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This question copied from "Linear Algebra - Friedberg". Can anyone explain the procedure, i.e. the strategy, of how to prove the statement you're asked to prove.

The question is:

Let W1 and W2 be subspaces of a vector space V . Prove that V is the direct sum of W1 and W2 if and only if each vector in V can be uniquely written as x1 + x2 where x1 ∈ W1 and x2 ∈ W2.

My swing at it: $V = W_1 \oplus W_2 \ \ \ \ \ <=> \ \ \ \ \ V = \{x_1 + x_2: x_1 \in W_1, x_2 \in W_2\}$

I don't know how to proceed. In reality, the answer seems so obvious to me, I just don't know how to put it down on paper.

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    @DonAntonio Yes, I had mistaken the definition. I stand corrected.2012-12-09

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$(1)\;\;\;\;\;V=W_1\oplus W_2\Longrightarrow W_1\cap W_2=\{0\}\,\,,\,\,V=W_1+W_2$

Supose that for some vector $\,v\in V\,$ we have two expressions -- [directly, no reduction ad absurdum as we don't assume the expressions are different] --

$v=w_1+w_2=u_1+u_2\,\,,\,\,w_i,u_i\in W_i\,\,,\,i=1,2\Longrightarrow\,\,\text{-- [gather similar terms] --} $

$w_i-u_1=u_2-w_2\in W_1\cap W_2=\{0\}\Longrightarrow w_1=u_1\,\,,\,w_2=u_2$

and the expression is unique

$(2)\;\;\;\;\;\;\;\;V=W_1+W_2\,\,\,\text{and every vector in}\,\,V\,\,\text{ has a unique expression} \,\,v=w_1+w_2$

$w_i\in W_i\,\,,\,i=1,2$

$x\in W_1\cap W_2\Longrightarrow x=0+x=x+0\,\,\text{are two expressions for this vector}\Longrightarrow x=0$

[Whatever is in the intersection gives us a straightforward contradiction to the assumption of trivial intersection ...unless... it is the zero vector]

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    Ok, I'll try to add something between square parentheses [].2012-12-09