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This is a question that I'm trying to solve since yesterday.

Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a linear transformation such that

$$\begin{equation} \langle u,v\rangle = 0, \langle Tu,v\rangle >0\quad\Rightarrow\quad \langle u,Tv\rangle >0 \end{equation}$$

We have to proof the following

1) $\langle u,v\rangle = 0$, $\langle Tu,v\rangle =0\quad\Rightarrow\quad \langle u,Tv\rangle =0$;

2) There exists an orthonormal basis for $T$;

3) $T$ is symmetric.

Im stuck in the first really looks easy but i just cant prove this. I used Cauchy-Schwarz inequality to see that if we have $\langle u,v\rangle=0$ and $\langle Tu,v\rangle=0$, then $|\langle u,Tv\rangle + \langle v,Tv\rangle| = 0$, in that case I want to show that $\langle v,Tv\rangle=0$ so I this implies $\langle u,Tv\rangle=0$. Also, with this $u$ and $v$, I have that $\langle u,v+T^\ast v\rangle=0$. I dont know what to do from here...I`m out of ideas, any help is very welcome.

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up vote 4 down vote accepted

Suppose that $\langle u,v\rangle=0$. Then, of course, $\langle -v,u\rangle=0$.

If $\langle u,Tv\rangle >0$, this is $\langle Tv,u\rangle >0$, and so $\langle v,Tu\rangle >0$, i.e. $\langle Tu,v\rangle >0$.

Similarly, if $\langle u,Tv\rangle <0$, then $\langle u,T(-v)\rangle >0$ and so by the previous paragraph $\langle Tu,-v\rangle >0$, i.e. $\langle Tu,v\rangle <0$.

The last two paragraphs show that if $\langle u,Tv\rangle\ne0$, then $\langle Tu,v\rangle\ne0$. So, if $\langle Tu,v\rangle=0$, then $\langle u,Tv\rangle =0$.

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it really is $>$ – Integral Dec 13 '12 at 13:59
Ok, I've modified the answer to use $>$. – Martin Argerami Dec 13 '12 at 15:15
I dont understand your proof. Remember you dont know yet that T is symmetric. If you are not using this fact, please explain me your argument. – Integral Dec 13 '12 at 22:31
I'm not using that $T$ is symmetric. The argument explains itself. If you don't understand some step, tell me and I'll try to explain. – Martin Argerami Dec 13 '12 at 22:53
ok, thanks! In line 2, how do you go from $\langle Tv, u\rangle >0$ to $\langle v, Tu\rangle >0$ ? – Integral Dec 14 '12 at 0:13

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