- #1

Mr Davis 97

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## 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?