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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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What's the definition of "direct sum" that you have? –  DonAntonio Dec 9 '12 at 4:14
    
@DonAntonio Two vectors being added together: $(1,2) \oplus (2,3) = (3, 5)$ –  Imray Dec 9 '12 at 4:17
    
That's not direct sum of vector spaces, that's vector addition. The definition I've usually seen is that we say that $V = W_1 \oplus W_2$ if $V = W_1 + W_2$ and $W_1 \cap W_2 = \{0\}$. –  Javier Badia Dec 9 '12 at 4:23
    
That's not any definition of direct sum but only the definition of sum. Check your notes. –  DonAntonio Dec 9 '12 at 4:23
    
@DonAntonio Yes, I had mistaken the definition. I stand corrected. –  Imray Dec 9 '12 at 4:34
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1 Answer 1

up vote 3 down vote accepted

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

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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On the fourth line, did you mean $u_2-w_2$ rather than $u_2-W_2$? –  Imray Dec 9 '12 at 4:36
    
Yup, typo. I fix it now –  DonAntonio Dec 9 '12 at 4:43
    
Thanks! Can you explain how you came up with the proof? What is your approach to this sort of question? –  Imray Dec 9 '12 at 4:51
1  
Well, I'm extremely intelligent and resourceful and also...no, just kidding: this is a common exercise in basic linear algebra, and I think about 115% of all first year students of mathematics deal with it in this or that fashion. I happen to have some experience since I graduated long ago, and I've even taught this stuff several times, so... –  DonAntonio Dec 9 '12 at 4:55
1  
Ok, I'll try to add something between square parentheses []. –  DonAntonio Dec 9 '12 at 5:07
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