Let $H$ be a Hilbert space, $<,>$ be its norm, and $A:H\to H$ be an operator. Is it true $A=0 \Leftrightarrow\langle\Psi, A \Psi\rangle=0\ \left(^\forall \Psi\in H\right)$?
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2False for real Hilbert spaces, standard fact for complex Hilbert spaces. – geetha290krm Jan 13 '24 at 12:14
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If $A$ is rotation by $\pi/2$ in $\mathbb R^2$, then isn't $\langle v, Av\rangle=0$ for all $v\in \mathbb R^2$? – Al.G. Jan 13 '24 at 12:19
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Doesn't the Related tab give a dupe: https://math.stackexchange.com/questions/2576899/langle-tu-u-rangle-0-forall-u-in-e-longrightarrow-t-0 ? – Al.G. Jan 13 '24 at 12:21
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Thank you for all comments. I know that it is true for complex Hilbert spaces and false for real Hilbert spaces. But I don't find the proof for complex Hilbert spaces. – neconoco Jan 13 '24 at 14:03
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I found the answer https://math.stackexchange.com/questions/1839427/let-h-be-hilbert-and-t-a-blo-such-that-th-rightarrow-h-prove-that-lan?noredirect=1&lq=1. – neconoco Jan 13 '24 at 14:55
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1Does this answer your question? Let $H$ be hilbert and $T$ a BLO, such that $T:H\rightarrow H$. Prove that $\langle T(x),x \rangle = 0$ implies $T = 0$. – Chad K Jan 13 '24 at 15:04
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Yes, thank you for your comment. – neconoco Jan 14 '24 at 02:35
1 Answers
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Theorem:Let $H$ be a complex Hilbert space, $<,>$ be its inner product, and $A$ be an operator. If $<x,Ax>=0$ (for any $x\in H$), then $A=0$.
Proof: For any $x,y\in H$, $<x+y,T(x+y)>=<x,Tx>+<x,Ty>+<y,Tx>+<y,Ty>$ $=<x,Ty>+<y,Tx>$.
By the assumption, $<x,Ty>+<y,Tx>=0$.
We replace $x$ by $ix$ and get
$-i<x,Ty>+i<y,Tx>=0$
By the two equation, we get
$<x, Ty>=0$.
We substitute $x=Ty$ and get $||Ty||^2=0$. It means $Ty=0$ for any $y\in H$.
neconoco
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