# Showing that $X=U\oplus W$

Let $X$ be a vector space, $V$ and $W$ are subspaces of $X$. Then I would like somebody to help on showing that $X$ is a direct sum of $V$ and $W$ if and only if $x\in X$ has a unique representation $x=v+w$, for some $v\in V$ and $w\in W$. Thanks in advance!

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What definition of direct sum do you have? That $X = V + W$ et $V \cap W = \{0\}$? What have you tried? – nik May 27 '12 at 10:51

1) Suppose $\,X=V\oplus W\,$ , so in particular $\,\forall x\in X\,\,\exists v_x\in V\,,\,w_x\in W\,\,s.t.\,\,x=v_x+w_x\,$ .If also $\,x=a+b\,,\,a\in V\,,\,b\in W\,$ , then $\,v_x+w_x=a+b\Longrightarrow v_x-a=b-w_x\in V\cap W$ , and since by assumption this intersection is trivial, we get $\,v_x-a=0=b-w_x\Longrightarrow a=v_x\,,\,b=w_x$ and the representation above is unique.
2) Suppose now that $\,\forall x\in X\,\,\exists!v_x\in V\,,\,\exists!w_x\in W\,\,s.t.\,\,x=v_x+w_x$ . Then clearly $\,X=V+W\,$ , and if $\,y\in V\cap W\,$ then $\,0_X=0_V+0_W=y+(0-y)\Longrightarrow\,$ by uniqueness of representation, we get that $\,y=0_V=0_W\,$ (since, of course, $\,0=0_X=0_V=0_W\,$ . This was just a matter of notation to make things clearer), so that in fact the sum $\,V+W=X\,$ is direct.