Hi guys.

i need help to prove the following:

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

Prove that $\displaystyle W$ must be a trivial subspace."

this is what i've tried so far:

let's suppose there exist only one subspace $\displaystyle W'$.

and let's suppose it isnottrivial.

let $\displaystyle B_w=\left \{ w_1, w_2... w_n \right \}$ be the base for $\displaystyle W$.

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

let $\displaystyle v\in V$,

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

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

according to the assumption, there exist only one $\displaystyle W'$, but there are infinite number of bases for $\displaystyle W$.

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!