Answer
See solution
Work Step by Step
The characteristic polynomial is found by calculating the determinant of $A-\lambda I$ and setting it equal to 0 for a given matrix A.
The eigenvalues of A satisfy the equation $det(A-\lambda I)=0$. But we know $detA=detA^T$, which means $det(A-\lambda I)=det(A-\lambda I)^T=0$. Moreover, $(A-\lambda I)^T=A^T-\lambda I^T=A^T-\lambda I$. Therefore, $det(A-\lambda I)=det(A^T-\lambda I)$, which means $A$ and $A^T$ have the same characteristic equation.
