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.

 

[Dick]

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

 

[Dick]

(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?

 

[Dick]

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?

It doesn't prove W1+W2=V. It proves if W1+W2=V then the decomposition is unique. You have to also prove if the decomposition is unique then W1+W2=V.

 

[zcd]

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\}