Problem Suppose $U$ is a subspace of $V$. What is $U + U$? Solution By directly applying the definition of sums of subspaces: $$U + U \stackrel{D1.36}{=} \lbrace 2u | u \in U \rbrace$$ Now let $v = 2u$, then $v \in U$, because $U$ is a subspace and thus closed under multiplication. Therefore: $$U + U = \lbrace v | v \in U \rbrace = U$$

Problem Verify the assertion in Example 1.38 Solution By directly applying the definition of sums of subspaces we obtain: $$U + W \stackrel{D1.36}{=} \lbrace (x_1, x_1, y_1, y_1) + (x_2, x_2, x_2, y_2) | x_1,y_1,x_2,y_2 \in \mathbb{F} \rbrace \stackrel{D1.12}{=} \lbrace (x_1 + x_2, x_1 + x_2, y_1 + x_2, y_1 + y_2) | x_1,y_1,x_2,y_2\rbrace$$ Now let $x = x_1 + x_2$, $y = x_1 + y_2$ and $z = y_1 + y_2$ (notice that the right hand side does not put any constraints on $x,y,z$ as each equation above has a variable that is not in any of the other equations e....

Problem Prove that the union of three subspaces of $V$ is a subspace of $V$if and only if one of the subspaces is contained in the other. Solution $(\leftarrow)$ The case where one of the three subsets contain one of the other was handled in P1C12. Therefore we can assume that the three subspaces, denoted $U_1$, $U_2$ and $U_3$ from here on out, each have elements that aren’t in the union of the other two sets - see this figure: Let now $v \in U_2$ be such that $v$ $\cancel{\in}$ $U_3$ and let $w \in U_3$ such that $w$ $\cancel{\in}$ $U_2$ and choose nonzero $a,b \in \mathbb{F}$ such that $a - b = 1$....

Problem Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other. Solution Assume that two subspaces $U_1$ and $U_2$ of $V$ are have non-overlapping elements i.e. that there exists $v \in U_1$ with $v$ $\cancel{\in}$ $U_2$ and $u \in U_2$ with $u$ $\cancel{\in}$ $U_1$. Then $v,u \in U_1 \cup U_2$, but if $v + u \in U_1$, then $u = (v + u) - v \in U_1$ a contradiction and similarly if $v + u \in U_2$, then $v = (v + u) - u \in U_2$ a contradiction....

Problem Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$ Solution Let $U_1, U_2, \dots, U_k$ be a collection of subspaces in $V$, then: $0 \in \bigcap_{i=1}^{k} U_i$ because $U_1, U_2, \dots, U_k$ are all subspaces i.e. they all contain $0$. Let $v,u \in \bigcap_{i=1}^{k} U_i$, then $v + u \in \bigcap_{i=1}^{k} U_i$ because $v,u \in U_1 \to v + u \in U_1$, $v,u \in U_2 \to v + u \in U_2$, $\dots$, $v,u \in U_k \to v + u \in U_k$, by the fact that $U_1, U_2, \dots, U_k$ are all subspaces....