Do intersections commute with direct sum? This is just a basic linear algebra question without that much context to it.  I'm wondering if the following identity holds for vector spaces:
$$ (A \oplus  B)  \cap C  = (A \oplus 0 )\cap C  + (0 \oplus B)\cap C. $$
My intuition tells me it's always true, but I could be wrong.
 A: This is so wrong...
Let $A=Span(1,0), B=Span(0,1), C=Span(1,1)$.
Then
$$(A\oplus B)\cap C=\mathbb{R}^2\cap C=C$$ and
$$(A\cap C)\oplus (B\cap C)=0\oplus 0 = 0.$$
A: No, vector spaces don't verify that:
$$(A \oplus  B)  \cap C  = (A\cap C)  \oplus (B\cap C).$$
To prove it, we can find a counterexpample:
Consider the vector subspaces $\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle$, $\langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle$ and $\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle$. Notice that
$$\left(\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle\oplus \langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle\right)\cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle=\mathbb{R}^2\cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle=\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle,$$
but
$$\left(\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle  \cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\right)\oplus\left(\langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle  \cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\right)=0\oplus 0=0,$$
and clearly $\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\neq 0$.
A: Let $\mathbf{A,B,C}$ are subspaces of finite-dimension vector space, then
$(A\bigoplus B)\bigcap C = (A\bigcap C)\bigoplus (B\bigcap C)$.
Proof:


*

*For all $x \in (A\bigoplus B)\bigcap C$, we have $x\in (A\bigoplus
        B)$ and $x\in C$. Then $x$ can be represented by $x=x_1 + x_2, x_1
        \in A, x_2 \in B$ (for unique representation in direct sum),
    $x_1,x_2 \in  (A\bigoplus B)\bigcap C$ (closed additive). So $x_1
    \in C, x_2 \in C$. Then  $x_1 \in A \bigcap C, x_2 \in B \bigcap C$,
    it concludes that $ x=x_1 + x_2 \in (A\bigcap C)\bigoplus (B\bigcap
    C)$, which implies that $(A\bigoplus B)\bigcap C \subseteq (A\bigcap
    C)\bigoplus (B\bigcap C)$.

*For all $x \in (A\bigcap C)\bigoplus (B\bigcap C)$, we have $x = x_1
        + x_2, x_1 \in (A\bigcap C), x_2 \in (B\bigcap C) $.
             Then $x=x_1 + x_2 \in (A \bigoplus B)$ and $x=x_1 + x_2 \in C$(closed additive). It concludes that $(A\bigcap C)\bigoplus
        (B\bigcap C) \subseteq (A\bigoplus B)\bigcap C$.

*From (1),(2) , it concludes that $(A\bigoplus B)\bigcap C =
(A\bigcap C)\bigoplus (B\bigcap C)$.

