Let A a diagonalizable matrix. Show that $A$ and $A^{t}$ are similar.

Question

Let A a diagonalizable matrix. Show that $A$ and $A^{t}$ are similar.

Answer 1

This result holds true without the hypothesis that $A$ is diagonalizable. However, given that this hypothesis is nevertheless given, one may assume you are supposed to show this using an auxiliary result, namely that if $A$ is similar to$~B$, then $A^t$ is similar to$~B^t$, which is not hard to show. This will then allow you to reduce the proof to a special case where you replace $A$ by a special matrix similar to $A$, and for which the result is easier. Can you guess what that special matrix might be?

Answer 2

Using the fact shown by Omnomnomnom :

Recall : A square matrix and its transpose have the same determinant.

$\det(A^{T} - \lambda I) = \det((A - \lambda I)^{T}) = \det (A - \lambda I)$

It's easy to see they have the same eigenvalue.

Answer 3

Note that any two diagonalizable matrices are similar if and only if they have the same eigenvalues (including multiplicity). Show that if $\lambda$ is an eigenvalue of $A$, then it is an eigenvalue of $A^T$.