Prove $A^{T}$ is diagonalizable

Question

Let $A$ be an $n \times n$ matrix. Prove that if $A$ is diagonalizable, then $A^T$ is also diagonalizable.

By definition, since $A$ is diagonalizable,

$A = P^{-1} B P$, where $B$ is a diagonal matrix and $P$ is an invertible matrix.

$A^T = P^T B^T (P^{-1})^T$

$B^T$ is a diagonal matrix by definition, we have $(P^{-1})^T = (P^T)^{-1}$, let $D = P^{-1}$ then, $A^T = DB^T D^{-1}$ where $B^T$ is diagonal, thus $A^T$ is diagonalizable by definition

Does this work?

Answer 1

This works.

Here is another approach. Recall that a matrix $A$ is diagonalizable if and only if the minimal polynomial $\mu_A(t)$ is a product of distinct linear factors. Since $\mu_{A}(t)=\mu_{A^\top}(t)$, it follows that $A$ is diagonalizable if and only if $A^\top$ is diagonalizable.

Answer 2

Consider determinant:

$\det(A)=\det(A^T)$

$(A-\lambda I)^T=(A^T-\lambda I)$ since $I$ is symmetric.

So we get: $\det(A-\lambda I)^T=\det(A^T-\lambda I)=\det(A-\lambda I)$,

Which means that $A$ and $A^T$ have the same eigenvalues, $\lambda$.