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Let $W_1, W_2, W_3,..., W_k$ be subspaces of a finite-dimensional vector space V.

Claim

There $\exists $ $\gamma_1 \cup \gamma_2\cup..\cup\gamma_k$ which is an ordered basis for $V$ ($\gamma_i$ is an ordered basis for $W_i$)

$\rightarrow$ $\;\;$$V = W_1\oplus W_2...\oplus W_k$,

Proof

For each $i$, let $\gamma_i$ be an ordered basis for $W_i$ s.t. $\gamma_1\cup\gamma_2\cup..\cup\gamma_k$is an ordered basis for V.

Then $V= span(\gamma_1\cup\gamma_2\cup..\cup\gamma_k)= span(\gamma_1)+span(\gamma_2)+\cdot\cdot\cdot+span(\gamma_k)=\sum_{i=1}^{k}W_i$

Fix $j\;(1\le j\le k)$, and suppose that, for some nonzero vector $v\ \in V$,

$v\in W_j\cap\sum_{i\neq j}W_i$.

Then

$v\in W_j=span(\gamma_j)$ and $v\in \sum_{i \neq j}W_i = span(\cup_{i \neq j} \gamma_i)$

Hence $v$ is a nontrivial linear combination of both $\gamma_j$and$\cup_{i \neq j} \gamma_i$,

so that $v$ can be expressed as a linear combination of $\gamma_1\cup\gamma_2\cup..\cup\gamma_k$

in more than one way. But it contradicts to $\gamma_1\cup\gamma_2\cup..\cup\gamma_k$ is a basis for $V$

Thus we conclude that $ W_j\cap\sum_{i\neq j}W_i = \{0\}$

Beverlie
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  • Maybe your $\gamma_i$ should be disjoint? – User May 14 '17 at 08:22
  • @tommyxu3 at first I supposed like you but it wasn't needed to be. secret is on the statement that "$\beta$ is basis for V if and only if $v \in V$ can be uniquely expressed as a linear combination of $\beta$" – Beverlie May 14 '17 at 08:25
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    Let $V=\Bbb{R}^3,~W_1=x-y$ plane, $W_2=y-z$ plane and $\gamma_1={(1,0,0),(0,1,0)},~\gamma_2={(0,1,0),(0,0,1)}.$ Then $W_1+W_2$ is not an internal direct sum. – User May 14 '17 at 08:29
  • @tommyxu3 it's interesting.. could you pick up the absurd point from my proof? – Beverlie May 14 '17 at 08:40
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    If some $\gamma_i$ intersect, then your conclusion that $v$ has more than one expressions may fail, for the two may be identical. – User May 14 '17 at 09:01
  • @tommyxu3 this claim is about just existence not about unique existence. I think so my proof is correct – Beverlie May 15 '17 at 10:43
  • What I pointed out is exactly on your condition of existence, and you can see why it doesn't work by the counterexample. – User May 15 '17 at 13:29

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