Let $A$, $B$ and $C$ be three subspaces in a vector space V. Is the following true or false?
A ∩ (B + C) = (A ∩ B) + (A ∩ C)
I think that this is false but I am struggling to find a proper counter example to prove it. Say we let $A$ be the $x$-axis, $B$ be the $y$-axis, and $C$ be the $z$-axis. I can't seem to think of any way to get one side not to equal the other. It seems like the coordinates $(0, 0, 0)$ can work on either side. For example $A$ $∩$ $((0, 3, 0) + (0, 0, 2))$ for the left side of the equation still gives $(0, 0, 0)$. And a coordinate that intersects both the $x$- axis and $y$- axis as well as the $x$- axis and $z$-axis (for the right side of the equation) is also $(0, 0, 0)$.
Any help?
Let $A = \langle(0,1) \rangle, B = \langle(1,1) \rangle, C = \langle(-1,1) \rangle$ be subspaces of $\mathbb{R}^2$. Now,
$$ B+C = \{(a-b,a+b) : a,b \in \mathbb{R}\} = \mathbb{R}^2 $$
so $A \cap (B+C) = A$. However, $A \cap B = A \cap C = 0$ and so $(A \cap B) + (A \cap C) =0$.