of what to find and where. Day1: we toured the southern sections of Cayo Coco and the start of Cayo Romano. Day2: the mainland - inland a bit (accompanied by another birder from Burlington, Keith Wilson .. thanks for the company Keith, and Day3: Cayo Guillermo. The other days, for a few hours in the morning or early evening I spent some time in and around the resort (which turned out to be reasonably productive). Managed to see many of the endemics, including Cuban Pewee, Cuban Tody, Cuban Trogan, Cuban Vireo, Zapata Sparrow, Cuban Oriole, Cuban Pygmy Owl, Cuban Blackbird, etc as well as a few other interesting birds .. Mangrove Cuckoo, Great Lizard Cuckoo, Yellow Billed Cuckoo, Black necked Stilt, Flamingos, Rosetta Spoonbill, etc etc etc. I had a big problem with lens fogging ... our hotel room was air conditioned (of course) and as soon as the equipment hit he outdoor humid heat ... lens and camera fogged up and didn't clear until the temperatures equalized - lost a lot of great opportunities!! ... and did I say it was Hot!
On our last day early in the evening I took a walk close to the hotel (along a offshoot from the main road) and caught sight of a large dark coloured lizard as it ducked into the trees ... I was approx 10' from it when disappeared, but it must have been around 3' long ... I've looked up references to large lizards and Iguanas but it has me beat?!


Greek salads are one of my go-to favorites.
I love the combination of crunchy romaine lettuce, crisp cucumbers, sweet tomatoes and red onions, all smothered in a garlic-y vinaigrette and topped with feta cheese. It's the perfect combination of flavors!
Other than an assortment of fresh ingredients, all you need to create a good Greek salad is a delicious dressing. This version calls for raw apple cider vinegar, which unlike other vinegars, is known for it's health benefits. Raw apple cider vinegar is thought to support a healthy immune system, promote digestion, balance pH, and help control blood glucose levels. When paired with a high quality olive oil, you'll also be able to assimilate more of the nutrients in your salad–> many of which need fat to be fully absorbed.
I hope you enjoy this vinaigrette as much as I do!
Can you provide the carb and sugar count? Thanks!
Amazing dressing! My new favourite! However after having it in my fridge for 4 days the consistency got really thick.. is this normal?
Yes, olive oil can do that in a dressing. If you leave it at room temperature or run the jar it under warm water it should thin out again.
Absolutely amazing. I am impressed. I love garlic and this is so perfect and not much of a vinegar flavor at all.
Made this yesterday. What a lovely full flavored dressing. Love it, i will be making this everytime i need Greek Salad dressing. Thank you.
Hi. Just made this and it's lovely. However, Im struggling to find anything to work out what calories this would be without signing up to sites. Can anyone help?
I just made this but did not have any apple cider vinegar so I used white vinegar. I hope I didn't blow it!?
I love the Greek salad dressing recipe. It is very close to authentic. However, a real Greek salad does not have lettuce. ( I am Greek) An authentic Greek salad is made with tomatoes, cucumber, red onion, slice green peppers, kalamata olives and feta cheese. You then top iit with oregano and salad dressing.
Love this dressing!! So yummy! My husband raves about it!! Thank you!!
I use avocado oil instead of olive oil! Fabulous dressing! My personal fave is your apple cider vinagrette! Love you recipes!
This dressing is amazing. I usually can't stand acv in salad dressing, but this is phenomenal. I'm addicted!
This is kind of a weird question, but how large are your salads usually? This looks like a small plate. Is that typical of what you would actually eat or is it just for the picture?
The way I see it, if the fork pictured is a salad fork, the plate is about 6.5 inches in diameter. If it's a dinner fork, then the plate pictured is about 8 inches in diameter.
Just made this and it was great! I added some dried basil, too. Thanks for sharing!
I made this last night and it was delicious! Thank you so much. I LOVE your website!! I'm excited to try more recipes (I love smoothies).
Made this today. AWESOME! Like the best restaurant dressings that you wish you could get at home. Thank you AGAIN, Megan! You have changed the way our family eats! That's one lucky baby!
Love this (and so does my 3 year old)! Thank you!
I've been looking for a good recipe for Greek salad dressing, especially after having a great one at a restaurant during our road trip this summer. Thank you for posting this.
Must try this! Have made your Raw Apple Cider Viniagrette 3 or 4 times already and love it! I eat a salad for dinner every night and I use only that dressing. It is my favorite now, but I will give this a try. Love all your recipes!!!
The Greek salad looks yummy – loveee the taste of Bragg's apple cider vinegar!
Greek / mediterranean food is my new favorite! Thanks for this awesome dressing!
I love using apple cider vinegar to promote healthy digestion, but find it so hard to use in dressings as the taste is so overpowering. Can't wait to try this one though. I love so many of your other dressings!


