Eigenvalues of a Matrix and its Transpose are the Same

Problem 508

Let $A$ be a square matrix.
Prove that the eigenvalues of the transpose $A^{\trans}$ are the same as the eigenvalues of $A$.

Proof.

Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.
Hence if the matrices $A$ and $A^{\trans}$ have the same characteristic polynomial, then they have the same eigenvalues.

So we show that the characteristic polynomial $p_A(t)=\det(A-tI)$ of $A$ is the same as the characteristic polynomial $p_{A^{\trans}}(t)=\det(A^{\trans}-tI)$ of the transpose $A^{\trans}$.

We have
\begin{align*}
&p_{A^{\trans}}(t)=\det(A^{\trans}-tI)\\
&=\det(A^{\trans}-tI^{\trans}) && \text{since $I^{\trans}=I$}\\
&=\det\left(\, (A-tI)^{\trans} \,\right)\\
&=\det(A-tI) && \text{since $\det(B^{\trans})=\det(B)$ for any square matrix $B$}\\
&=p_A(t).
\end{align*}

Therefore we obtain $p_{A^{\trans}}(t)=p_A(t)$, and we conclude that the eigenvalues of $A$ and $A^{\trans}$ are the same.

Remark: Algebraic Multiplicities of Eigenvalues

Remark that since the characteristic polynomials of $A$ and the transpose $A^{\trans}$ are the same, it furthermore yields that the algebraic multiplicities of eigenvalues of $A$ and $A^{\trans}$ are the same.

Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse...