text: OpenStudy (anonymous):
  
  The subspace X of the real vector space R^3 is given as the following span of a set of vectors: X = span(1; 1; 0); (2; 5; 14); (0; 1; 2); (5; 0; 10) Determine a basis for X as well as the dimension of X. (d) Let V and W be two real vector spaces and let T : V > W be a linear transformation. Define the kernel of T and prove that it is a subspace of V .
  
  7 years ago
  OpenStudy (anonymous):
  
  For the first part, make those vectors the columns of a matrix, and row reduce the matrix to the reduced echelon form. Count the pivots. Thats the dimension of X. the columns that contain the pivots will tell you which columns form the basis.
  
  7 years ago
  OpenStudy (anonymous):
  
  For the second, they just want a general definition of the Null Space (or Kernel) of a Linear Transformation. In general, the Null Space, which i'll call N, is: $N = \left\{ \space x\space|\space T(x) = 0 \right\}$ Its all the vectors in the vector space such that T(x) = 0. It is a subspace because it is closed under addition and scalar multiplication. If u and v are in the Null Space, and c is a scalar, then (using the properties of a Linear Transformation) we obtain: $T(cu+v) = T(cu)+T(v) = cT(u)+T(v) = c*0+0 = 0$ Also, 0 is in the Null Space: $T(0) = 0$
  
  7 years ago
  OpenStudy (anonymous):
  
  thanks alot!
  
  7 years ago
  OpenStudy (anonymous):
  
  Reduced row echelon form: 1 -2 0 -5 0 1 1/7 5/7 0 0 0 0 so dim=4 basis:{(1,0),(-2,1),(0,1/7),(-5,5/7)}
  
  7 years ago
  OpenStudy (anonymous):
  
  dont know how to type a matrix!
  
  7 years ago