How to show $A$ and $A^T$ have same eigenvalues if $A$ is a square matrix?

Question

Is there a way to show it through determinant way rather than standard equation $T(x)=\lambda x$? For instance $\det (A-\lambda I)=\det(A^T-\lambda I)$. I don't know how to correctly express the proof.

Answer 1

You can demostrate that \begin{equation} \mbox{det}\left[A^T-\lambda I\right]=\mbox{det}\left[(A-\lambda I)^T\right]=\mbox{det}\left[A-\lambda I\right] \end{equation} From there, it follows that they eigenvalues are the same

Answer 2

If you already know that $\det(A-\lambda I) = \det(A^T-\lambda I)$, then you're done, because this means that $A$ and $A^T$ have the same characteristic polynomial. Since eigenvalues are roots of that polynomial, it follows that $A$ and $A^T$ have the same eigenvalues (though not necessarily the same eigenvectors).

To prove that, using that transpositions preserve the determinant, note that: $$\det(A-\lambda I) = \det((A-\lambda I)^T) = \det(A^T - (\lambda I)^T) = \det(A^T - \lambda I^T) = \det(A^T-\lambda I).$$

Answer 3

Choose a basis so that $A$ is upper triangular. Then the determinant of $A$ is the product of its eigenvalues, which are its diagonal elements. $A^T$ is lower triangular, so its diagonal elements are its eigenvalues also. And the diagonal elements of $A$ and $A^T$ are the same.

Note: the whole thing depends on the ability to choose a nice basis at the start!