--- COMMENTARY: $\textbf{Exercise 27 Commentary:}$ This exercise shows that if $P \in \mathcal{L}(V)$ and $P^2 = P$, then $V$ can be decomposed as a direct sum of $\text{null}(P)$ and $\text{range}(P)$. Such linear operators $P$ are called projection operators, as they "project" vectors in $V$ onto $\text{range}(P)$ along $\text{null}(P)$. The key steps are: (1) Show that $\text{null}(P) \cap \text{range}(P) = \{0\}$, since if $Pv = v \neq 0$, then applying $P$ again yields $P^2v = Pv \neq 0$. (2) Show that $V = \text{null}(P) + \text{range}(P)$ by writing an arbitrary $v \in V$ as $v = (v - Pv) + Pv$, where $v - Pv \in \text{null}(P)$ and $Pv \in \text{range}(P)$. $\textbf{Exercise 27 Examples:}$ 1. Let $V = \mathbb{R}^3$, and define $P: V \to V$ by $P(x,y,z) = (x,0,0)$. Then $P^2 = P$, and we have $\text{null}(P) = \text{span}\{(0,1,0), (0,0,1)\}, \text{range}(P) = \text{span}\{(1,0,0)\}$. 2. Let $V = \mathcal{P}_3(\mathbb{R})$, and define $P: V \to V$ by $P(a + bx + cx^2 + dx^3) = a + bx$. Then $P^2 = P$, and we have $\text{null}(P) = \text{span}\{1, x^2, x^3\}, \text{range}(P) = \mathcal{P}_1(\mathbb{R})$. 1. Let $V = \mathbb{F}_2^4$, and define $P: V \to V$ by $P(x_1, x_2, x_3, x_4) = (x_1, x_2, 0, 0)$. Then $P^2 = P$, and we have $\text{null}(P) = \text{span}\{(0,0,1,0), (0,0,0,1)\}, \text{range}(P) = \text{span}\{(1,0,0,0), (0,1,0,0)\}$.