$\textbf{Exercise 3.}$ Suppose $T \in \mathcal{L}(V)$. Prove that the intersection of every collection of subspaces of $V$ invariant under $T$ is invariant under $T$. $\textbf{Solution 3.}$ Suppose $\{V_\alpha\}_{\alpha \in \Gamma}$ is a collection of subspaces of $V$ invariant under $T$; here $\Gamma$ is an arbitrary index set. We need to prove that $\bigcap_{\alpha \in \Gamma} V_\alpha$, which equals the set of vectors that are in $V_\alpha$ for every $\alpha \in \Gamma$, is invariant under $T$. To do this, suppose $u \in \bigcap_{\alpha \in \Gamma} V_\alpha$. Then $u \in V_\alpha$ for every $\alpha \in \Gamma$. Thus $Tu \in V_\alpha$ for every $\alpha \in \Gamma$ (because every $V_\alpha$ is invariant under $T$). Thus $Tu \in \bigcap_{\alpha \in \Gamma} V_\alpha$, which implies that $\bigcap_{\alpha \in \Gamma} V_\alpha$ is invariant under $T$. ---  --- Commentary: Problem 3 (Section 5A): $\textbf{Exercise 3.}$ Suppose $T \in \mathcal{L}(V)$. Prove that the intersection of every collection of subspaces of $V$ invariant under $T$ is invariant under $T$. $\textbf{Solution 3.}$ Suppose $\{V_\alpha\}_{\alpha \in \Gamma}$ is a collection of subspaces of $V$ invariant under $T$; here $\Gamma$ is an arbitrary index set. We need to prove that $\bigcap_{\alpha \in \Gamma} V_\alpha$, which equals the set of vectors that are in $V_\alpha$ for every $\alpha \in \Gamma$, is invariant under $T$. To do this, suppose $u \in \bigcap_{\alpha \in \Gamma} V_\alpha$. Then $u \in V_\alpha$ for every $\alpha \in \Gamma$. Thus $Tu \in V_\alpha$ for every $\alpha \in \Gamma$ (because every $V_\alpha$ is invariant under $T$). Thus $Tu \in \bigcap_{\alpha \in \Gamma} V_\alpha$, which implies that $\bigcap_{\alpha \in \Gamma} V_\alpha$ is invariant under $T$. $\textit{Commentary:}$ This exercise shows that the intersection of invariant subspaces is invariant. More precisely, if $T$ is a linear operator on $V$ and $\{V_\alpha\}_{\alpha \in \Gamma}$ is any collection of subspaces of $V$ that are each invariant under $T$, then their intersection $\bigcap_{\alpha \in \Gamma} V_\alpha$ is also invariant under $T$. The proof is again straightforward: if $u$ is in the intersection, then it's in every $V_\alpha$. Since each $V_\alpha$ is invariant, $Tu$ is also in every $V_\alpha$, so it's in their intersection. This result, together with the previous one about sums, forms the basis for the lattice structure of invariant subspaces. In many applications, one is interested in finding the "smallest" or "largest" invariant subspace satisfying certain properties, and these are often obtained by taking intersections or sums of other invariant subspaces. $\textit{Example:}$ Let $V = \mathbb{R}^3$ and define $T: \mathbb{R}^3 \to \mathbb{R}^3$ by $T(x, y, z) = (x, y, 0)$. Let $V_1 = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (the $xy$-plane), $V_2 = \{(x, 0, z) : x, z \in \mathbb{R}\}$ (the $xz$-plane), and $V_3 = \{(0, y, z) : y, z \in \mathbb{R}\}$ (the $yz$-plane). Each of these subspaces is invariant under $T$. Their intersection $V_1 \cap V_2 \cap V_3 = \{(0, 0, 0)\}$ (the origin) is also invariant under $T$, because $T(0, 0, 0) = (0, 0, 0)$.