Do A and A^t always have the same eigenvectors?

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I am trying to show that a matrix A and A^t have the same eigenvalues. Well I am sure its because det(a) = det(A^t), the characteristic polynomial is det(A-xI). But this question I am trying to solve also asks for an example of a 2x2 matrix A where A and A^t have different eigenvectors. I am kinda lost of this one. Wouldn't they have the same eigenvectors because they have the same eigenvalues?
 
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Originally posted by gimpy
I am trying to show that a matrix A and A^t have the same eigenvalues. Well I am sure its because det(a) = det(A^t), the characteristic polynomial is det(A-xI). But this question I am trying to solve also asks for an example of a 2x2 matrix A where A and A^t have different eigenvectors. I am kinda lost of this one. Wouldn't they have the same eigenvectors because they have the same eigenvalues?

no. consider the eivenvalue equation:

[tex]A\mathbf{v}=\lambda\mathbf{v}[/tex]
then take the transpose of that equation:

[tex]\mathbf{v}^TA^T=\lambda\mathbf{v}^T[/tex]

so the eigenvector of A becomes something you might call a "left eigenvector" of A^T, but there is no reason to think that it should also be a regular eigenvector.

the eigenvalues must be the same, however, for the reason you stated.

i imagine, that to find an example of a matrix whose transpose has different eigenvectors, you should just grab any old generic nonsymmetric matrix, it will probably do.
 
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thanks lethe,

I feel kinda stupid that i overlooked that equation. Its really fundamental stuff[b(]