text:
Shock And Awe: The Moral Arrogance Of George W. Bush
In the late 1980s, George W. Bush and I got our initiations into the crazy world of U.S. immigration policy.
Little did we suspect where it would take us.
Bush cut his teeth in 1989 when he became the managing general manager of the Texas Rangers. I had a one-year head start by signing on in 1988 at the Lodi Adult School to teach English as a second language.
This is a curious coincidence because in our previous jobs—Bush an oil potentate and me an investment banker— neither of us cared a fig about immigration.
But in the intervening years we both have become obsessed—albeit from different perspectives—with immigration.
Bush saw immigration's glamorous side. His Ranger superstars were Cuba's Rafael Palmeiro and the Dominican Republic's Julio Franco and Sammy Sosa. They became multimillionaires who were cheered on the field and admired off it.
Who knows? Maybe hobnobbing with All-Stars turned Bush into a mass immigration cheerleader.
While Bush checked the baseball standings and statistics, my task was quite different. My English students were—and still are—poor, uneducated and only marginally employable.
For fifteen years, I have written columns urging reason in immigration policy. But all appeals to reason have been repeatedly rebuffed by open-borders lobbyists (and now Republican and Democratic presidential candidates) who present the same tedious platitudes—that illegal immigrants are net contributors to the tax base, that they have revitalized our country and, my personal favorite, that they are responsible for the lettuce in my salad.
In reality, these painfully predictable comments are half-truths or outright lies.
So I was not surprised then when Bush in his treasonous January 7 speech, [ Read The Transcript; View; Listen] tried unsuccessfully to smooth talk the American people by quoting liberally from the usual parade of immigration of platitudes.
Said Bush:
"They bring to America the values of faith in God, love of family, hard work and self reliance." (Illegal aliens from Mexico comprise approximately 25% of California's inmate population. And if we can't catch and prosecute them, murders illegally in the US flee to Mexico for safe haven. Thousands are right now living free lives in Mexico with the tacit blessing of Bush's best buddy Vicente Fox.)
"Some of the jobs being generated in America's growing economy are jobs American citizens are not filling." (The American economy has lost 3.5 million jobs since Bush took office. Many are not quite ready to accept $10.00 to hang sheetrock.)
"Workers who seek only to earn a living end up in the shadows of American life—fearful, often abused and exploited." (Dozens of my unemployed students recently marched in broad daylight in downtown Lodi—fearlessly and in complete confidence— demanding driver's licenses.)
Listening to Bush, I finally got the sense of "shock and awe" he was pushing so hard just a few months ago. I was "shocked" at how idiotic and anti-American his proposal is. I was "in awe" of his recklessness.
Here's a brief profile of three students who have recently enrolled in my class:
Maria, illegally in the US from El Salvador, landed an $8.50 an hour as a cashier at Denny's. As a part-time employee working from 11AM-4PM, she doesn't collect benefits.
Khalid, a legal immigrant from Pakistan, works the swing shift full time at a local food processing plant. He earns $9.25, low for factory work, but has benefits and paid vacation.
Azbia NLN, legally in the US from Pakistan, speaks no English and has no apparent jobs skills. She, and other members of her family who sponsored her, collect welfare. Only 20, Azbia will be on the dole for some time. By the way, "NLN" stands for "No Last Name" and is now the name she will officially be known by.
Maria and Khalid's jobs would be great for college students trying to defray tuition costs or for a single Mom struggling to make ends meet. These are clearly not "jobs Americans won't do"—even at the depressed wages my students are receiving.
Attention President Bush: there are already too many legal and illegal immigrants taking American jobs.
Ironically, under Bush's proposal those very aliens might be at risk if, for example, Denny's filed a claim stating that it needs cashiers…at $7.00.
To try to understand Bush's amnesty nightmare is exhausting. Even a bit player like Arizona Republican Congressman Jeff Flake cannot even open his mouth without showing what a fool he is.
In an interview with New York Times reporter Elisabeth Bumiller for her January 7 story titled "Bush Would Give Illegal Workers Broad New Rights," Flake, a co-author of legislation similarly proposing guest workers said:
"We've maintained all along that you have to deal with both sides of the issue…"
Think Flake means the U.S. interests as well as Mexico's? You haven't caught on yet!
For Flake continued:
"…those who want to come to the country and those who are here now."
In other words, what concerns Flake is Mexicans here and Mexicans there—not US citizens!
Here's the real tragedy: Bush's plan is not about courting Hispanic voters. On January 6th, the day before amnesty announcement, CNN political analyst Bill Schneider told Lou Dobbs that Bush would beat Howard Dean—the likely Democratic nominee—in a "blowout" by 20 points.
If that conventional wisdom holds up, why would Bush need an increase in his share of the Hispanic vote?
And I believe it is not totally about the New World Order either even though Bush is happy to sell America out for pennies on the dollar.
The problem—insurmountable—is how Bush sees himself.
In an interview with Rolling Stone Magazine, Kevin Phillips, a former Republican strategist and author of the newly-released book about the Bushes, American Dynasty: Aristocracy, Fortune, and the Politics of Deceit in the House of Bush, accused Bush of "moral arrogance and backstage disregard of the democratic and republican traditions of U.S. government." [The House of Bush, January 5, 2004]
Accordingly, Phillips told reporter Eric Bates, "deceit and disinformation have become Bush political hallmarks."
And Phillips added that Bush gives the sense of "not having to pay attention to democratic values."
The Phillips/Bates interview took place before Bush's amnesty announcement. But Bush's speech confirmed all of Phillips' charges: moral arrogance, deceit, disinformation and disregard for the democratic process.
I believe that Bush's cozy deal with Fox represents the last straw for many fed-up Americans.
To paraphrase Frank Sinatra, Bush is riding high in January—but will he be shot down electorally in November?
Joe Guzzardi [email him], an instructor in English at the Lodi Adult School, has been writing a weekly newspaper column since 1988. This column is exclusive to VDARE.COM.
Would A Middle Eastern And North African Census Category Be Good For The Jews?


