--- Commentary: $\textbf{Exercise 22.}$ Suppose $U = {(x, y, x+y, x-y, 2x) \in \mathbb{F}^5 : x, y \in \mathbb{F}}.$ Find three subspaces $W_1, W_2, W_3$ of $\mathbb{F}^5$, none of which equals ${0}$, such that $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3.$ $\textbf{Solution 22.}$ Let $\begin{align*} W_1 &= {(0, 0, a, 0, 0) \in \mathbb{F}^5 : a \in \mathbb{F}}, \\ W_2 &= {(0, 0, 0, b, 0) \in \mathbb{F}^5 : b \in \mathbb{F}}, \\ W_3 &= {(0, 0, 0, 0, c) \in \mathbb{F}^5 : c \in \mathbb{F}}. \end{align*}$ Then $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$, as is easy to verify. $\textit{Commentary:}$ This exercise extends the concept of a direct sum decomposition to more than two subspaces. It asks to decompose $\mathbb{F}^5$ into a direct sum of the given subspace $U$ and three other nonzero subspaces. The solution chooses $W_1, W_2, W_3$ to be subspaces where all coordinates are 0 except for the third, fourth, and fifth coordinates respectively. This ensures that each $W_i$ intersects $U$ and the other $W_j$ only at 0, because $U$ requires specific relationships between the coordinates that are never satisfied by the vectors in the $W_i$. Moreover, any vector in $\mathbb{F}^5$ can be uniquely written as a sum of vectors from $U, W_1, W_2, W_3$, by using the same approach as in the previous exercises to get the $U$ component, and then distributing the remaining third, fourth, and fifth coordinates among $W_1, W_2, W_3$. This exercise demonstrates that a vector space can be decomposed into a direct sum of more than two subspaces, and provides practice in finding such a decomposition. $\textit{Additional Examples:}$ In $\mathbb{R}^5$, if $U = {(x, y, x+y, x-y, 2x) : x, y \in \mathbb{R}}$, then $W_1 = {(0, 0, a, 0, 0) : a \in \mathbb{R}}$, $W_2 = {(0, 0, 0, b, 0) : b \in \mathbb{R}}$, $W_3 = {(0, 0, 0, 0, c) : c \in \mathbb{R}}$ satisfy $\mathbb{R}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$. In $\mathbb{F}_2^5$, if $U = {(x, y, x+y, x+y, 0) : x, y \in \mathbb{F}_2}$, then $W_1 = {(0, 0, a, 0, 0) : a \in \mathbb{F}_2}$, $W_2 = {(0, 0, 0, b, 0) : b \in \mathbb{F}_2}$, $W_3 = {(0, 0, 0, 0, c) : c \in \mathbb{F}_2}$ satisfy $\mathbb{F}_2^5 = U \oplus W_1 \oplus W_2 \oplus W_3$. Note that in $\mathbb{F}_2$, $-1 = 1$, so $x-y = x+y$. In the space of polynomials $\mathbb{Q}[x,y,z]$, if $U = {ax+by+c(x+y)+d(x-y)+e(2x) : a,b,c,d,e \in \mathbb{Q}}$, then $W_1 = {f : f \in \mathbb{Q}[x,y,z]}$, $W_2 = {gy : g \in \mathbb{Q}[x,y,z]}$, $W_3 = {hz : h \in \mathbb{Q}[x,y,z]}$ satisfy $\mathbb{Q}[x,y,z] = U \oplus W_1 \oplus W_2 \oplus W_3$. In the space of continuous functions $C([0,1], \mathbb{C})$, if $U = {ax+by+c(x+y)+d(x-y)+e(2x) : a,b,c,d,e \in \mathbb{C}}$, then $W_1 = {f \in C([0,1], \mathbb{C}) : f(0) = f(1) = 0}$, $W_2 = {g \sin(2\pi x) : g \in \mathbb{C}}$, $W_3 = {h \cos(2\pi x) : h \in \mathbb{C}}$ satisfy $C([0,1], \mathbb{C}) = U \oplus W_1 \oplus W_2 \oplus W_3$. These examples illustrate the concept of a direct sum decomposition into more than two subspaces in a variety of contexts, including over the real numbers, over a finite field, in a polynomial ring, and in a function space. In each case, the key is to choose the $W_i$ in a way that ensures they intersect each other and $U$ only at 0, and that every vector can be uniquely written as a sum of components from $U$ and each $W_i$.