$\textbf{Exercise 21.}$ Suppose $T \in \mathcal{L}(V)$ is invertible. (a) Suppose $\lambda \in \mathbb{F}$ with $\lambda \neq 0$. Prove that $\lambda$ is an eigenvalue of $T$ if and only if $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$. (b) Prove that $T$ and $T^{-1}$ have the same eigenvectors. $\textbf{Solution 21.}$ (a) First suppose $\lambda$ is an eigenvalueof $T$. Thus there exists a nonzero vector $v \in V$ such that $\begin{equation*} Tv = \lambda v. \end{equation*}$ Applying $T^{-1}$ to both sides of the equation above, we get $v = \lambda T^{-1}v$, which is equivalent to the equation $T^{-1}v = \frac{1}{\lambda}v$. Thus $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$. To prove the implication in the other direction, replace $T$ with $T^{-1}$ and $\lambda$ by $\frac{1}{\lambda}$ and then apply the result from the paragraph above. (b) The proof of (a) shows that if $v$ is an eigenvector of $T$, then $v$ is also an eigenvector of $T^{-1}$. Replacing $T$ by $T^{-1}$, we also see that if $v$ is an eigenvector of $T^{-1}$, then $v$ is also an eigenvector of $T$ [which equals $(T^{-1})^{-1}$]. Thus $T$ and $T^{-1}$ have the same eigenvectors. ---  --- Commentary: $\textbf{Exercise 21.}$ Suppose $T \in \mathcal{L}(V)$ is invertible. (a) Suppose $\lambda \in \mathbb{F}$ with $\lambda \neq 0$. Prove that $\lambda$ is an eigenvalue of $T$ if and only if $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$. (b) Prove that $T$ and $T^{-1}$ have the same eigenvectors. $\textbf{Solution 21.}$ (a) First suppose $\lambda$ is an eigenvalue of $T$. Thus there exists a nonzero vector $v \in V$ such that $\begin{equation*} Tv = \lambda v. \end{equation*}$ Applying $T^{-1}$ to both sides of the equation above, we get $v = \lambda T^{-1}v$, which is equivalent to the equation $T^{-1}v = \frac{1}{\lambda}v$. Thus $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$. To prove the implication in the other direction, replace $T$ with $T^{-1}$ and $\lambda$ by $\frac{1}{\lambda}$ and then apply the result from the paragraph above. (b) The proof of (a) shows that if $v$ is an eigenvector of $T$, then $v$ is also an eigenvector of $T^{-1}$. Replacing $T$ by $T^{-1}$, we also see that if $v$ is an eigenvector of $T^{-1}$, then $v$ is also an eigenvector of $T$ [which equals $(T^{-1})^{-1}$]. Thus $T$ and $T^{-1}$ have the same eigenvectors. $\textit{Commentary:}$ This exercise explores the relationship between the eigenvalues and eigenvectors of an invertible linear operator $T$ and its inverse $T^{-1}$. Part (a) shows that if $\lambda$ is a non-zero eigenvalue of $T$, then $1/\lambda$ is an eigenvalue of $T^{-1}$, and vice versa. This is because if $Tv = \lambda v$, then multiplying both sides by $T^{-1}$ gives $v = \lambda T^{-1}v$, or equivalently, $T^{-1}v = (1/\lambda)v$. Part (b) shows that $T$ and $T^{-1}$ have exactly the same eigenvectors. This is because the equations $Tv = \lambda v$ and $T^{-1}v = (1/\lambda)v$ have the same non-zero solutions $v$. These results are useful in the study of invertible linear operators and their spectral properties. They show that the eigenvalues of $T^{-1}$ are the reciprocals of the eigenvalues of $T$, and that the eigenvectors are unchanged. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (2x + y, x + 2y)$. The matrix of $T$ with respect to the standard basis is $\begin{equation*} [T] = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}. \end{equation*}$ The eigenvalues of $T$ are $\lambda_1 = 3$ and $\lambda_2 = 1$, with corresponding eigenvectors $v_1 = (1, 1)$ and $v_2 = (1, -1)$. The inverse of $T$ is given by $T^{-1}(x, y) = (\frac{2x-y}{3}, \frac{-x+2y}{3})$, with matrix $\begin{equation*} [T^{-1}] = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}. \end{equation*}$ The eigenvalues of $T^{-1}$ are $1/\lambda_1 = 1/3$ and $1/\lambda_2 = 1$, with the same eigenvectors $v_1 = (1, 1)$ and $v_2 = (1, -1)$. Indeed: $\begin{align*} T^{-1}v_1 &= T^{-1}(1, 1) = (\frac{1}{3}, \frac{1}{3}) = \frac{1}{3}(1, 1) = \frac{1}{\lambda_1}v_1, \ T^{-1}v_2 &= T^{-1}(1, -1) = (1, -1) = \frac{1}{\lambda_2}v_2. \end{align*}$ ---