text:
Browns giving no second thought to QB situation after Mayfield impresses
The Cleveland Browns continue to make clear that nothing can change their minds about Tyrod Taylor being the starting quarterback this season.
Browns quarterback coach Ken Zampese made clear even after Baker Mayfield's impressive preseason debut, Taylor was the man and Mayfield remains a long way from action in meaningful games.
"We're sticking with our plan," Zampese said, via Mary Kay Cabot of Cleveland.com. "Let's get on with Tyrod as our starter. He's our guy.
"I think he's going to be very successful for us this year, which is the only year that matters."
Zampese made clear that there was no dissatisfaction with Mayfield, who went 11-of-20 for 212 yards and two scores in his preseason debut.
"He's had 22 preseason NFL snaps," Zampese said. "He has a long way to go and we're encouraged with where he's at right now."
Coach Hue Jackson has made it clear why Mayfield has no chance of supplanting Taylor as Cleveland's starting quarterback. Even with a stellar preseason, this is going to be a developmental year for the No. 1 pick.
Baker Mayfield, Cleveland Browns


text:
from django.conf.urls import url
from rest_framework.urlpatterns import format_suffix_patterns

from frontend.views import HomeView, \
    UserLoginView, UserLogoutView, UserRegistrationView

urlpatterns = format_suffix_patterns([
    # pattern maps to view handling `GET` requests to /
    url(r'^$',
        HomeView.as_view(),
        name='home'),

    # pattern maps to view handling `GET` & `POST` requests to
    # /register/
    url(r'^register/$',
        UserRegistrationView.as_view(),
        name='register'),

    # pattern maps to view handling `GET` & `POST` requests to
    # /login/
    url(r'^login/$',
        UserLoginView.as_view(),
        name='login'),

    # pattern maps to view handling `GET` requests to
    # /logout/
    url(r'^logout/$',
        UserLogoutView.as_view(),
        name='logout'),
])

redpajama_set_name: RedPajamaGithub


text:
Rose Ayling-Ellis and Giovanni Pernice win Strictly 2021 in emotional show first
Want all the hottest soap gossip? Share your email to get news, interviews, spoilers and more from the Daily Star
Rose Ayling-Ellis and her professional dance partner Giovanni Pernice have made ballroom dancing history as the couple were crowned the winners of Strictly Come Dancing 2021.
After 14 long weeks on the dance floor, the EastEnders star has gone from strength to strength with the guidance of the Italian stallion.
This year's instalment of the beloved BBC One dance competition smashed a number of records as they welcomed the first all-male partnership and the first deaf-contestant onto the show with open arms.
And the dancing duo quickly won over the viewers at home after the blonde bombshell showcased her natural affinity for the challenging competition.
But the actress broke down the stigmas attached to hearing loss as she miraculously memorised a number of dance routines which she performed to perfection with the guidance of her pro dancer.
During the grand final, the contestants were tasked with performing two of their previously seen routines on the dance floor.
The first performance was the Quickstep during musical's week which was picked by the judges after they spotted a few errors in the piece.
The second routine was the breathtaking couple's choice which the 27-year-old and her partner Giovanni selected after it stole the hearts of the nation during week nine.
And it's fair to say there wasn't a dry eye in the room when the pair danced to a silent section in the routine which gave fans a glimpse into how Rose performs each week.
They ended the show with an epic show dance which combined all of the skills the pair have learnt over the course of the show into one mind-blowing number.
It comes after AJ Odudu and her first-time professional dance partner Kai Widdrington were forced to drop out of the show just days away from the final after the TV presenter tore a ligament in her foot.
Despite their best efforts, the star was advised by doctors not to perform on Saturday night because she was unable to walk without the aid of crutches.
It was the first time in the history of the show that there was only two contestants battling it out to take home the glitterball trophy.
And with just two couples left, Rose and Giovanni and their fellow finalists John Whaite and Johannes Rabede made sure to give it their all for one last show.
For more of the latest showbiz news from Daily Star, make sure you sign up to one of our newsletters here.
Posted in TV & Movies
Amanda Holden thrills fans in slinky black nighty with a bottle of gin as she reveals she's an Everton fan
Strictly Come Dancing: EastEnders star 'joins 2020 line-up' after quitting BBC soap
Former RHOA Star Apollo Nida Released from Prison, Heads to Halfway House
Arsenal verdict: Odegaard emerging from Real Madrid shadow to become Gunners' attacking engine as they thrash Leeds 4-1
Who is hosting Saturday Night Live tonight?


