Do Eigenvalues of A and A^T have the same Eigenvectors?

[arpon]

Homework Statement

If $AA^T=A^TA$, then prove that $A$ and $A^T$ have the same eigenvectors.

Homework Equations

The Attempt at a Solution

$Ax=\lambda x$
$A^TAx=\lambda A^Tx$
$AA^Tx=\lambda A^Tx$
$A(A^Tx)=\lambda (A^Tx)$
So, $A^Tx$ is also an eigenvector of $A$.
What should be the next steps?

 

[Mark44]

Homework Statement

If $AA^T=A^TA$, then prove that $A$ and $A^T$ have the same eigenvectors.

Homework Equations

The Attempt at a Solution

$Ax=\lambda x$
$A^TAx=\lambda A^Tx$
$AA^Tx=\lambda A^Tx$
$A(A^Tx)=\lambda (A^Tx)$
So, $A^Tx$ is also an eigenvector of $A$.

What you have to show is that if x is an eigenvector of A, then x, not A^Tx, is also an eigenvector of A^T.

What should be the next steps?

 

[Mark44]

A fact that might be pertinent here is that for square matrices A and B, if AB = BA, then each one is the inverse of the other.

Edit: As pointed out by Ray, this is not true. Mea culpa.

 

[Ray Vickson]

A fact that might be pertinent here is that for square matrices A and B, if AB = BA, then each one is the inverse of the other.

If
$$A = \pmatrix{0&1\\1&0} \; \text{and} \; B = \pmatrix{1&0\\0&1}$$
then $AB = BA = A$, but $B \neq A^{-1}$ and $A \neq B^{-1}$.

However, if you mean that $AB = (BA)^{-1}$, etc., then that seems OK, in this example at least.

 

[Mark44]

If
$$A = \pmatrix{0&1\\1&0} \; \text{and} \; B = \pmatrix{1&0\\0&1}$$
then $AB = BA = A$, but $B \neq A^{-1}$ and $A \neq B^{-1}$.

However, if you mean that $AB = (BA)^{-1}$, etc., then that seems OK, in this example at least.

No, I meant the first, so my "fact" was incorrect. I guess I had that confused with if AB = I, then A and B are inverses.

 

[pasmith]

However, if you mean that $AB = (BA)^{-1}$, etc., then that seems OK, in this example at least.

I don't think we are entitled to assume that 0 is not an eigenvalue of $A$.