# Show that A and its transpose have the same eigenvalues

1. Dec 9, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Show that $A$ and $A^T$ have the same eigenvalues.

2. Relevant equations

3. 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?

2. Dec 9, 2016

### Staff: Mentor

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 $B=A-\lambda I$.)

3. Dec 9, 2016

### Mr Davis 97

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

4. Dec 9, 2016

### Mr Davis 97

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

5. Dec 10, 2016