TITLE: Prove that if $\langle Tx,x\rangle =0$ for all $x \in X$, then $T = 0$
QUESTION [9 upvotes]: This is exercise 10 in section 3.2 of Kreyszig's Introductory Functional Analysis with Applications:

(Zero Operator): Let $T:X \to X$ be a bounded linear operator on complex inner product space $X$. If $\langle Tx,x\rangle =0$ for all $x \in X$, show that $T=0$. 

I had solved little kindly guide me further     
Solution: 
Let $\langle Tx,x\rangle =0$ for all $x \in X$. Let $x=u+av$ where $v,u$ belong to $X$ and $a$ be the scalar then 
$$
\langle Tx,x\rangle =\langle T(u+av), u+av\rangle
$$
since $T$ is linear then 
\begin{align}
\langle Tx,x\rangle &=\langle Tu+aTv, u+av\rangle \\
    & =\langle Tu,u\rangle +\overline{a}\langle Tu,v\rangle +a\langle Tv,u\rangle +a\overline{a}\langle Tv,v\rangle \\
    & =\overline{a} \langle Tu,v\rangle +a\langle Tv,u\rangle 
\end{align}

REPLY [11 votes]: Hint: The fact that $X$ is a complex inner product space is significant; this fails to hold for real inner product spaces.
You've done well to consider $\langle T(u + av), (u + av) \rangle$.  Now, note the following: since your equality is true for all $a$, it is true for any single value of $a$.  In particular, we have

$\langle Tu,v \rangle + \langle Tv,u \rangle = 0$ (set $a = 1$)
$\langle Tu,v \rangle - \langle Tv,u \rangle = 0$ (set $a = i$)