$\textbf{Exercise 26.}$ Suppose $T \in \mathcal{L}(V)$ is such that every nonzero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator. $\textbf{Solution 26.}$ For each $v \in V \setminus \{0\}$, there exists $\lambda_v \in \mathbb{F}$ such that $\begin{equation*} Tv = \lambda_v v. \end{equation*}$ To show that $T$ is a scalar multiple of the identity, we must show that $\lambda_v$ is independent of $v$ for each $v \in V \setminus \{0\}$. To do this, suppose $v, w \in V \setminus \{0\}$. We want to show that $\lambda_v = \lambda_w$. First consider the case where $v, w$ is linearly dependent. Then there exists $b \in \mathbb{F}$ such that $w = bv$. We have $\begin{align*} \lambda_w w &= Tw \\ &= T(bv) \\ &= bTv \\ &= b(\lambda_v v) \\ &= \lambda_v w, \end{align*}$ which shows that $\lambda_v = \lambda_w$, as desired. Finally, consider the case where $v, w$ is linearly independent. We have $\begin{align*} \lambda_{v+w}(v + w) &= T(v + w) \\ &= Tv + Tw \\ &= \lambda_v v + \lambda_w w, \end{align*}$ which implies that $\begin{equation*} (\lambda_{v+w} - \lambda_v)v + (\lambda_{v+w} - \lambda_w)w = 0. \end{equation*}$ Because $v, w$ is linearly independent, this implies that $\lambda_{v+w} = \lambda_v$ and $\lambda_{v+w} = \lambda_w$, so again we have $\lambda_v = \lambda_w$, as desired. ---  --- Commentary: $\textbf{Exercise 26.}$ Suppose $T \in \mathcal{L}(V)$ is such that every nonzero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator. $\textbf{Solution 26.}$ For each $v \in V \setminus \{0\}$, there exists $\lambda_v \in \mathbb{F}$ such that $\begin{equation*} Tv = \lambda_v v. \end{equation*}$ To show that $T$ is a scalar multiple of the identity, we must show that $\lambda_v$ is independent of $v$ for each $v \in V \setminus \{0\}$. To do this, suppose $v, w \in V \setminus \{0\}$. We want to show that $\lambda_v = \lambda_w$. First consider the case where $v, w$ is linearly dependent. Then there exists $b \in \mathbb{F}$ such that $w = bv$. We have $\begin{align*} \lambda_w w &= Tw \\ &= T(bv) \\ &= bTv \\ &= b(\lambda_v v) \\ &= \lambda_v w, \end{align*}$ which shows that $\lambda_v = \lambda_w$, as desired. Finally, consider the case where $v, w$ is linearly independent. We have $\begin{align*} \lambda_{v+w}(v + w) &= T(v + w) \\ &= Tv + Tw \\ &= \lambda_v v + \lambda_w w, \end{align*}$ which implies that $\begin{equation*} (\lambda_{v+w} - \lambda_v)v + (\lambda_{v+w} - \lambda_w)w = 0. \end{equation*}$ Because $v, w$ is linearly independent, this implies that $\lambda_{v+w} = \lambda_v$ and $\lambda_{v+w} = \lambda_w$, so again we have $\lambda_v = \lambda_w$, as desired. $\textit{Commentary:}$ This exercise provides a characterization of scalar multiples of the identity operator. If every non-zero vector is an eigenvector of $T$, then $T$ must be a scalar multiple of the identity, i.e., $T(v) = \lambda v$ for some fixed scalar $\lambda$ and all vectors $v$. The proof involves showing that the eigenvalue $\lambda_v$ is the same for all non-zero vectors $v$. This is done by considering two cases: if $v$ and $w$ are linearly dependent, then the result follows by the linearity of $T$; if $v$ and $w$ are linearly independent, then considering $v + w$ (which is also an eigenvector) leads to the result. This theorem is a fundamental result in the study of linear operators and their spectral properties. It shows that the only linear operators for which every vector is an eigenvector are the scalar multiples of the identity. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (3x, 3y)$. Then every non-zero vector in $\mathbb{R}^2$ is an eigenvector of $T$ with eigenvalue 3. Indeed, for any $(x, y) \neq (0, 0)$, we have $\begin{equation*} T(x, y) = (3x, 3y) = 3(x, y). \end{equation*}$ So $T$ is the scalar multiple of the identity operator with $\lambda = 3$. On the other hand, if we define $S: \mathbb{R}^2 \to \mathbb{R}^2$ by $S(x, y) = (x, 0)$, then $(1, 0)$ is an eigenvector of $S$ but $(0, 1)$ is not (because $S(0, 1) = (0, 0)$ which is not a multiple of $(0, 1)$). So $S$ is not a scalar multiple of the identity. ---