# Proof that $V = W_1 \oplus W_2$

One of many assignments is:

Let $W_1$ and $W_2$ be subspaces of a vector space $V$. Prove that $V$ is the direct sum of $W_1$ and $W_2$ if and oly if each vector in $V$ can be uniquely written as $x_1 + x_2$, where $x_1 \in W_1$ and $x_2 \in W_2$.

I'm not quite sure about the uniquely in the question. Does it actually mean that any $v \in V$ can only be written by one certain pair of vectors from $W_1$ and $W_2$? If so, I don't see why it is necessarily only a one pair of vectors that, added to each other, gives that certain $v \in V$.

Thank you for any explanation !

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Yes, this is what the uniqueness mean.

Assume $v=w_{1}+w_{2}$ but also $v=w_{1}'+w_{2}'$ then $$w_{1}+w_{2}=w_{1}'+w_{2}'\implies w_{1}-w_{1}'=w_{2}'-w_{2}$$

but $W_{1},W_{2}$are subspaces so $w_{1}-w_{1}'\in W_{1}$ and $w_{2}'-w_{2}\in W_{2}$ but those two elements are equal so $$w_{2}'-w_{2},w_{1}'-w_{1}\in W_{1}\cap W_{2}=\{0\}$$

Now deduce that both are the exact same representation.

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Yes. If $x=x_1+x_2=x'_1+x'_2$, then $x_1-x'_1 = x'_2-x_2$. Hence either $x_j=x'_j$ for $j=1,2$, or $W_1 \cap W_2 \neq \{0\}$.

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thank you. It's so obvious now... It seems I still have to get used to the right way of thinking. – foaly Sep 27 '12 at 12:03

Yes, this is equivalent to saying that $V=W_1+W_2$ (i.e., without the uniqueness assumption) and $W_1\cap W_2=\{0\}$.

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The uniqueness requirement is part of the definition of a direct sum of subspaces. Without it you have the definition of a sum of subspaces.

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the definition in my book (linear algebra by friedberg, insel, spence) does not directly mention uniqueness. maybe it's supposed to be implied. – foaly Sep 27 '12 at 11:58
@Marc The OP's book probably defines a sum of subspaces to be a direct sum if the intersection of the two subspaces is trivial, and wants the OP to prove uniqueness of representation as a consequence. – Ragib Zaman Sep 27 '12 at 12:03