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Let $H$ be hilbert and $T$ a BLO, such that $T:H\rightarrow H$. Prove that $\langle T(x),x \rangle = 0$ implies $T = 0$.

Any hints to tackle this problem?

i tried writing x as $x = u + v$ where $u \in Y$ and $v \in Y^T$ for some closed linear subspace of $H$, but i did not see somethins smart. Is it maybe smart to try using the contra positive?

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    $\begingroup$ Is it true? What if $T$ is a rotation? $\endgroup$
    – Ningxin
    Jun 25, 2016 at 17:12
  • $\begingroup$ hmmm, i saw H must be a complex space for this to be true. So it probably goes wrong for real spaces somewhere. $\endgroup$
    – Kees Til
    Jun 25, 2016 at 17:19

1 Answer 1

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since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\ \end{eqnarray} Then $$ \langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad (1) $$ so we replace $y$ by $iy$ in the last equality we get : $$ -i\langle T x,y\rangle +i\langle Ty,x\rangle=0 \qquad (2) $$ multiplying $(2)$ by $i$ and add to $(1)$ we get $$ \langle Tx,y\rangle=0 \qquad \forall x,y\in H $$ then we put $y=Tx$ we get $\|Tx\|^2=0$ for all $x\in H$ so $T=0$.

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    $\begingroup$ use the linearity of scalar product and your assumption $\endgroup$
    – Hamza
    Jun 25, 2016 at 17:17
  • $\begingroup$ @QiyuWen in 1 he just writes out the equation as $<T(x),x> + <T(y),y> + <T(x),y> + <T(y),x>$. And the first two terms are 0 by the assumption $\endgroup$
    – Kees Til
    Jun 25, 2016 at 17:21
  • $\begingroup$ Is there some kind of trick solving these kind of questions? How do you for example come up with using $<T(x+y),x+y>$, is it because you know a few terms will cancel out? $\endgroup$
    – Kees Til
    Jun 25, 2016 at 17:30
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    $\begingroup$ yeah and because we have an Polarization identity which use the sum of vectors $\endgroup$
    – Hamza
    Jun 25, 2016 at 17:34
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    $\begingroup$ Somewhat similar idea to Hamza: A common trick to proving something when knowing certain statement is true for all $x\in X$ is to introduce one degree of freedom by setting $x=u+\alpha v$ for $u,v\in X$ and $\alpha$ scalar. $\endgroup$
    – Chee Han
    Jun 25, 2016 at 19:45

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