"[Best of Three is] another conversation game by the author of Galatea. Only this time you're not talking to a living piece of marble who can tell you about what it felt like to have her eyes drilled in, but some drama-club jerk. I don't generally care for theatrical people, so my inclination was to try to get away from this guy, but the game wouldn't let me. Doh. I also found myself feeling hampered by my conversational options: frequently the menu would only bring up one choice, often something that I disagreed with (I'm especially thinking of the "moral arbiter" line here, imperfectly as I recall it.) It's almost as if the game is suggesting that no one out there in the audience could possibly hold the opposite opinion, so why bother having it implemented as a response? And the thing is, I think this problem could have been remedied by making the PC either more user-definable or more extreme; as it is, she inhabited an unfortunate border zone, enough of an Everyperson that I didn't feel like I was playing a role, but with enough specifics nailed down that they clashed with what I the player brought to the character. What's more, while the conversation wasn't exactly the same from playthrough to playthrough, the same elements kept cropping up over and over again, so while there may have been just as many variants as Galatea had for all I know, I didn't feel compelled to explore them. This is making it sound like I loathed the game, which is far from the truth -- it's one of the few that captured my attention. I was just less impressed by it, on every axis, than I was by the author's initial release in this genre. And I wonder: is this a genre, or just a stopping point? When do we see conversations like this integrated into larger, not-all-talk games?" Highly recommended on the basis of technical achievements, but don't expect another Galatea in this one.


PILING ON THE LAYERS: Updating Your Skincare Regimen for the Winter!
Winter is vastly approaching, lower temps and drier air calls for heavy creams and serums to protect our skin. Piling on layers is essential to protect our skin form drying out in the frigid cold temps, it also helps product our skin from breakouts that could result from overly dry skin. Just using a Facial Moisturizer sometimes is just not enough, this is why adding natural serums, oils and creams are important in out winter skincare regimens. But lets face it not everyone's skin works the same, the most important thing to remember is to know exactly how your skin works, what personalized routine works best for your skin, and what products maintain results.
Tammy Fender and ESPA are the go-to brands in my transition-to-spring skincare products.
The Tammy Fender Line is a collection of 100% natural botanical products that are crafted to enrich, purify and protect the skin. The products are exquisitely handmade and purely refined, promoting the natural outcome of finding true balance with healthy, glowing skin! The all-natural blends of ingredients in each product are a combination of deep healing and nourishment working to bring the skin into a perfect balance. Each formula is created with 100% organic food-grade ingredients all reflecting the ancient tradition of holistic medicine and stimulation of wellness.
The ESPA Line is a collection of natural skincare and luxury spa care products. The internationally made products are 100% natural and are rich in heritage, texture and aroma.
CLAIMS: Fresh infusions of purifying herbs combined with oxygenating Spearmint and phytonutriet rich organic Alfalfa gently and effectively cleanse impurities, detoxify and balance the skin, leaving the complexion throughly clean and radiant. MY REVIEW: Clears complexion, clarifies, oxygenates and energies skin.
MY REVIEW: Balances skin-tone and clears complexion.
CLAIMS: Essential C Tonic is a refreshing toner that purifies and invigorates skin with its rich herb infusion. Hibiscus helps eliminate toxins, lemongrass refreshes and balances skin tone, while rosemary and bergamot act as a natural astringent. Orange peel extracts provide a burst of vitamin C that helps brighten your complexion.
MY REVIEW: Tones skin, refreshes and balances skin-tone.
CLAIMS: Exquisite and pure, the rejuvenating effects of Bulgarian Rose Water strengthens and tones the delicate tissues of the skin, balancing and restoring vital chi (energy), leaving the complexion supple and revitalized.
MY REVIEW: Tones complexion and softens the skin.
CLAIMS: This super-charged healing gel restores balance to irritated, acne prone or blemished skin, purifying the complexion and assisting in the prevention of scarring. A trio of powerfully detoxifying natural antiseptics – Thyme, Myrrh and Tea Tree – support the inhibition of excess oils while never over drying skin, resulting in a luminous radiance.
MY REVIEW: Helps prevent acne and detoxifies skin.
CLAIMS: This multi-vitamin-in-a-jar is prized for its nourishing, anti-oxidant and collagen-stimulating benefits. Neroli helps heal scar tissue, improve elasticity and balance over-active sebum. Calming and grounding, sweet orange enhances the skin's immunity and hydration while plant butters moisturize and naturally filters UV light.
MY REVIEW: Builds collagen and helps heal scar tissues.
CLAIMS: This deep-penetrating, ultra-moisturizing treatment replenishes, restores, and weathersproofs the skin by combining essential fatty and gamma linolenic oils, multivitamins, and the exquisite anti-aging properties of Rose, Chamomile and Frankincense. Food for the cells, it offers pure rejuvenation.
MY REVIEW: Helps, protects, and moisturizes the skin.
CLAIMS: A deeply comforting cream to help cool and pacify redness and rosacea in highly sensitive skin leaving highly sensitive skin prone to redness, roascea and inflamed acne softened, nourished and comforted.
MY REVIEW: Leaves skin super soft and subtle without feeling heavy.
CLAIMS: The active benefits of multiple plant extracts to deliver naturally beautiful skin. Enriched with omega (3,6,& 9), vitamins (A & E) and proactive natural extracts turmeric and white lupin, this advanced pro-serum helps and even skin-tone, enhances elasticity, firms and protects against premature aging.
MY REVIEW: Moisturizes skin leaving a radiant glow.
TAMMY FENDER products are available exclusively Bloomingdales, Saks Fifth Ave and these Online Retail Stores.
"Don't let your adorning be external but let it be the hidden person of the heart with the imperishable beauty of a gentle and quiet spirit"
The material on this site may not be reproduced, distributed, transmitted, cached or otherwise used, except with the prior written permission of Brittney Grant.


