--- $\textbf{Exercise 30 Commentary:}$ This exercise shows that if $\phi \in \mathcal{L}(V, \mathbb{F})$ is a non-zero linear functional on $V$, and $u \in V$ is not in $\text{null}(\phi)$, then $V$ can be decomposed as a direct sum of $\text{null}(\phi)$ and the 1-dimensional subspace spanned by $u$. The key steps are: (1) Show that $\text{null}(\phi) \cap \text{span}\{u\} = \{0\}$, since if $au \in \text{null}(\phi)$, then $\phi(au) = a\phi(u) = 0$ implies $a = 0$ (because $\phi(u) \neq 0$). (2) Show that every $v \in V$ can be written as $v = (v - \frac{\phi(v)}{\phi(u)}u) + \frac{\phi(v)}{\phi(u)}u$, where the first term is in $\text{null}(\phi)$ and the second term is in $\text{span}\{u\}$. $\textbf{Exercise 30 Examples:}$ 1. Let $V = \mathbb{R}^3$, and define $\phi: V \to \mathbb{R}$ by $\phi(x,y,z) = x+2y-z$. Take $u = (1,1,1)$. Then $\text{null}(\phi) = \text{span}\{(2,1,1), (-1,0,1)\}$, and $V = \text{null}(\phi) \oplus \text{span}\{(1,1,1)\}$. 2. Let $V = \mathcal{P}_2(\mathbb{R})$, and define $\phi: V \to \mathbb{R}$ by $\phi(a+bx+cx^2) = a$. Take $u = 1$. Then $\text{null}(\phi) = \text{span}\{x, x^2\}$, and $V = \text{null}(\phi) \oplus \text{span}\{1\}$. 1. Let $V = \mathbb{F}_4^2$, and define $\phi: V \to \mathbb{F}_4$ by $\phi(x,y) = x+2y$. Take $u = (1,1)$. Then $\text{null}(\phi) = \text{span}\{(2,1)\}$, and $V = \text{null}(\phi) \oplus \text{span}\{(1,1)\}$.