Show that A and its transpose have the same eigenvalues

[Mr Davis 97]

Homework Statement

Show that $A$ and $A^T$ have the same eigenvalues.

Homework Equations

The Attempt at a Solution

If they have the same eigenvalues, then $Ax = \lambda x$ iff $A^T x = \lambda x$. In other words, we have to show that $|A - \lambda I| = 0$ iff $|A^T - \lambda I| = 0$. From the properties of transpose, we see that $(A - \lambda I)^T = A^T - \lambda I$. It's a property of transposes that $A^T$ is invertible iff $A$ is also invertible. So we have shown that $A - \lambda I$ is invertible iff $A^T - \lambda I$ is also invertible. So this shows that they have the same eigenvalues.

I think that this is the correct solution, but I am a little confused about the beginning part of the proof. Does this imply that A and its transpose also have the same eigenvectors? Or do they have the same eigenvlues but different eigenvectors?

 

[fresh_42]

Homework Statement

Show that $A$ and $A^T$ have the same eigenvalues.

Homework Equations

The Attempt at a Solution

If they have the same eigenvalues, then $Ax = \lambda x$ iff $A^T x = \lambda x$. In other words, we have to show that $|A - \lambda I| = 0$ iff $|A^T - \lambda I| = 0$. From the properties of transpose, we see that $(A - \lambda I)^T = A^T - \lambda I$. It's a property of transposes that $A^T$ is invertible iff $A$ is also invertible. So we have shown that $A - \lambda I$ is invertible iff $A^T - \lambda I$ is also invertible. So this shows that they have the same eigenvalues.

I think that this is the correct solution, but I am a little confused about the beginning part of the proof. Does this imply that A and its transpose also have the same eigenvectors? Or do they have the same eigenvlues but different eigenvectors?

You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of $A$?
What are the eigenvalues of $A^T$?
And how are $\det B$ and $\det B^T$ related for any matrix $B$? (Esp. for $B=A-\lambda I$.)

 

[Mr Davis 97]

You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of $A$?
What are the eigenvalues of $A^T$?
And how are $\det B$ and $\det B^T$ related for any matrix $B$? (Esp. for $B=A-\lambda I$.)

$det(A) = det(A^T)$ for all square $A$. So we have that $det(A - \lambda I) = det((A - \lambda I)^T) = det(A^T - \lambda I)$

 

[Mr Davis 97]

You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of $A$?
What are the eigenvalues of $A^T$?
And how are $\det B$ and $\det B^T$ related for any matrix $B$? (Esp. for $B=A-\lambda I$.)

So as a last note do $A$ and $A^T$ also have the same eigenvectors?

 

[fresh_42]

So as a last note do $A$ and $A^T$ also have the same eigenvectors?

Good question. I don't think so, but I don't have an example at hand.
I've found this example:
https://www.physicsforums.com/threads/q1-does-a-and-its-transpose-have-the-same-eigenspace.215613/