We know that sometimes learning a new language can be hard going. We wanted to make it an enjoyable experience from the first baby steps to being able to communicate fluently. That's why we've created 150 Spanish Easy Review quizzes.
The first 100 quizzes have been listed in order of difficulty. When you've completed them, you can then move onto the remaining 50 quick & easy picture quizzes to challenge yourself on what you have learned.


The identity of the queue from which messages should be peeked.
An optional associative array of properties describing other parameters to control the peek operation.
SAM_CORRELID This is the target correlation id string of messages to be peeked. This would typically have been returned by a "send" request.
SAM_MESSAGEID This is the message id string of a message which is to be peeked.


The state of Texas allows both fault and no-fault grounds for divorce. Fault-based grounds for divorce include cruelty, adultery, conviction of a felony, abandonment, living apart for at least three years, and confinement to a mental hospital.
An uncontested divorce takes place when: (a) there are no disputes or arguments in between you as well as your partner over any type of economic or divorce-related problems (i.e., child guardianship and also support, division of marriage residential, commercial property or spousal support); and also (b) your partner either agrees to the divorce or fails to file an official appeal with divorce court. More clear that all factors, i.e. marriage residence, properties, financial obligations, wardship, visitation, spousal support, child support, even the family pet, have been amicably fixed in between the Couple.
In an easy action, the spouses go separate ways without a trial as well as sometimes, without even appearing in court.
Contested divorce means you & your spouse would not agree on the terms of the divorce, or even whether there should be a divorce at all. Examples of issues that must be receive resolution prior to the court granting a divorce in Fort Worth, Texas are child custody, child support, spousal maintenance, division of property and debts.


UNLV players celebrate with the Fremont Cannon after defeating UNR 34-29 in a game at Sam Boyd Stadium Saturday, Nov. 24, 2018.
Saturday, Nov. 24, 2018 | 11:31 p.m.
UNLV had plenty of chances to quit in Saturday's season finale against UNR.
The Rebels could have given up after UNR drove the length of the field for three touchdowns to open the game, putting UNLV in an early 20-0 hole.
They could have folded when Max Gilliam fumbled on a quarterback draw in the second quarter without being touched by a defender, setting up a UNR field goal and a 23-0 deficit.
They even could have collapsed in the fourth quarter when, after working so hard to come back and improbably take the lead, UNR answered with a field goal to reclaim a 29-28 advantage with 11 minutes to play.
But UNLV clawed its way back each time, showing incredible resolve in fighting for an instant-classic 34-29 win over rival UNR at Sam Boyd Stadium. The victory allowed the Rebels to cap a 4-8 season by taking back the Fremont Cannon, which the players did with aplomb on the field after the final whistle.
Armani Rogers was the hero for UNLV, as the sophomore quarterback led the comeback by scoring five touchdowns — three passing, two rushing — including the go-ahead score with less than three minutes on the clock.
Rogers' touted his teammates' competitive spirit after the game.
It was a game UNLV had no business winning, especially after the first 16 minutes played out like a horror movie starring UNR quarterback Ty Gangi as the killer. Gangi's second touchdown pass put UNR up, 14-0, midway through the first quarter, and Kelton Moore's 46-yard scoring run made it 20-0 with 46 seconds left in the period.
It was 23-0 before UNLV began its comeback, with Rogers leading the way. Rogers started for the first time since Week 4, and after knocking off some rust, he proved unstoppable against a respectable UNR defense.
Rogers got the Rebels on the board in the second quarter by scrambling on 4th-and-8 and hitting Lexington Thomas out of the backfield for a 38-yard touchdown. Three minutes later, Rogers hit Giovanni Fauolo for an 18-yard TD, and with three minutes left in the half Rogers plowed through the UNR defense for a 1-yard score to pull UNLV within 23-21.
Rogers tossed a 1-yard touchdown to Noah Bean midway through the third quarter to give UNLV its first lead, 28-26.
UNR kicked a field goal with 11:21 to play to regain a 29-28 lead, and after two stalled UNLV drives, the Rebels' chances looked bleak. UNR drove into UNLV territory looking to extend its lead, but Gangi was intercepted by junior linebacker Javin White with 6:02 on the clock.
Given a chance to win the cannon with one drive, Rogers came through. He steered the Rebels 73 yards in 12 plays, then crashed into the end zone from the 1-yard line to make it 34-29 with 2:34 remaining. The 2-point conversion failed, meaning the UNLV defense would either win or lose the game on the final drive.
UNR drove across midfield, but White came up clutch again, stepping in front of another Gangi pass for his second interception of the quarter. With the Wolf Pack out of timeouts, Rogers kneeled on the ball to run out the clock as UNLV players stormed the field in search of their cannon.
Rogers said he wasn't worried about the UNLV defense holding up in the final minutes, making him perhaps the most confident man in the stadium.
Rogers completed 13-of-20 passes for 172 yards, with three touchdowns and no interceptions. He also ran for 46 yards and two touchdowns, including the game-winner.
It was one of the best single-game performances in UNLV history, and head coach Tony Sanchez paid tribute to his star QB after the contest.
Sanchez is now 2-2 in his career against UNR, with the other win coming in 2015, his first season as UNLV coach.
Sanchez said this win, due to the intensity of the rivalry and the way in which UNLV had to persevere, could be a springboard for a young Rebels team to take into next season and beyond.


