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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}

Ben Grossmann
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maria
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    I've made a quick edit to make your question readable. In the future, please format your mathematical expressions using MathJax, as is described on this page. – Ben Grossmann Jun 01 '15 at 12:33
  • If $b(x,y)$ is a sesquilinear form (linear in the first coordinate, conjugate linear in the second) then $b(x,y) = \frac{1}{4}\sum_{n=0}^{3}b(x+i^{n}y,x+i^{n}y)$. Apply this to $b(x,y)=\langle Tx,y\rangle$, knowing that $\langle Tx,x\rangle =0$. – Disintegrating By Parts Jun 01 '15 at 15:39
  • How do you go from the 2nd line to the 3rd line when calculating $\langle Tx,x\rangle$? – James Sep 16 '19 at 13:20
  • @James Using that, on a complex inner product space, the inner product is conjugate-linear in the first variable and linear in the second by definition. – wueb Oct 23 '20 at 13:25

1 Answers1

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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 $a = i$)
Ben Grossmann
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