Direct sum of trivial subspaces

**Stormey** · January 14th 2013, 10:42 AM

Hi guys.
i need help to prove the following:

"Let $W$ be subspace of $V$ , such that there exist one and only one subspace $W'$ for wich $W\oplus W'=V$ .
Prove that $W$ must be a trivial subspace."

this is what i've tried so far:

let's suppose there exist only one subspace $W'$ .
and let's suppose it is not trivial.

let $B_w=\left \{ w_1, w_2... w_n \right \}$ be the base for $W$ .
let $B_{w}'=\left \{ w'_1, w'_2... w'_m \right \}$ be the base for $W'$ .

let $v\in V$ ,
since $W\oplus W'=V$ , there exist one and only one linear combination such that:

$v=\alpha_1w'_1+\alpha_2w'_2+...+\alpha_mw'_m+\alph a_{m+1}w_1+\alpha_{m+2}w_2+...+\alpha_{n}w_n$

according to the assumption, there exist only one $W'$ , but there are infinite number of bases for $W$ .
therefor, i can define new base for W and use other scalares...

now, i'm not sure if i'm in the right way or where, if at all, lies the contradiction.
i could really use some direction here.

thanks in advanced!

**johng** · January 14th 2013, 03:58 PM

Suppose $W\ne V$ and $W\ne <0>$ .
let $B_w=\left \{ w_1, w_2... w_n \right \}$ be a base for $W$ and $\left \{ y_1, y_2... y_m \right \}$ a basis for $W^{\prime}$ . Set $W_1^{\prime}=<w_1+y_1, y_2, ... y_m>$ .
You can easily show, I hope, that $V=W\oplus W_1^{\prime}$ and $W^{\prime}\ne W_1^{\prime}$

**Stormey** · January 15th 2013, 02:50 AM

Hi johng, thanks for your help.

don't i need to show first that $W_1^{\prime}=<w_1+y_1, y_2, ... y_m>$ is linear independent?

**johng** · January 17th 2013, 07:56 AM

Yes, that's part of the proof.
Suppose $a_1(w_1+y_1)+a_2y_2+...a_my_m=0$ , i.e., $a_1w_1+a_1y_1+...a_my_m=0$ . Since $w_1,y_1,...y_m$ are part of a basis of $V$ , they are linearly independent. Hence all a's are 0 and so $w_1+y_1, y_2,...y_m$ are linearly independent.

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