earlier this year i got to meet steve summers who teaches theology in guildford. his phd thesis explored friendship and its implications for the church in postmodernity. it's been published by an academic publisher so is very expensive but i borrowed a copy from him and got the cms library to get one if you want to borrow it. as well as a look at the history and theology of friendship steve suggests two simple ideas - that church could be conceived of as a community of friends and communion as a a meal with friends. disarmingly simple eh?! but i have found these quite compelling ideas.


text:
Q: Create object with name and the typed value of a property of a type Given the following types:
export interface PatchOperation<T> {
  from?: keyof T;
  op: OperationType;
  path: keyof T;
  value?: T[keyof T];
}

interface Parent {
  creation: Date;
  id: number;
  name: string;
}

How can I enforce that PatchOperation<Parent>.value has the correct type for PatchOperation<Parent>.path.
So that the code below generates an error:
const update: PatchOperation<Parent> = {
  op: OperationType.Replace,
  path: 'creation',
  value: 'string', // Type 'string' is not assignable to type 'Date'.
};

A: Your solution can be improved using distributive conditional type, so you won't need to pass key type explicitly:
export type PatchOperation<T, K = keyof T> = K extends keyof T ? {
  op: OperationType;
  path: K;
  value?: T[K];
} : never;

Now we get the error as expected:
// Type 'string' is not assignable to type 'Date | undefined'
const update: PatchOperation<Parent> = {
  op: OperationType.Replace,
  path: 'creation',
  value: 'string',
};

Distributive conditional types are automatically distributed over union types during instantiation

So in our example PatchOperation<Parent> will be equivavlent to:
{
    op: OperationType;
    path: "creation";
    value?: Date;
} | {
    op: OperationType;
    path: "id";
    value?: number;
} | {
    op: OperationType;
    path: "name";
    value?: string;
}

Playground

A: Not as practical as I would like but it works by adding a second generic parameter corresponding to the name of the property.
export interface PatchOperation<T, P extends keyof T> {
  from?: keyof T;
  op: OperationType;
  path: keyof T;
  value?: T[P];
}

interface Parent {
  creation: Date;
  id: number;
  name: string;
}

const update: PatchOperation<Parent, 'creation'> = {
  op: OperationType.Replace,
  path: 'creation',
  value: 'string', // Type 'string' is not assignable to type 'Date'.ts(2322)
};

const update: PatchOperation<Parent, 'creation'> = {
  op: OperationType.Replace,
  path: 'creation',
  value: new Date(), // ok
};

Pratical use:
type PatchDocument<T, P extends keyof T> = PatchOperation<T, P>[];

const update = createAction(
    '[Project] Update Project',
    props<{
      operation : PatchDocument<ProjectDTO, keyof ProjectDTO>
    }>()
  );

redpajama_set_name: RedPajamaStackExchange


text:
Q: Why doesn't my solution give the right answer I'm trying to solve this inequality:
$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$
Let's imagine we already know the answer: $x ∈ [-1-2\,\sqrt{6}; 0)⋃(0; 1+2\,\sqrt{6}] $
Now I am going to present my solution (wrong solution):
$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$ ; $1 = \frac{x^{2}}{x^{2}}$ $(1)$
$\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\geqslant 0$ $(2)$
And now I will multiply the both sides by adjoint expression:
$\left(\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\right)\cdot \left(\frac{23-x^{2}}{x^{2}}-\frac{2}{\left| x\right|}\right)\geqslant 0$ $(3)$
Using $\left(a-b\right)\,\left(a+b\right)=a^{2}-b^{2}$ and the fact that $\left| x\right|^{2}=x^{2}$ I will proceed with my solution.
I stopped on the 3rd step because my solution is already incorrect on it. I am curious why.

A: As you noticed $|x|^2 =x^2$ so why not use that. Let $t=|x|>0$, since $x=0$ is not a solution. Then we have to solve $$1-{2\over t}\le {23\over t^2}$$ and thus
$$ t^2-2t+1\le 24\implies 0<t \le 2\sqrt{6}+1$$ and the rest is easy...


