# Direct sum of trivial subspaces

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• Jan 14th 2013, 10:42 AM
Stormey
Direct sum of trivial subspaces
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 is not trivial.

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!
• Jan 14th 2013, 03:58 PM
johng
Re: Direct sum of trivial subspaces
Suppose $\displaystyle W\ne V$ and $\displaystyle W\ne <0>$.
let $\displaystyle B_w=\left \{ w_1, w_2... w_n \right \}$ be a base for $\displaystyle W$ and $\displaystyle \left \{ y_1, y_2... y_m \right \}$ a basis for $\displaystyle W^{\prime}$. Set $\displaystyle W_1^{\prime}=<w_1+y_1, y_2, ... y_m>$.
You can easily show, I hope, that $\displaystyle V=W\oplus W_1^{\prime}$ and $\displaystyle W^{\prime}\ne W_1^{\prime}$
• 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 $\displaystyle W_1^{\prime}=<w_1+y_1, y_2, ... y_m>$ is linear independent?
• Jan 17th 2013, 07:56 AM
johng
Re: Direct sum of trivial subspaces
Yes, that's part of the proof.
Suppose $\displaystyle a_1(w_1+y_1)+a_2y_2+...a_my_m=0$, i.e., $\displaystyle a_1w_1+a_1y_1+...a_my_m=0$. Since $\displaystyle w_1,y_1,...y_m$ are part of a basis of $\displaystyle V$, they are linearly independent. Hence all a's are 0 and so $\displaystyle w_1+y_1, y_2,...y_m$ are linearly independent.