# Linear question

1. Dec 15, 2003

### gimpy

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

2. Dec 16, 2003

### lethe

no. consider the eivenvalue equation:

$$A\mathbf{v}=\lambda\mathbf{v}$$
then take the transpose of that equation:

$$\mathbf{v}^TA^T=\lambda\mathbf{v}^T$$

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.

Last edited: Dec 16, 2003
3. Dec 16, 2003

### gimpy

thanks lethe,

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