--- Commentary: $\textbf{Exercise 23.}$ Prove or give a counterexample: If $V_1, V_2, U$ are subspaces of $V$ such that $V = V_1 \oplus U \text{ and } V = V_2 \oplus U,$ then $V_1 = V_2$. $\textbf{Solution 23.}$ To construct a counterexample for the assertion above, let $V = \mathbb{F}^2$, let $V_1 = {(x, 0) : x \in \mathbb{F}}$, let $V_2 = {(0, y) : y \in \mathbb{F}}$, and let $U = {(z, z) : z \in \mathbb{F}}$. Then $\mathbb{F}^2 = V_1 \oplus U \text{ and } \mathbb{F}^2 = V_2 \oplus U,$ as is easy to verify, but $V_1 \neq V_2$. $\textit{Commentary:}$ This exercise asks whether two subspaces that form a direct sum decomposition of a vector space with the same third subspace must be equal. The answer is no, and the counterexample demonstrates why. The counterexample takes place in $\mathbb{F}^2$ and chooses $V_1$ to be the $x$-axis, $V_2$ to be the $y$-axis, and $U$ to be the line $y=x$. Then both $V_1$ and $V_2$ form a direct sum decomposition of $\mathbb{F}^2$ with $U$, because every vector in $\mathbb{F}^2$ can be uniquely written as a sum of a vector from the $x$-axis (or $y$-axis) and a vector from the line $y=x$. However, $V_1$ and $V_2$ are clearly different. This counterexample works because $U$ is "diagonal" in a sense, intersecting both $V_1$ and $V_2$ only at 0, and its vectors have equal $x$ and $y$ components, allowing it to "fill in" the missing component when added to vectors from $V_1$ or $V_2$. This exercise highlights that the direct sum decomposition of a vector space is not unique in general. There can be multiple ways to decompose a vector space into a direct sum of subspaces. $\textit{Additional Examples:}$ In $\mathbb{R}^3$, let $V_1 = {(x, 0, 0) : x \in \mathbb{R}}$, $V_2 = {(0, y, 0) : y \in \mathbb{R}}$, and $U = {(z, z, z) : z \in \mathbb{R}}$. Then $\mathbb{R}^3 = V_1 \oplus U$ and $\mathbb{R}^3 = V_2 \oplus U$, but $V_1 \neq V_2$. In $\mathbb{C}^2$, let $V_1 = {(z, 0) : z \in \mathbb{C}}$, $V_2 = {(0, w) : w \in \mathbb{C}}$, and $U = {(u, iu) : u \in \mathbb{C}}$. Then $\mathbb{C}^2 = V_1 \oplus U$ and $\mathbb{C}^2 = V_2 \oplus U$, but $V_1 \neq V_2$. In the space of polynomials $\mathbb{F}_3[x]$, let $V_1 = {ax : a \in \mathbb{F}_3}$, $V_2 = {bx^2 : b \in \mathbb{F}_3}$, and $U = {c(1+x+x^2) : c \in \mathbb{F}_3}$. Then $\mathbb{F}_3[x] = V_1 \oplus U$ and $\mathbb{F}_3[x] = V_2 \oplus U$, but $V_1 \neq V_2$. In the space of functions from $\mathbb{R}$ to $\mathbb{R}$, let $V_1$ be the subspace of constant functions, $V_2$ be the subspace of functions $f$ satisfying $f(0) = 0$, and $U$ be the subspace of odd functions. Then the space of functions = $V_1 \oplus U$ and the space of functions = $V_2 \oplus U$, but $V_1 \neq V_2$. These examples demonstrate the same principle in various contexts, including over the real and complex numbers, over a finite field, and in a function space. In each case, $U$ is chosen to "complement" both $V_1$ and $V_2$ in a way that ensures a direct sum decomposition, but $V_1$ and $V_2$ are different subspaces.