$W=\left( W\cap V_1 \right) \oplus \left( W\cap V_2 \right) $ if and only if $W=\left( W+V_1 \right) \cap \left( W+V_2 \right) $ Let $V_1,V_2$ be two subspaces of the finite dimensional vector space $V$, such that $V=V_1\oplus V_2$. Let $W$ be any subspace of $V$.
Prove that, $W=\left( W\cap V_1 \right) \oplus \left( W\cap V_2 \right) $ if and only if $W=\left( W+V_1 \right) \cap \left( W+V_2 \right) $.
I have showed that $\left( W+V_1 \right) \cap \left( W+V_2 \right)$ is a subspace of $W$. But how to proceed further?
 A: “$\Rightarrow$”: Given $W=(W\cap V_1)\oplus (W\cap V_2)$ we want to show $W=(W+V_1)\cap (W+V_2)$. Let $W_i=W\cap V_i$ and choose complements $U_i$ so that $V_i=W_i\oplus U_i$. Thus we have decomposed $V$ as
$$
V = \underbrace{W_1 \oplus U_1}_{V_1} \oplus \underbrace{W_2\oplus U_2}_{V_2} = \underbrace{W_1\oplus W_2}_W \oplus \underbrace{U_1 \oplus U_2}_U.
$$
Hence, every vector $v\in V$ has a unique decomposition as
$$
v = w_1 + u_1 + w_2 + u_2
$$
with $w_i\in W_i$ and $u_i\in U_i$. The elements of $W+V_1$ are those with $u_2=0$ and the elements of $W+V_2$ are those with $u_1=0$. Hence the intersection consists of those elements with $u_1=u_2=0$, which are exactly the elements of $W$.
“$\Leftarrow$”: In the other direction, we are given $V=V_1\oplus V_2$ and $W=(W+V_1)\cap(W+V_2)$. We want to show that $W=W_1\oplus W_2$, where $W_i=W\cap V_i$ as before. Since $W_1,W_2\subseteq W$, we have $W_1+W_2\subseteq W$. Furthermore, the sum is direct since
$$
W_1\cap W_2 \subseteq V_1\cap V_2 = \{0\}.
$$
To show that $W_1\oplus W_2$ is not a proper subspace of $W$ but in fact equal to $W$, we need the finite dimensionality of the involved spaces: From $W_i=W\cap V_i$ we have the following identities of dimension:
\begin{align*}
\dim(W+V_i) = \dim(W) + \dim(V_i) - \dim(W_i).
\end{align*}
Since we are given $W=(W+V_1)\cap(W+V_2)$ this yields
\begin{align*}
\dim(W) &= \dim(W+V_1) + \dim(W+V_2) - \dim(\underbrace{W+V_1+W+V_2}_V)
\\&= \dim(W) + \dim(V_1) - \dim(W_1)
\\&\quad+ \dim(W) + \dim(V_2) - \dim(W_2)
-\dim(V)
\\&= \dim(W) - \dim(W_1)
+ \dim(W)  - \dim(W_2).
\end{align*}
This is equivalent to $\dim(W) = \dim(W_1)+\dim(W_2) = \dim(W_1\oplus W_2)$ and hence $W=W_1\oplus W_2$.
