Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 3 / 7F (Linear Mathematics)

True. Both vector subspaces contain zero, and if $u,v\in U\cap W$, every linear combination $\alpha u+\beta v$ belongs to each of $U$ and $W$. Thus it belongs to their intersection. This proves the intersection of vector subspaces is a vector subspace.

False. Distinct vector subspaces may be nested: in $V={\mathbb{R}}^2$, take $U=\{0\}$ and $W=\mathrm{span}(1,0)$. Their union is $W$, a subspace. More precisely, $U\cup W$ is a subspace exactly when one is contained in the other. If neither contains the other, choose $u\in U\setminus W$ and $w\in W\setminus U$. Were $u+w$ in the union, subtracting $u$ or $w$ in the appropriate subspace would contradict those choices.

False. In ${\mathbb{R}}^2$ let $U=\mathrm{span}(1,1)$, $W=\mathrm{span}(1,0)$ and $X=\mathrm{span}(0,1)$. Then $W+X={\mathbb{R}}^2$, so $U\cap (W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the sum of vector subspaces. The right side is always contained in the left; this example makes the inclusion strict.

True. Choose a basis $u_1,\dots ,u_r$ of $U$ and extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of the finite-dimensional vector space $V$. Put $W=\mathrm{span}(w_1,\dots ,w_s)$. Every basis vector belongs to $U+W$, so $U+W=V$. Linear independence of the complete basis makes a vector belonging to both spans necessarily zero. ThereforeThis constructs a direct-sum complement; it need not be unique.