$\textbf{Exercise 2.}$ Suppose that $T \in \mathcal{L}(V)$ and $V_1, \ldots, V_m$ are subspaces of $V$ invariant under $T$. Prove that $V_1 + \cdots + V_m$ is invariant under $T$. $\textbf{Solution 2.}$ Consider a vector $u \in V_1 + \cdots + V_m$. There exist $v_1 \in V_1, \ldots, v_m \in V_m$ such that $\begin{equation*} u = v_1 + \cdots + v_m. \end{equation*}$ Applying $T$ to both sides of this equation, we get $\begin{equation*} Tu = Tv_1 + \cdots + Tv_m. \end{equation*}$ Because each $V_k$ is invariant under $T$, we have $Tv_1 \in V_1, \ldots, Tv_m \in V_m$. Thus the equation above shows that $Tu \in V_1 + \cdots + V_m$, which implies that $V_1 + \cdots + V_m$ is invariant under $T$. ---  --- Commentary: $\textbf{Exercise 2.}$ Suppose that $T \in \mathcal{L}(V)$ and $V_1, \ldots, V_m$ are subspaces of $V$ invariant under $T$. Prove that $V_1 + \cdots + V_m$ is invariant under $T$. $\textbf{Solution 2.}$ Consider a vector $u \in V_1 + \cdots + V_m$. There exist $v_1 \in V_1, \ldots, v_m \in V_m$ such that $\begin{equation*} u = v_1 + \cdots + v_m. \end{equation*}$ Applying $T$ to both sides of this equation, we get $\begin{equation*} Tu = Tv_1 + \cdots + Tv_m. \end{equation*}$ Because each $V_k$ is invariant under $T$, we have $Tv_1 \in V_1, \ldots, Tv_m \in V_m$. Thus the equation above shows that $Tu \in V_1 + \cdots + V_m$, which implies that $V_1 + \cdots + V_m$ is invariant under $T$. $\textit{Commentary:}$ This exercise shows that the sum of invariant subspaces is invariant. More precisely, if $T$ is a linear operator on $V$ and $V_1, \ldots, V_m$ are subspaces of $V$ that are each invariant under $T$, then their sum $V_1 + \cdots + V_m$ is also invariant under $T$. The proof is straightforward: if $u$ is in the sum, then it's a sum of vectors $v_1, \ldots, v_m$ from the individual subspaces. Since each subspace is invariant, $Tv_1, \ldots, Tv_m$ are also in their respective subspaces, so their sum $Tu$ is in the sum of the subspaces. This result is useful in the study of operator theory and representation theory, where decompositions of a space into invariant subspaces play a key role. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (x, 0)$. Let $V_1 = {(x, 0) : x \in \mathbb{R}}$ (the $x$-axis) and $V_2 = {(0, y) : y \in \mathbb{R}}$ (the $y$-axis). Both $V_1$ and $V_2$ are invariant under $T$, because $T(x, 0) = (x, 0) \in V_1$ for all $(x, 0) \in V_1$ and $T(0, y) = (0, 0) \in V_2$ for all $(0, y) \in V_2$. The sum $V_1 + V_2 = \mathbb{R}^2$ is also invariant under $T$, because for any $(x, y) \in \mathbb{R}^2$, we have $T(x, y) = (x, 0) \in \mathbb{R}^2$. --