--- new commentary: $\textbf{Exercise 22 Commentary:}$ This exercise bounds the dimension of the null space of the composition $ST$ of two linear maps in terms of the null space dimensions of $S$ and $T$ individually. Specifically, it shows that $\text{dim null}(ST) \leq \text{dim null}(S) + \text{dim null}(T)$. The idea is to construct a linear map $R : \text{null}(ST) \to V$ whose range is contained in $\text{null}(S)$, and then use dimension formulas. This dimension inequality helps analyze compositions of linear maps. --- $\textbf{Exercise 22 Examples:}$ - Let $V = \mathbb{R}^4, U = \mathbb{R}^3, W = \mathbb{R}^2$, and define $S : V \to W$ and $T : U \to V$ by: $S\begin{pmatrix}x_1\x_2\x_3\x_4\end{pmatrix} = \begin{pmatrix}x_1+x_2\x_3+x_4\end{pmatrix}, \quad T\begin{pmatrix}y_1\y_2\y_3\end{pmatrix} = \begin{pmatrix}y_1\y_2\y_3\0\end{pmatrix}$ Then $\text{null}(S) = \text{span}{\begin{pmatrix}1\-1\0\0\end{pmatrix}, \begin{pmatrix}0\0\1\-1\end{pmatrix}}$, so $\text{dim null}(S) = 2$. Also, $\text{null}(T) = {0}$, so $\text{dim null}(T) = 0$. One can check that $\text{null}(ST) = {0}$, so $\text{dim null}(ST) = 0 \leq 2 + 0 = \text{dim null}(S) + \text{dim null}(T)$. - Let $V = \mathcal{P}_3(\mathbb{R}), U = \mathcal{P}_2(\mathbb{R}), W = \mathbb{R}^2$, and let $S : V \to W$ and $T : U \to V$ be differentiation maps. Then $\text{null}(S) = \text{span}{1}$, so $\text{dim null}(S) = 1$. Also, $\text{null}(T) = {0}$, so $\text{dim null}(T) = 0$. One can check that $\text{null}(ST)$ consists of constant polynomials, so $\text{dim null}(ST) = 1 \leq 1 + 0 = \text{dim null}(S) + \text{dim null}(T)$. - Let $V = \mathbb{F}_2^4, U = \mathbb{F}_2^3, W = \mathbb{F}_2^2$, and let $S,T$ be linear maps with matrices: $S = \begin{pmatrix}1&0&0&1\0&1&1&0\end{pmatrix}, \quad T = \begin{pmatrix}1&0&1\0&1&1\1&1&0\0&0&0\end{pmatrix}$ Then $\text{dim null}(S) = 2, \text{dim null}(T) = 1$, and one can compute that $\text{dim null}(ST) = 3 \leq 2 + 1 = \text{dim null}(S) + \text{dim null}(T)$.