R^4 subspace question

there is U,W,V subspaces of $\displaystyle R^{4}$ and dimU=dimV=dimW=3
prove that $\displaystyle U\cap V\cap W\neq{0}$
how i tried to solve it:

dim(U+V)=dimU+dimV-$\displaystyle dim(U\cap V)$
because U+V is a subspace of $\displaystyle R^{4}$then $\displaystyle dim(U+V)\leq4$
so $\displaystyle dim(U\cap V)\geq10$
$\displaystyle dim(W+(U\cap V))=dimW+dim(U\cap V)-dim(W\cap U\capV)$
because $\displaystyle W+(U\cap V)$ is a subspace of $\displaystyle R^{4}$then
$\displaystyle dim(W+(U\cap V))\leq4$
by inputing the previos data i get
$\displaystyle dim(W\cap U\cap V)\geq17$
is it correct?

Quote:

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Originally Posted by transgalactic

$\displaystyle dim(W\cap U\cap V)\geq17$ is it correct?

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Impossible, $\displaystyle W\cap U\cap V\subset \mathbb{R}^4$ so, $\displaystyle \dim (W\cap U\cap V)\leq 4$ .

indeed, we have 4 = dim(R^4) ≥ dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 6 - dim(U∩V).

this means that dim(U∩V) ≥ 6-4 = 2.

now we also have 4 ≥ dim((U∩V)+W) = dim(U∩V) + dim(W) - dim((U∩V)∩W) ≥ 5 - dim(U∩V∩W).

this means that dim(U∩V∩W) ≥ 5-4 = 1.

ok i understand now
thanks