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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.

ANYN11
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  • Anything at all you tried before posting here? – Hetebrij Sep 14 '16 at 18:48
  • Yes. I can write up my thought if it would help in some way? – ANYN11 Sep 14 '16 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 $Tx$. – Hetebrij Sep 14 '16 at 19:00
  • I need to show that it is neither injective or surjective? Why is that? – ANYN11 Sep 14 '16 at 19:08

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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.

Landon Carter
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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.

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 $T=0$.

Hetebrij
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  • 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 $y$? – ANYN11 Sep 14 '16 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. – Hetebrij Sep 14 '16 at 19:29
  • Okay. Thank you very much :) – ANYN11 Sep 14 '16 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! – DonAntonio Sep 14 '16 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. – Hetebrij Sep 14 '16 at 21:14
  • @Hetebrij In my opinion, all the first two lines are at least very misleading, if not wrong, since you say that "Since $;T;$ is bijective I know that...", but what comes then does not depend on $;T;$ being bijective or not, and this part is weird as you say "for all $;Tx;$ there's some $;y;$...", which is, imo, very unclear. What you wrote thereafter is more seriously wrong, imo, since you wrote $;\langle Txc,y\rangle=\langle Tx,Tx\rangle;$ ...and this implies you believe $;T;$ is surjective... – DonAntonio Sep 14 '16 at 21:59
  • @Hetebrij Cont. .... (and even, why would $;y=Tx;$ and not $;y=Tz,,z\neq x;$ ? I think this part is very confusing, to say the least) , which of course it shouldn't...or at least you didn't prove it. – DonAntonio Sep 14 '16 at 21:59
  • @DonAntonio If $T$ were surjective, the wording would be "for all $x \in X$ there is some $y$ such that $Ty =x$." Which is different from $y =T x \in X$, which is my text paraphrased. And if $\langle Tx , y \rangle =0$ for all $x,y \in X$, then in particular we can take $y =Tx$, as $x \in X$ and $T: X \to X$, which indeed implies $\langle Tx , y \rangle = | Tx |^2$. Surjectivity would be choosing $x$ such that $Tx =y$, i.e. $\langle Tx , y \rangle = | y |^2$. Furthermore, $y =Tx$ by our choice of $y$, we could have also chosen $\tilde{y} = Tz$, but then we cannot conclude $Tx = 0$. – Hetebrij Sep 15 '16 at 16:52