$\textbf{Exercise 6.}$ Define $T \in \mathcal{L}(\mathbb{F}^2)$ by $T(w, z) = (z, w)$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 6.}$ Suppose $\lambda$ is an eigenvalue of $T$. For this particular operator, the eigenvalue-eigenvector equation $T(w, z) = \lambda(w, z)$ becomes the system of equations $\begin{align*} z &= \lambda w \\ w &= \lambda z. \end{align*}$ Substituting the value for $z$ from the first equation into the second equation gives $w = \lambda^2 w$. Thus $1 = \lambda^2$ (we can ignore the possibility that $w = 0$ because if $w = 0$, then the first equation above implies that $z = 0$). Thus $\lambda = 1$ or $\lambda = -1$. The set of eigenvectors corresponding to the eigenvalue 1 is $\begin{equation*} \{(w, w) : w \in \mathbb{F} \text{ and } w \neq 0\}. \end{equation*}$ The set of eigenvectors corresponding to the eigenvalue -1 is $\begin{equation*} \{(w, -w) : w \in \mathbb{F} \text{ and } w \neq 0\}. \end{equation*}$ ---  --- Commentary: $\textbf{Exercise 6.}$ Define $T \in \mathcal{L}(\mathbb{F}^2)$ by $T(w, z) = (z, w)$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 6.}$ Suppose $\lambda$ is an eigenvalue of $T$. For this particular operator, the eigenvalue-eigenvector equation $T(w, z) = \lambda(w, z)$ becomes the system of equations $\begin{align*} z &= \lambda w \\ w &= \lambda z. \end{align*}$ Substituting the value for $z$ from the first equation into the second equation gives $w = \lambda^2 w$. Thus $1 = \lambda^2$ (we can ignore the possibility that $w = 0$ because if $w = 0$, then the first equation above implies that $z = 0$). Thus $\lambda = 1$ or $\lambda = -1$. The set of eigenvectors corresponding to the eigenvalue 1 is $\begin{equation*} \{(w, w) : w \in \mathbb{F} \text{ and } w \neq 0\}. \end{equation*}$ The set of eigenvectors corresponding to the eigenvalue -1 is $\begin{equation*} \{(w, -w) : w \in \mathbb{F} \text{ and } w \neq 0\}. \end{equation*}$ $\textit{Commentary:}$ This exercise provides a simple example of finding eigenvalues and eigenvectors for a specific linear operator. The operator $T$ swaps the coordinates of a vector in $\mathbb{F}^2$. To find the eigenvalues, we set up the eigenvalue-eigenvector equation $T(w, z) = \lambda(w, z)$ and solve for $\lambda$. This gives a quadratic equation $\lambda^2 = 1$, which has solutions $\lambda = \pm 1$. For each eigenvalue, we then find the corresponding eigenvectors by solving the system of equations. The eigenvectors for $\lambda = 1$ are the non-zero vectors on the line $z = w$, while the eigenvectors for $\lambda = -1$ are the non-zero vectors on the line $z = -w$. This example illustrates the general process of finding eigenvalues and eigenvectors, which involves solving a characteristic equation and then solving systems of linear equations. $\textit{Example:}$ Let's take $\mathbb{F} = \mathbb{R}$. For $\lambda = 1$, the eigenvectors are all non-zero vectors of the form $(w, w)$, which geometrically represent all points on the line $y = x$ except the origin. For example, $(1, 1)$, $(2, 2)$, $(-3, -3)$ are all eigenvectors corresponding to $\lambda = 1$. For $\lambda = -1$, the eigenvectors are all non-zero vectors of the form $(w, -w)$, which geometrically represent all points on the line $y = -x$ except the origin. For example, $(1, -1)$, $(2, -2)$, $(-3, 3)$ are all eigenvectors corresponding to $\lambda = -1$.