$\textbf{Exercise 13.}$ Suppose $T \in \mathcal{L}(V)$. Suppose $S \in \mathcal{L}(V)$ is invertible. (a) Prove that $T$ and $S^{-1}TS$ have the same eigenvalues. (b) What is the relationship between the eigenvectors of $T$ and the eigenvectors of $S^{-1}TS$? $\textbf{Solution 13.}$ Suppose $v \in V$ and $\lambda \in \mathbb{F}$. Then clearly $\begin{equation*} Tv = \lambda v \iff (S^{-1}TS)(S^{-1}v) = \lambda S^{-1}v \end{equation*}$ Thus we see that $T$ and $S^{-1}TS$ have the same eigenvalues, and furthermore $v$ is an eigenvector of $T$ if and only if $S^{-1}v$ is an eigenvector of $S^{-1}TS$. --- $. Suppose $S \in \mathcal{L}(V)$ is invertible. (a) Prove that $T$ and $S^{-1}TS$ have the same eigenvalues. (b) What is the relationship between the eigenvectors of $T$ and the eigenvectors of $S^{-1}TS$? $\textbf{Solution 13.}$ Suppose $v \in V$ and $\lambda \in \mathbb{F}$. Then clearly $\begin{equation*} Tv = \lambda v \iff (S^{-1}TS)(S^{-1}v) = \lambda S^{-1}v \end{equation*}$ Thus we see that $T$ and $S^{-1}TS$ have the same eigenvalues, and furthermore $v$ is an eigenvector of $T$ if and only if $S^{-1}v$ is an eigenvector of $S^{-1}TS$. $\textit{Commentary:}$ This exercise explores how eigenvalues and eigenvectors behave under a similarity transformation. If $T$ is a linear operator and $S$ is an invertible linear operator, then $S^{-1}TS$ is called the similarity transform of $T$ by $S$. The exercise shows that $T$ and $S^{-1}TS$ have the same eigenvalues, and that their eigenvectors are related by the transformation $S^{-1}$. The proof is a straightforward verification using the definitions. This result is fundamental in the theory of matrix similarity and diagonalization. It shows that the eigenvalues of a linear operator are invariant under a change of basis (represented by $S$), and that the eigenvectors simply transform according to the change of basis. In particular, if $S$ is chosen to diagonalize $T$, then the eigenvectors of $T$ are the columns of $S$. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (3x + 4y, 4x + 3y)$. The eigenvalues of $T$ are $\lambda_1 = 7$ and $\lambda_2 = -1$, with corresponding eigenvectors $v_1 = (1, 1)$ and $v_2 = (1, -1)$. Now let $S: \mathbb{R}^2 \to \mathbb{R}^2$ be the rotation by $45^\circ$ counterclockwise, i.e., $S(x, y) = (\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}})$. Then $S^{-1}(x, y) = (\frac{x+y}{\sqrt{2}}, \frac{-x+y}{\sqrt{2}})$. The similarity transform $S^{-1}TS$ is given by: $\begin{align*} (S^{-1}TS)(x, y) &= S^{-1}(T(S(x, y))) \\ &= S^{-1}(T(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}})) \\ &= S^{-1}(\frac{7x-y}{\sqrt{2}}, \frac{x+7y}{\sqrt{2}}) \\ &= (7x, -y). \end{align*}$ The eigenvalues of $S^{-1}TS$ are also $\lambda_1 = 7$ and $\lambda_2 = -1$. The corresponding eigenvectors are: $\begin{align*} S^{-1}v_1 &= S^{-1}(1, 1) = (\sqrt{2}, 0), \\ S^{-1}v_2 &= S^{-1}(1, -1) = (0, \sqrt{2}). \end{align*}$ Indeed, $(S^{-1}TS)(\sqrt{2}, 0) = 7(\sqrt{2}, 0)$ and $(S^{-1}TS)(0, \sqrt{2}) = -1(0, \sqrt{2})$. ---