# Direct sum of trivial subspaces

• Jan 14th 2013, 10:42 AM
Stormey
Direct sum of trivial subspaces
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.

• Jan 14th 2013, 03:58 PM
johng
Re: Direct sum of trivial subspaces
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 $W_1^{\prime}=$.
You can easily show, I hope, that $V=W\oplus W_1^{\prime}$ and $W^{\prime}\ne W_1^{\prime}$
• Jan 15th 2013, 02:50 AM
Stormey
Re: Direct sum of trivial subspaces
Hi johng, thanks for your help.

don't i need to show first that $W_1^{\prime}=$ is linear independent?
• Jan 17th 2013, 07:56 AM
johng
Re: Direct sum of trivial subspaces
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.