--- Commentary: $\textbf{Exercise 15.}$ Suppose $U$ is a subspace of $V$. What is $U + U$? $\textbf{Solution 15.}$ By definition, $U + U = {u + v : u, v \in U}$. Clearly $U \subseteq U + U$ because if $u \in U$, then $u$ equals $u + 0$, which expresses $u$ as a sum of two elements of $U$. Conversely, $U + U \subseteq U$ because the sum of two elements of $U$ is an element of $U$ (because $U$ is a subspace of $V$). Conclusion: $U + U = U$. $\textit{Commentary:}$ This exercise explores the sum of a subspace with itself. The result is that $U + U = U$ for any subspace $U$. The proof relies on two key observations. First, any element of $U$ can be written as a sum of two elements of $U$ (one of which can be 0), so $U$ is a subset of $U + U$. Second, the sum of any two elements of $U$ is in $U$, because $U$ is a subspace. This shows that $U + U$ is a subset of $U$. Together, these observations imply that $U + U = U$. This result is a special case of the more general fact that the sum of a subspace with itself any number of times is equal to the original subspace. This is because a subspace is closed under vector addition, so adding elements of the subspace together always yields another element of the subspace. $\textit{Example:}$ In $\mathbb{R}^3$, if $U$ is the $xy$-plane, then $U + U = U$. In $\mathbb{Q}_p^2$, if $U = {(x, px) : x \in \mathbb{Q}_p}$, then $U + U = U$. In the space of continuous functions $C([0,1], \mathbb{R})$, if $U$ is the subspace of polynomials of degree at most 3, then $U + U = U$. In any vector space $V$, the trivial subspaces ${0}$ and $V$ satisfy ${0} + {0} = {0}$ and $V + V = V$. These examples illustrate that the result $U + U = U$ holds for any subspace $U$ in any vector space $V$. It's a fundamental property of subspaces that stems from the closure of subspaces under vector addition.