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$V=X \oplus Y$ if every vector in $V$ can be written uniquely as $v=x+y$ for $x \in X$ and $y \in Y$.

Suppose $V=X \oplus Y$. We want to show $X \cap Y =\{0\}$. Suppose not. Let $v \in X \cap Y$. Then $v=v+0$ where $v \in X$ and $0 \in Y$. Also, $v=0+v$ where $0 \in X$ and $v \in Y$. Thus, $v$ cannot be written uniquely.

Is this proof sufficient?

user5826
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  • Is $0\in X\cap Y$? You seem to be assuming it is, which is not valid. – The Count Mar 18 '19 at 03:52
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    @TheCount We know it is since $X$ and $Y$ are subspaces. – user5826 Mar 18 '19 at 03:54
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    Indeed, if $X$ and $Y$ are subspaces, then $X\cap Y={}$ is impossible. – Gerry Myerson Mar 18 '19 at 03:58
  • @GerryMyerson I meant to write ${0}$. Edited – user5826 Mar 18 '19 at 04:01
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    @JonathanZ on second thought, you're entirely right. I'll delete my comment – rubikscube09 Mar 18 '19 at 05:18
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    This is really a direct proof, so you shouldn’t disguise it as a proof by contradiction. You already know that ${0}\subseteq X\cap Y$, because the intersection of subspaces always contains $0$. So now you take something in $X\cap Y$; the fact that $v=v+0=0+v$, but the expression is unique, tells you something about the two expressions, from which you can conclude that.... – Arturo Magidin Mar 18 '19 at 05:20
  • @ArturoMagidin: I'm used to seeing this done by contradiction, like in the question. But your answer is really slick and clever. Some might say it's too clever, but I'd give it an upvote if it was an answer. – JonathanZ Mar 18 '19 at 05:24
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    @JonathanZ: It's a "fake proof by contradiction". See here and here – Arturo Magidin Mar 18 '19 at 14:39
  • If you want them to be subspaces, you should call them subspaces. Since you didn't, I didn't assume it. – The Count Mar 20 '19 at 02:14

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The only thing missing is specifying that $v \ne 0$, so instead of saying "Let $v \in X \cap Y$" I'd say "Pick a non-zero $v$ in $X \cap Y$". Other than that it's fine.

JonathanZ
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