text: # Finding eigenvalues and eigenvectors 2x2 matrix
  
  • andrey21
  
  #### andrey21
  
  Find the eigenvalues and corresponding eigenvectors of the following matrix.
  
  1,1
  1,1
  
  Here is my attempt to find eigenvalues:
  
  1-lambda 1
  1 1-lambda
  
  Giving me:
  
  (Lambda)^2 -2(lambda) = 0
  
  lambda = 0 lambda = 2
  
  Is this correct??
  
  Yes, this is good!
  
  Great from there I get a little confused finding the eigenvectors. I sub the lambda values back into matrix but don't know where to go there.
  
  lambda = 2 1-2 1
  1 1-2
  
  giving me: -1 1
  1 -1
  
  Where do I go next?
  
  So you sub lambda into your matrix. This gives you a matrix A. Now you need to find a vector x such that Ax=0. This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
  
  Rite so I have the two following matrices from lambda values:
  
  lambda = 2
  -1, 1
  1,-1
  
  so that means -x1 + X2 = 0 AND x1 -x2 = 0 therefore x1 = x2
  
  Am i on the rite track?
  
  Yes, so all vectors of the form $$(x_1,x_1)$$ are eigenvectors. Specifically, (1,1) is an eigenvector with eigenvalue 2...
  
  Great so when eigenvalue = 2 eigenvector = (1,1)
  
  When lambda = 0 I get the following matrix:
  
  1 1
  1 1
  
  Now is this just saying x1 + x2 = 0 so x1 = -x2
  
  eigenvector be (1,-1) ?
  
  Yes, that is correct! (1,-1) is indeed an eigenvector of the matrix!
  
  Fantastic thank you micromass.
  
  Can I just ask you if this is correct?
  
  matrix A = 2, 1
  0, -1
  
  Eigenvalues I get are:
  
  2 and -1:
  
  when lambda = 2:
  
  0, 1
  0, 0 So x2 = 0 therefore eigenvector is (1,0)
  
  when Lambda = -1
  
  3,1
  0,0 n so 3x1 + x2 = 0 3x1 = -x2 so eigenvector is (-1/3, 1)
  
  Correct?
  
  Correct! It seems you've got the idea!
  
  thnx micromass:)
  
  So you sub lambda into your matrix. This gives you a matrix A.
  Actually that gives you A - $\lambda$I, which is different from A (unless $\lambda$ happens to be zero).
  Now you need to find a vector x such that Ax=0.
  I know what you mean, but what you really want is to find a solution of (A - $\lambda$I)x = 0.
  This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
  
  Just quick question micromass for the post 7 I've sed eigenvector is (1,-1) should it be (-1,1) or are they both acceptable?
  
  Both are good. If x is an eigenvector and if $$\lambda\neq 0$$, then $$\lambda x$$ is an eigenvector as well. In this case: (1,-1) is an eigenvector. So take $$\lambda=-1$$, then (-1,1) is an eigenvector as well!
  
  Ah rite I see so when lambda = 0 eigenvector is (1,-1)? Its just confusing as when matrix wa:
  
  3,1
  0,0
  
  eigenvector is (-1/3,1) ?
  
  Ah rite I see so when lambda = 0 eigenvector is (1,-1)? Its just confusing as when matrix wa:
  
  3,1
  0,0
  
  eigenvector is (-1/3,1) ?
  
  Assuming this is A - $\lambda$I, yes, <-1/3, 1> is an eigenvector, and so are <-1, 3>, <1/3, -1>, <1, -3>, and many more. You can easily check whether a vector x is an eigenvector with associated eigenvalue $\lambda$ by verifying that Ax = $\lambda$x, or equivalently, that (A - $\lambda$I)x = 0.