Question

Show that square matrix $A$ and its transpose $A^T$ have the same eigen values.

Answer 1

Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.

Hence if the matrices A and $A^T$ have the same characteristic polynomial, then they have the same eigenvalues.

So we show that the characteristic polynomial pA(t)=det(A−tI) of A is the same as the characteristic polynomial $p{A}^T\left(t\right)=det(A^T-tI)$of the transpose $A^T$.

We have

$p{A}^T\left(t\right)=det(A^T-tI)$

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

$=det\left(\right(A-tI{)}^T)$

$=det(A-tI)$

$=pA\left(t\right)$

.since $I^T=I$

since $det\left(B^T\right)=det\left(B\right)$for any square matrix B

Therefore we obtain $p{A}^T\left(t\right)=pA\left(t\right)$, and we conclude that the eigenvalues of A and $A^T$ are the same.