$\textbf{Exercise 25.}$ Suppose $T \in \mathcal{L}(V)$ and $u, w$ are eigenvectors of $T$ such that $u + w$ is also an eigenvector of $T$. Prove that $u$ and $w$ are eigenvectors of $T$ corresponding to the same eigenvalue. $\textbf{Solution 25.}$ Suppose $\begin{align*} Tu &= \alpha u \\ Tv &= \beta w \\ T(u + w) &= \gamma(u + w). \end{align*}$ Adding the first two equations above, we have $T(u + w) = \alpha u + \beta w$, which when combined with the third equation above gives $\begin{equation*} (\alpha - \gamma)u + (\beta - \gamma)w = 0. \end{equation*}$ If $\alpha \neq \beta$ then $u, w$ is linearly independent (by 5.11) and the equation above then implies that $\alpha - \gamma = \beta - \gamma = 0$, which then implies that $\alpha = \beta$. Thus we see that $\alpha = \beta$. ---  --- Commentary: $\textbf{Exercise 25.}$ Suppose $T \in \mathcal{L}(V)$ and $u, w$ are eigenvectors of $T$ such that $u + w$ is also an eigenvector of $T$. Prove that $u$ and $w$ are eigenvectors of $T$ corresponding to the same eigenvalue. $\textbf{Solution 25.}$ Suppose $\begin{align*} Tu &= \alpha u \\ Tv &= \beta w \\ T(u + w) &= \gamma(u + w). \end{align*}$ Adding the first two equations above, we have $T(u + w) = \alpha u + \beta w$, which when combined with the third equation above gives $\begin{equation*} (\alpha - \gamma)u + (\beta - \gamma)w = 0. \end{equation*}$ If $\alpha \neq \beta$ then $u, w$ is linearly independent (by 5.11) and the equation above then implies that $\alpha - \gamma = \beta - \gamma = 0$, which then implies that $\alpha = \beta$. Thus we see that $\alpha = \beta$. $\textit{Commentary:}$ This exercise provides a condition under which two eigenvectors must correspond to the same eigenvalue. If $u$ and $w$ are eigenvectors of $T$ and their sum $u + w$ is also an eigenvector, then $u$ and $w$ must have the same eigenvalue. The proof is by contradiction. If $u$ and $w$ had different eigenvalues $\alpha$ and $\beta$, then they would be linearly independent (by a previous exercise). But then the equation $(\alpha - \gamma)u + (\beta - \gamma)w = 0$ would imply that $\alpha = \beta = \gamma$, a contradiction. This result is useful in the study of the eigenspace structure of a linear operator. It shows that the sum of two eigenvectors is an eigenvector if and only if they have the same eigenvalue, in which case the sum is in the same eigenspace. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (3x, 3y)$. The eigenvalues of $T$ are $\lambda_1 = 3$ with eigenspace $\mathbb{R}^2$ (every non-zero vector is an eigenvector with eigenvalue 3). Let $u = (1, 0)$, $w = (0, 1)$, and $v = u + w = (1, 1)$. Then: $\begin{align*} Tu &= (3, 0) = 3(1, 0) = 3u, \\ Tw &= (0, 3) = 3(0, 1) = 3w, \\ Tv &= (3, 3) = 3(1, 1) = 3v. \end{align*}$ So $u$, $w$, and $v$ are all eigenvectors with the same eigenvalue 3, as predicted by the theorem. On the other hand, if we define $S: \mathbb{R}^2 \to \mathbb{R}^2$ by $S(x, y) = (3x, 4y)$, then $u = (1, 0)$ is an eigenvector with eigenvalue 3, $w = (0, 1)$ is an eigenvector with eigenvalue 4, but their sum $v = (1, 1)$ is not an eigenvector at all, because $Sv = (3, 4)$ is not a multiple of $v$. ---