1
$\begingroup$

Let $Z$ be a vector space over some field $K$ and $U, V, W \subseteq Z$ linear subspaces. I am trying to prove or disprove the statement $$(U+W)\cap V = U\cap V + W\cap V,$$ where $X + Y := \{x+y\,|\, x\in X, \,y \in Y\}$.

Since I could not come up with counter-examples (I tried with vectors from $\mathbb R^3$, maybe there are examples that are more advanced I did not come up with), I started trying to prove it. I tried to start with the "$\subseteq$"-direction:

Let $x \in (U + W)\cap V.$ Then $x\in U+W$ and $x\in V$. We want to show that $x\in U\cap V + W\cap V$, i.e. find $u \in U\cap V$ and $w\in W \cap V$ with $x = u + w$. But here I already don't know how to proceed since I don't know how to find those $u, v$. Any help is appreciated.

EDIT: The "$\supseteq $"-dircetion should be true, I already proved that.

  • 0
    Try thinking about 1-dimensional subspaces of $\Bbb{R}^2$. – 2017-01-19

2 Answers 2

1

Take $Z=\Bbb{R}^2$ and $U$, $V$, $W$ to be three distinct one-dimensional subspaces. Then any one of $U$, $V$, and $W$ is contained in the sum of the other two, so, in particular, $V\cap(U+W)=V$. However, the intersection of any two of these subspaces is $\{0\}$, so $V\cap U+V\cap W=\{0\}+\{0\}=\{0\}$.

  • 0
    Oh of course, I somehow thought that $(1,1)\cap (1,0) = (1,0) $ – 2017-01-19
2

This is wrong. For example, let $U=(1,0), \:W=(0,1), \: V=(1,1)$. Then $U+W=\Bbb{R}^2$ and so $(U+W)\cap V = (1,1)$. But $U\cap V + W\cap V=0+0=0$. In fact we can prove in general $$ U\cap V + W\cap V\subset (U+W)\cap V $$