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Suppose $V$ is vector space and $V_i,i=1,2,...,n$ are subspaces of $V$.
We want to show if all vectors of $V$ have a unique representation of the form $v=v_1+v_2+...+v_n , v_i\in V_i$ then $V=V_1\oplus V_2\oplus ...\oplus V_n$.
I have proved it for the case when $n$ is two and for general I need to show that $V_i \cap (V_1+...+V_(i-1)+V_(i+1)+...V_n)=(0)$ for all $i$ ,but I don't know how.
And a second question arises in the proof of this latter mater; is it true that if $V_i\cap V_1=(0)$ and $V_i\cap V_2=(0)$, then $V_i\cap (V_1+V_2)=(0)$?

Gratefully waiting for your hints or solutions.

Aref
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    Consider $V=\mathbb{R}^2$, $V_1=\mathbb{R}(1,0)$, $V_2=\mathbb{R}(0,1)$ and $V_i=\mathbb{R}(1,1)$. – Julian Kuelshammer May 23 '14 at 14:12
  • http://math.stackexchange.com/questions/691237/proof-sum-i-1pe-i-doteq-bigoplus-i-1p-e-i-leftrightarrow-forall-i-i?lq=1 , http://math.stackexchange.com/questions/690492/proof-by-contradiction-e-1e-2-doteq-e-1-oplus-e-2-leftrightarrow-e-1-cap-e?lq=1 , http://math.stackexchange.com/questions/691203/proof-by-contradictione-1-e-2-doteq-e-1-oplus-e-2-leftrightarrow-forall-x-i – mle May 23 '14 at 15:39

1 Answers1

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The intersection can not be the empty set, since all of those subspaces contains the zero vector.

Hint: if there is a vector in the intersection then find two representation for it. Use this to find a representation for the zero vector. This would mean that all coefficients are zero which should prove that the vector that you started with is the zero vector

Belgi
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