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I've a doubt on the definition of direct sum of two vector subspaces. Is it possible to state that

$$W_1\oplus W_2=V \iff \dim(V)=\dim(W_1)+\dim(W_2) \wedge W_1\cap W_2={ \vec{o} }$$

In particular my doubts are about the fact that the condition $W_1 + W_2=V$ is not mentioned. Is it implied in the two conditions in the statement?

Thanks in advice for your help

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    $\begingroup$ For finite-dimensional vector spaces, your definition is equivalent. But the original definition holds for infinite-dimensional vector spaces as well. $\endgroup$ – Greg Martin Dec 4 '15 at 0:10
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    $\begingroup$ @Greg And by "original definition" I suppose you mean the one stating both $V=W_1 + W_2$ and the intersection is trivial... Otherwise the notion of dimension is not very useful $\endgroup$ – M Turgeon Dec 4 '15 at 4:20
  • $\begingroup$ Thanks for the answers! I forgot to mention that I was supposing a finite dimensional space. Could you give me an idea of how is the condition $V=W_1+W_1$ contained in the ones in the statement? $\endgroup$ – Gianolepo Dec 4 '15 at 23:42
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If $V \neq W_1 + W_2$, this means that there is a vector $v$ in $V$ which is not contained in the sum. But then it spans a subspace of dimension one, not contained in the sum $W_1 + W_2$, which contradicts the fact that $\dim V =\dim W_1 + \dim W_2$.

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  • $\begingroup$ Thanks for the answer, one question if I may, is the condition $W_1\cap W_2=\vec{o}$ necessary in this case? Can't it just be deduced from Grassman formula? $\endgroup$ – Gianolepo Dec 12 '15 at 21:48

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