# Upper div Linear Algebra subspace proof

Part A) Suppose that U,V,W are subspaces of a vector space. Prove that (U$\cap$V)+(U$\cap$W) $\subseteq$ U$\cap$(V+W)

Part B) Prove that if V$\subset$U then the equality holds.

Part C) Find subspaces of $\mathbb{R}^2$ for which the equality does not hold.

comments: Part B refers to the expression in Part A being an equality instead of a subspace. And part C, is just the opposite. I'm not really sure where to start with all these. How should I express U,V, and W to show the relation is true. professor claims it is true.

1. If $\mathbf{x} \in (U \cap V)+(U \cap W)$, then, by definition, we have $$\mathbf{x} \in \mathrm{span}\big((U \cap V) \cup (U \cap W)\big)=\mathrm{span}\big(U \cap (V \cup W)\big)\tag{1}.$$

Now we can use $(1)$ to show that $\mathbf{x} \in U$ and $\mathbf{x} \in V+W$.

We observe the following: The right-hand side of $(1)$ is a subspace of $U$, so we know $\mathbf{x} \in U$; From $(1)$, we also know $\mathbf{x} \in \mathrm{span}(V \cup W)=V+W$.

We conclude that $(U \cap V)+(U \cap W) \subseteq U \cap (V+W)$.

2. Now assume $V \subseteq U$. We can show that $$U \subseteq (U \cap V)+(U \cap W) \subseteq U \cap (V+W) \subseteq U.$$

We observe that $$U \cap (V+W)=U \cap \mathrm{span}(V,W)=U \cap \mathrm{span}(U,W)=U.$$

3. If $U=\mathrm{span}\{(0,1)\}$ and $V=\mathrm{span}\{(1,0)\}$ and $W=\mathrm{span}\{(1,1)\}$ then...

$$(U \cap V)+(U \cap W)=\{(0,0)\}$$ whereas $$U \cap (V+W)=U \cap \mathbb{R}^2=U=\mathrm{span}\{(0,1)\}.$$