# Linear algebra direct sum proof

zcd

## Homework Statement

Let W1 and W2 be subspaces of a vector space V. Prove that $$W_1\oplus{}W_2=V \iff$$ each vector in V can be uniquely written as x1+x2=v, where $$x_1\in W_1$$ and $$x_2\in W_2$$

## Homework Equations

$$W_1\oplus{}W_2=V$$ means $$W_1\cap W_2 =\{0\}$$, $$W_1 + W_2 =V$$ and W1 & W2 are subspaces of V

8 axioms defining vector space

## The Attempt at a Solution

I'm trying to assume that $$\exists x'_1,x'_2: x'_1+x'_2=v$$ and $$x'_1\in W_1, x'_2\in W_2$$ and then proving $$x'_1=x_1, x'_2=x_2$$, but I'm unsure of where to go from that step.

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## Answers and Replies

Homework Helper
If x1+x2=v and x1'+x2'=v then (x1-x1')+(x2-x2')=0. (x1-x1') is in W1. (x2-x2') is in W2. 0 is in W1. Is (x2-x2') in W1?

zcd
$$(x_2-x'_2)\in W_1\iff x_2-x'_2=0$$

Homework Helper
$$(x_2-x'_2)\in W_1\iff x_2-x'_2=0$$

So you've got it?

zcd
I can see how this proves $$W_1\oplus{}W_2=V$$, but I'm still unsure on how x1 and x2 are unique. Is it because if $$x_i\neq x'_i$$ then the two vectors are in different subspaces, while if $$x_i= x'_i$$ the two subspaces will overlap at zero?

I can see how this proves $$W_1\oplus{}W_2=V$$, but I'm still unsure on how x1 and x2 are unique. Is it because if $$x_i\neq x'_i$$ then the two vectors are in different subspaces, while if $$x_i= x'_i$$ the two subspaces will overlap at zero?
For the reverse argument, can I say that $$S:=W_1 \cap W_2$$ and $$\exists s\inS: x_1+x_2+s=v\implies s=0$$, therefore $$S=\{ 0\}$$