Does $\langle u,Tu \rangle=0$ imply that $T=0$?

Question

I have this simple question :

for an operator $T$ in a complex Hilbert space we have: $\langle u,Tu \rangle =0$ for all $u$ in this Hilbert space. So does this imply that $T=0$? If yes, how to prove it?

Consider a rotation by $\pi/2$ in $\mathbb{R}^2$ with the standard inner product. — Michael Burr, Mar 14 '16 at 01:13

score 3 · Answer 1 · answered Mar 14 '16 at 01:15

3

It is true when the Hilbert space is complex. You can see this via polarization. The others showed that it is false when the field is the reals.

answered Mar 14 '16 at 01:15

Friedrich Philipp

4,051

Friedrich Philipp how o show that in complexe ? – M.luffy Mar 14 '16 at 01:16
See Lemma 2 here: https://www.math.ksu.edu/~nagy/func-an-2007-2008/hs-1.pdf – Friedrich Philipp Mar 14 '16 at 01:20

Tsemo Aristide · Answer 2 · 2016-03-14T01:49:33.017

1

You have $<u+v,T(u+v)>=<u,T(u)>+<u,T(v)>+<v,T(u)>+<v,T(v)>=<u,T(v)>+<v,T(u)>=0$. Replace $v$ by $iv$, you have $<u,T(iv)>+<iv,T(u)>=-i<u,T(v)>+i<v,T(u)>$. This implies that $<u,T(v)>+<v,T(u)>=0$ and $<u,T(v)>-<v,T(u)>=0$. Thus $<v,T(u)>=0$ for every $v$. This implies $T(u)=0$.

edited Mar 14 '16 at 01:49

answered Mar 14 '16 at 01:14

Tsemo Aristide

87,475

it is a complexe hilbert space !! – M.luffy Mar 14 '16 at 01:17

Does $\langle u,Tu \rangle=0$ imply that $T=0$?

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