Answer
See solution
Work Step by Step
A has n linearly independent eigenvectors, so it is diagonalizable. If A is diagonalizable, $A^T$ is also diagonalizable:
$A=PDP^{-1}$
$A^T=(PDP^{-1})^T=(P^{-1})^TD^TP^T=(P^T)^{-1}D^TP^T$
Let $J=(P^T)^{-1}$ and $E=D^T$. Then $A^T=JEJ^{-1}$. This means $A^T$ must also have n linearly independent eigenvectors by the Diagonalization Theorem.
