Show that if $\langle Tx,y \rangle = 0$ for all $x,y \in X$, then $T$ is the zero operator.

Question

Let $X$ be a complex vector space with inner product and $T \, : \, X \rightarrow X$ a bounded linear operator.

Show that if $\langle Tx,y \rangle = 0$ for all $x,y \in X$, then $T$ is the zero operator.

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My current suggestion: Since $T$ is a bijective mapping I now that for all $Tx$ there is some $y$, that is $Tx = y$ for some $x$ and $y$. Since this should be true for all $x,y \in X$ i can just write $$\langle Tx,y \rangle = \langle Tx,Tx \rangle$$ This is only true if $Tx = 0$ if we should have this equality for all $x$, hence $T$ is the zero operator.

Answer 1

I believe this is going to be slightly different from whatever you are going to see as solutions:

Fix any $x\in X$ then $\langle Tx,y\rangle=0$ for every $y\in X$ shows that $Tx\in X^{\perp}=\{0\}$ and thus $x\in Null(T)$. But this is true for any $x\in X$ so $X=Null(T)$ showing $T$ is the zero operator.

Answer 2

Approach of ANYN11, added for convience of the reader:

Since $T$ is a bijective mapping I now that for all $Tx$ there is some $y$, that is $Tx = y$ for some $x$ and $y$. Since this should be true for all $x,y \in X$ i can just write $$\langle Tx,y \rangle = \langle Tx,Tx \rangle$$ This is only true if $Tx = 0$ if we should have this equality for all $x$, hence $T$ is the zero operator.

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So, there are a few points in you approach which should be rephrased.
First of all, if $T$ is indeed $0$, then $TX = \{ 0 \} \neq X$ and $Tx=Ty$ for all $x,y \in X$, thus $T$ is not bijective.
However, we have that $T : X \to X$, so clearly $Tx \in X$, so you know that for all $x \in X$ there is some $y \in X$, namely $Tx$, such that $Tx =y$.
Then you can use the assumption on $T$, $\langle Tw , z \rangle = 0$ for all $w,z \in X$, to conclude \begin{align} \| Tx \|^2 = \langle Tx, Tx \rangle = \langle Tx , y \rangle =0, \end{align} so $Tx = 0$.
As this holds for all $x \in X$, we have $T=0$.