$\textbf{Exercise 23.}$ Suppose $V$ is finite-dimensional and $S, T \in \mathcal{L}(V)$. Prove that $ST$ and $TS$ have the same eigenvalues. $\textbf{Solution 23.}$ Suppose $\lambda \in \mathbb{F}$ is an eigenvalue of $ST$. We want to prove that $\lambda$ is an eigenvalue of $TS$. Because $\lambda$ is an eigenvalue of $ST$, there exists a nonzero vector $v \in V$ such that $\begin{equation*} (ST)v = \lambda v. \end{equation*}$ Now $\begin{align*} (TS)(Tv) &= T(STv) \\ &= T(\lambda v) \\ &= \lambda Tv. \end{align*}$ If $Tv \neq 0$, then the equation above shows that $\lambda$ is an eigenvalue of $TS$, as desired. If $Tv = 0$, then $\lambda = 0$ (because $S(Tv) = \lambda v$) and furthermore $T$ is not invertible, which implies that $TS$ is not invertible (by Exercise 11 in Section 3D), which implies that $\lambda$ (which equals 0) is an eigenvalue of $TS$. Regardless of whether or not $Tv = 0$, we have shown that $\lambda$ is an eigenvalue of $TS$. Because $\lambda$ was an arbitrary eigenvalue of $ST$, we have shown that every eigenvalue of $ST$ is an eigenvalue of $TS$. Reversing the roles of $S$ and $T$, we conclude that every eigenvalue of $TS$ is also an eigenvalue of $ST$. Thus $ST$ and $TS$ have the same eigenvalues. ---  --- Commentary: Problem 23 (Section 5A): $\textbf{Exercise 23.}$ Suppose $V$ is finite-dimensional and $S, T \in \mathcal{L}(V)$. Prove that $ST$ and $TS$ have the same eigenvalues. $\textbf{Solution 23.}$ Suppose $\lambda \in \mathbb{F}$ is an eigenvalue of $ST$. We want to prove that $\lambda$ is an eigenvalue of $TS$. Because $\lambda$ is an eigenvalue of $ST$, there exists a nonzero vector $v \in V$ such that $\begin{equation*} (ST)v = \lambda v. \end{equation*}$ Now $\begin{align*} (TS)(Tv) &= T(STv) \\ &= T(\lambda v) \\ &= \lambda Tv. \end{align*}$ If $Tv \neq 0$, then the equation above shows that $\lambda$ is an eigenvalue of $TS$, as desired. If $Tv = 0$, then $\lambda = 0$ (because $S(Tv) = \lambda v$) and furthermore $T$ is not invertible, which implies that $TS$ is not invertible (by Exercise 11 in Section 3D), which implies that $\lambda$ (which equals 0) is an eigenvalue of $TS$. Regardless of whether or not $Tv = 0$, we have shown that $\lambda$ is an eigenvalue of $TS$. Because $\lambda$ was an arbitrary eigenvalue of $ST$, we have shown that every eigenvalue of $ST$ is an eigenvalue of $TS$. Reversing the roles of $S$ and $T$, we conclude that every eigenvalue of $TS$ is also an eigenvalue of $ST$. Thus $ST$ and $TS$ have the same eigenvalues. $\textit{Commentary:}$ This exercise proves a fundamental result in linear algebra: if $S$ and $T$ are linear operators on a finite-dimensional vector space, then $ST$ and $TS$ have the same eigenvalues. The proof is a bit tricky. It starts by assuming that $\lambda$ is an eigenvalue of $ST$ with eigenvector $v$, and then shows that either $Tv$ is an eigenvector of $TS$ with eigenvalue $\lambda$, or $\lambda = 0$ and $TS$ is not invertible (and hence has 0 as an eigenvalue). The key steps are to apply $T$ to the equation $STv = \lambda v$ and to use the result that the product of two operators is invertible if and only if each operator is invertible. By reversing the roles of $S$ and $T$, we see that every eigenvalue of $TS$ is also an eigenvalue of $ST$. This result is important in the study of commuting operators and simultaneous diagonalization. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $S, T: \mathbb{R}^2 \to \mathbb{R}^2$ by $S(x, y) = (x + y, x)$ and $T(x, y) = (x, x + y)$. The matrices of $S$ and $T$ with respect to the standard basis are $\begin{equation*} [S] = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \quad [T] = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}. \end{equation*}$ The product $ST$ is given by $ST(x, y) = (2x + y, x + y)$, with matrix $\begin{equation*} [ST] = [S][T] = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}. \end{equation*}$ The eigenvalues of $ST$ are $\lambda_1 = 3$ and $\lambda_2 = 0$, with corresponding eigenvectors $v_1 = (1, 1)$ and $v_2 = (-1, 1)$. The product $TS$ is given by $TS(x, y) = (x + y, 2x + y)$, with matrix $\begin{equation*} [TS] = [T][S] = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}. \end{equation*}$ The eigenvalues of $TS$ are also $\lambda_1 = 3$ and $\lambda_2 = 0$, but with different eigenvectors $w_1 = (1, 2)$ and $w_2 = (1, -1)$. This example illustrates that $ST$ and $TS$ have the same eigenvalues, even though they may have different eigenvectors.