Do intersections commute with direct sum?

Question

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.

Answer 1

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 $\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\neq 0$.

Answer 2

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:

1. 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 $(A\bigoplus B)\bigcap C \subseteq (A\bigcap C)\bigoplus (B\bigcap C)$.

2. 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 $(A\bigcap C)\bigoplus (B\bigcap C) \subseteq (A\bigoplus B)\bigcap C$.

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

Answer 3

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 $$(A\cap C)\oplus (B\cap C)=0\oplus 0 = 0.$$