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Let $V$ be the vector space and $W_1, W_2, W_3$ are the subspaces in $V$. Decide, if $(W_1 \lor W_2) \cap W_3 = W_1 \lor (W_2 \cap W_3)$.

I think it isn't right.

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    Do you have more information about the subspaces? Are the orthogonal to each other, for example? – 2017-02-07
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    No, unfortunately. They are arbitrary. It is from basic linear algebra book. – 2017-02-07
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    The "subspaces" that you've defined are not spaces at all, at least not for arbitrary vectors $w_i$. Each subspace should include the origin, for example. – 2017-02-07
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    It's quite rare that a subspace contains just one vector. – 2017-02-07
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    Yes, it was a nonsense, so I deleted it, but I think I came up with the answer. – 2017-02-07

2 Answers 2

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I think I came up with the solution, I found the contradiction.

If $W_1, W_2, W_3$ are the subspace and their direct sum exists, then

$ W_3 \cap (W_1 \lor W_2) = 0$ by definition,

and because we are working with direct sum, $W_2 \cap W_3$ = 0, but

$W_1 \lor (W_2 \cap W_3) \neq 0$ if $W_1\neq \{0\}$

So that is the contradiction.

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    I need a contradiction so I assume that none of my subspaces intersects with another, hence direct sum exists. – 2017-02-07
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The equality is not true in general, but not for the reason you give.

A subspace consists of a single vector only if it is $\{0\}$. If you want to find a good counterexample, your idea is nonetheless good.

Suppose $\{w_1,w_2,w_3\}$ is a linearly independent set and let $W_i=\langle w_i\rangle$, for $i=1,2,3$.

Then it's easy to see that $(W_1\vee W_2)\cap W_3=\{0\}$, whereas $W_1\vee(W_2\cap W_3)=W_1$.

Note that $W_1\vee W_2$ consists of all linear combinations of the form $\alpha_1w_1+\alpha_2w_2$. On the other hand, if $v=\alpha_1w_1+\alpha_2w_2=\alpha_3w_3\in(W_1\vee W_2)\cap W_3$, then $$ \alpha_1w_1+\alpha_2w_2-\alpha_3w_3=0 $$ so $\alpha_1=\alpha_2=\alpha_3=0$ and $v=0$.

Prove similarly that $W_2\cap W_3=\{0\}$.