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It's a question from a test I didn't knew how to solve:

An inner product space in $V=R^n$, such as $$\left\langle \pmatrix{ x_1\\ \vdots \\ x_n}\pmatrix{ y_1\\ \vdots\\ y_n} \right\rangle = x_1y_1 + x_2y_2 +\ldots +x_ny_n $$ (standard)

$ v , u $ are not linear dependent in $V$. And $W = \operatorname{sp}d \left \{ (u+v),(u-v) \right \}$; now I am asked to prove that if exists a $w$ in $W^\perp$, $w\neq0$ then : $3 \leq \dim(V)$.

I tried to check for dimensions $2$, $1$ and to prove that $w=0$ but with no luck...

Thanks

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"sp" stands for "span", right? So $W$ is in fact the vector space generated by $u$ and $v$. Then show that $\{u,v,w\}$ is linearly independent. –  Davide Giraudo Feb 19 '12 at 11:03
    
yes it is :) I will try to solve it now. thank you! –  YNWA Feb 19 '12 at 11:10
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1 Answer

up vote 4 down vote accepted

Since $2u=(u+v)+(u-v)$ and $2v=(u+v)-(u-v)$, we have $\operatorname{span}\{u+v,u-v\}=\operatorname{span}(u,v)$. If we have $au+bv+cw=0$ where $a,b,c\in\mathbb R$ then $$0=\langle au+bv+cw,w\rangle=a\langle u,w\rangle+b\langle v,w\rangle+c\langle w,w\rangle=c\langle w,w\rangle.$$ Since $w\neq 0$ we have $\langle w,w\rangle\neq 0$ so $c=0$ and $au+bv=0$. Since $u$ and $v$ are linearly independent, we have $a=b=0$ so finally the set $\{u,v,w\}$ is linearly independent.

Note that the fact that $\langle\cdot,\cdot\rangle$ was the usual inner product was not necessary; it would work for example if $\langle x,y\rangle=\sum_{j=1}^na_jx_jy_j$ where $a_j$ are positive real numbers.

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thank you, simple and well written... :) –  YNWA Feb 19 '12 at 11:33
    
You're welcome. –  Davide Giraudo Feb 19 '12 at 11:33
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