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

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.

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.

• Anything at all you tried before posting here? Sep 14, 2016 at 18:48
• Yes. I can write up my thought if it would help in some way? Sep 14, 2016 at 18:52
• It sure does, as you are on the right track. However, you don't know whether $T$ is bijective, in fact, you have to show that it is neither injective nor surjective. Still, $Tx$ is in $X$, so you can pick $y$ equal to $Tx$. Sep 14, 2016 at 19:00
• I need to show that it is neither injective or surjective? Why is that? Sep 14, 2016 at 19:08

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.

• I think this is nice, short and elegant answer. +1 Sep 14, 2016 at 20:38

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.

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 $T=0$.

• This makes sense. Thanks. However, one point I'm missing is why you suddenly use $w$ and $z$. Couldn't I just use $x$ and $y$? Sep 14, 2016 at 19:26
• For clarity, you should not. I choose $w$ and $z$ since I wanted a distinction between the $x$ and $y$ I used before and the $w$ and $z$ for the assumption. Sep 14, 2016 at 19:29
• Okay. Thank you very much :) Sep 14, 2016 at 19:43
• I can't understand why both the OP and this answer say $\;T\;$ is injective...In fact, if we're to prove $\;T\;$ is zero then it can't be neither injective nor surjective! Sep 14, 2016 at 20:36
• @DonAntonio Where do I say $T$ is injective except in the header which I copied verbatim from the OP? And two lines below that, I say that when $T$ is indeed $0$, that that is not injective nor surjective, except for the trivial case $X=\{0\}$ of course. Sep 14, 2016 at 21:14