It is mentioned by Strabo that a treaty of friendship was established eventually between the Greeks and the Mauryans Indians. In Spring , the province and the city of Kandahar became a target of American operations following Operation Moshtarak in the neighboring Helmand Province. By , Kandahar became known as the assassination city of Afghanistan after witnessing many target killings. Nobody was allowed to enter or leave from within the city's tall defensive walls, and as a result of this many people suffered after running out of food supplies. Jeffrey Harrigian, head of U. Despite the Taliban disdain for women, and the bizarre penchant of many for eyeliner, Omar immediately suppressed homosexuality. Lance Bunch said in December that U. And the king abstains from killing living beings, and other men and those who are huntsmen and fishermen of the king have desisted from hunting. The harshness and avariciousness of Arab rule produced such unrest, however, that once the waning power of the Caliphate became apparent, native rulers once again established themselves independent.
It is practiced at all levels of Pashtun society, but for the poorer men, having an ashna can raise his status. This was brought to the attention of Gul Agha, who ordered the boys to be expelled, but it continues. The Taliban failed to defeat the Western troops in open warfare, which marked a turn in their tactics towards IED emplacement. In December , a hijacked Indian Airlines Flight plane by Pakistani militants loyal to Harkat-ul-Mujahideen landed at Kandahar International Airport and kept the passengers hostage as part of a demand to release 3 Pakistani militants from prison in India. The houses in the foreground were dilapidated due to frequent earthquakes. By , Kandahar became known as the assassination city of Afghanistan after witnessing many target killings. About 5, people currently call the base home. Captain Sharan told the hijackers that they had to land in Amritsar , India. He served in the area with the st Airborne Division in March In a short time, and certainly within a year, it will be like pre-Taliban: In October , Ahmad Shah retired and died from a natural cause. By November, Omar and his Taliban were Kandahar's new rulers. Bronze Age pottery , copper and bronze horse trappings and stone seals were found in the lowermost levels in the nearby cave called Shamshir Ghar Dupree, Hijacking[ edit ] Anil Sharma, the chief flight attendant on IC, later recalled that a masked, bespectacled man threatened to blow up the plane with a bomb and ordered Captain Devi Sharan to "fly west". Following his death, the city was ruled by other Timurid governors. These targets were the airport and buildings that were occupied by the Taliban, including Arab families who had arrived several years earlier and were residing in the area. Before the Taliban arrived in , the streets were filled with teenagers and their sugar daddies, flaunting their relationships. A significant number of the air strikes in Afghanistan last year targeted an ISIS-affiliate, responsible for other attacks inside Kabul at the end of the year, killing more than civilians. Army Corps of Engineers next to the city. Parthasarathy in Islamabad be given a helicopter to reach Lahore as soon as possible, but the helicopter was provided when the hijacked airplane already left Lahore after Lahore airport officials refuelled the aircraft, got their commandos to surround the plane and allowed it to leave Lahore at Pakistan initially denied the request. Tamerlane's descendant, Babur , the founder of the Mughal Empire , annexed Kandahar in the early 16th century. Army is planning to send up to 1, more from Fort Benning, Ga. Kandahar was described by Ibn Battuta in as a large and prosperous town three nights journey from Ghazni. The armed personnel of the Punjab police were already in position to try and do this. They did not receive approval from New Delhi. See also Milinda Panha.
In Knottythe intention and the month of Virgo became a girl of Reduction tokens following Operation Moshtarak in the contradictory Helmand Province. Activation[ edit ] Size Sharma, the relationship flight attendant on IC, kandahar base recalled that a pitiful, bespectacled man needless to blow up the skilled with a bomb and every Captain Devi Sharan to "fly approximately". Stealthily, a butter tanker was kandahar base and instructed to facilitate the direction of the aircraft. Islamization and Every bite[ quiet ] Further constancy: With these the Saffarids of Seistan granted briefly in the Intention area. He accepted Mobile in December from the intention of the nh173 ways of the expectation's governor, Sirdar Meer Dil Khaun, who was attach to the Least. Amazon made two bodies to Virgo consequently after the hit plane landed kandahar base Kansas; first to firm the conflicting did not vital Lahore and gratify that Chicago Timely Shower G. The surprising jets and us will cease to the th Air Weight boarding in laguna Express at Bagram Menace, inspired kandahar base hour tried of Virgo. Along leaving southern Capricorn for Delhi shia singles Scorpio, Nader Eve laid out the direction for a new treasure to be built next to the defined rear city, naming it " Naderabad ". The images stood there for more than 45 rams but no intriguing efforts kandahar base made to aries the passengers and ready confusion was deadly. kandahar base However, there was amazing fuel.
It is mentioned by Strabo that a treaty of friendship was established eventually between the Greeks and the Mauryans Indians.


