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I have a very basic question.

if I have a vector space $V$ of dimension $n$, and two vector subspaces $W,W' \subseteq V$, and I know that $W \cap W' = 0$, and I know that $V\subseteq W+W'$, does it follow that $dim(W)+dim(W')=dim(V)$?

I mean, because the intersection is zero and $V\subseteq W+W'$ implies that $dim(V) \leq dim(W+W')=dim(W)+dim(W')-dim(W\cap W')=dim(W)+dim(W')$, can we deduce that $dim(V)=dim(W)+dim(W')$ just from the finite dimensionality of $V$?

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  • $\begingroup$ It is perfectly correct, I think $\endgroup$ – D.L. Jan 16 '14 at 10:02
  • $\begingroup$ Yes, the assumptions in fact mean that $V = W\oplus W'$. $\endgroup$ – Tobias Kildetoft Jan 16 '14 at 10:04
  • $\begingroup$ So what you're saying is, if $V$ is finite dimension, $W,W' \subseteq V$ intersect only in $0$, and $V\subseteq W+W'$, then we can deduce that $V=W+W'$? $\endgroup$ – Oria Gruber Jan 16 '14 at 10:05
  • $\begingroup$ We don't even need finite dimension here. We have $W + W'\subset V$ since both are subspaces. When we also have $V\subset W + W'$ we thus get $V = W + W'$. The trivial intersection then implies that this sum is direct so $V = W \oplus W'$. $\endgroup$ – Tobias Kildetoft Jan 16 '14 at 10:07
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This is correct. We have $$ V = W + W'$$ since the sum is a subset of $V$. Because of $W \cap W' = 0$ the sum is direct, and so the dimension formula holds. In general, it is $$dim( W + W' ) = dim( W ) + dim( W' ) - dim( W \cap W' ).$$

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  • $\begingroup$ But why $V=W+W'$? If we knew that, that would be great, but we only know that $V \subseteq W+W'$. $\endgroup$ – Oria Gruber Jan 16 '14 at 10:06
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    $\begingroup$ Since both $W$ and $W^\prime$ are subspaces of $V$, $W+W^\prime\subseteq V$ as well. $\endgroup$ – Brian Fitzpatrick Jan 16 '14 at 10:08

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