Question with inner product 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 item...it 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.

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 $\langle u,Tv\rangle =0$. 