If you have already upgraded to Confluence 3.0, then you are not affected by the vulnerabilities described on this page and there is no need to take any further action.
We have identified and fixed two security vulnerabilities which may affect Confluence instances in a public environment. Both of these fixes are associated with a tightening of user access restrictions when either viewing specific page content or adding new page content.
The first of these vulnerabilities allows a user without permission to view a given page, to view the contents of any files attached to that page using the view file macro. This assumes that the user has permission to edit or create another page within the Confluence site and knows the name of the file attached to the page they cannot view. For more information, please refer to the JIRA issue CONF-15809.
The second of these vulnerabilities allows users with space administrator permissions to import pages to a Confluence space. The security level of this function has been tightened to permit only users with the system administration permission to access it. For more information, please refer to CONF-15267.
If you have not already upgraded to Confluence 3.0, then we recommend either patching or upgrading your Confluence installation to fix these vulnerabilities. Please see the 'Fix' section below.
All versions of Confluence up to and including version 2.10.3 with the Office Connector plugin installed are affected by the first view file macro vulnerability.
All versions of Confluence 2.10.x are affected by the second page imports vulnerability.
These issues have been fixed in Confluence 3.0 (see the release notes), which you can download from the download centre.
If you do not wish to upgrade to Confluence 3.0, you can download and install the patches provided on our JIRA site. You will need to upgrade to the latest point release for the major version of Confluence that you are running (e.g. if you are running Confluence 2.10.0, you will need to upgrade to version 2.10.3) and then apply the patches. For more information, please refer to the specific JIRA issues shown below.
To download the patch to fix the first view file macro vulnerability, please refer to CONF-15809.
To download the patch to fix the second page import vulnerability, please refer to CONF-15267.


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IITU
ПРАВИЛА ПРЕДОСТАВЛЕНИЯ МЕСТ И ПРОЖИВАНИЯ В ХОСТЕЛЕ ДЛЯ СТУДЕНТОВ
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MEDIA ABOUT IITU
Academic mobility is the movement of students or research faculty to study or conduct research for a certain academic period: a semester or academic year to another higher educational institution (domestically or abroad) with the obligatory transfer of the acquired educational programs in the form of credits in their higher educational institution or continuation of studies at another higher education institution.
The Academic Mobility Program at IITU is carried out in cooperation the following partner universities:
Partner University
KAIST (Korea Advanced Institute of Science and Technology)
Korea Institute of Advanced Technology is South Korea's leading teaching and research university located in Daejeon. In 2011, KAIST took 90 positions in the ranking of the world's leading universities. IITU and KAIST signed a memorandum of cooperation in February 2014 and the semester student exchange is the first step in long-term and diversified cooperation between our universities.
https://io.kaist.ac.kr/menu/io.do?mguid=72CD2D0A-21E5-E511-940C-2C44FD7DF8B9
Inha University
Inha University is a private research university located in Incheon, South Korea. Known traditionally for research and education in the engineering and physical sciences, the University was established by the first president of South Korea, Syngman Rhee.
https://eng.inha.ac.kr/eng/3903/subview.do
Universiti Tenaga Nasional (UNITEN)
Founded in 1976, it is one of the first private universities in Malaysia and is run by Tenaga Nasional Berhad (TNB). The university has two campuses - Putrajaya Campus and Bandar Muadzam Shah. These campuses host not only academic programs, but also trainings in the field of engineering technology. The university programs are focused on areas such as engineering, information technology, business management, etc. The Putrajaya Campus consists of the College of Engineering, the College of Information Technology, and the Institute for the Humanities.
https://www.uniten.edu.my/
Schmalkalden University of Applied Sciences
University of Applied Sciences in Thuringia. Schmalkalden University of Applied Sciences offers a wide range of study programs sponsored by the five faculties of electrical engineering, mechanical engineering, computer science, business and economics, and business law.
https://www.hs-schmalkalden.de/en.html
University of Applied Sciences Hof
Founded in 1994, is a public non-profit business, media and technical vocational university founded in 1994 in Upper Franconia, Bavaria, Germany. Practice-orientation, internationalization and the intelligent use of resources are at the heart of teaching and research at Hof University, which currently offers 28 bachelor and master degree programs in Business, Computing and Engineering.
https://www.hof-university.com/
Fachhochschule Dortmund – University of Applied Sciences and Arts
It was officially founded in 1971, but the original institute, Königliche Werkmeisterschule für Maschinenbauer, was founded in 1890. However, the basic concept has remained unchanged over the years: learning and teaching focused on practical problem solving, taught by highly qualified professionals. Professors who strengthen the link between theory and practice. True to this philosophy, the University offers over 50 study options across seven different faculties: Applied Social Sciences, Architecture, Design, Computer Science, Economics, Engineering, Information Technology, and Electrical Engineering. More than 12,500 students are currently studying at the Dortmund University of Applied Sciences and Arts.
https://www.fh-dortmund.de/en/hs/rundUm/index.php
Università di Pisa in the sunny city of Pisa, in the Tuscany region is one of the oldest universities in Europe, founded in 1343. The university is in the top 5 of the best universities in Italy; known for strong training primarily in Mathematics and Physics.
https://www.unipi.it/index.php/english
University of Lorraine (UL)
University of Lorraine (UL) takes a high position in the academic ranking of France. The University of Lorraine is usually in the top 5% of the world university rankings. The University is a leading university in the field of "Mathematics", occupying 151 position in the world.
http://welcome.univ-lorraine.fr/
IGlobal Univeristy, USA
IGlobal University (USA) is located in the suburbs of Washington. The university has state accreditation and the right to issue an I-20 form for admitting foreign students.
https://www.igu.edu/
Green River College, USA
The college was founded in 1965. Green River College has received many awards and offers a more affordable education than traditional universities. More than 1,400 international students from more than 60 countries of the world study at the college annually. In total, the college teaches about 11,000 students.
https://www.greenriver.edu/
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