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I have been asked to prove the following:

If $V$ is a finite-dimensional inner product space over $\mathbb C$, and if $A:V→V$ satisfies $⟨Av,v⟩≥0$ for all $v∈V$, then $A$ is Hermitian.

A proof is available at the address Show that a positive operator is also hermitian but I have come with a different approach that seems a lot simpler (which suggests that it's likely very wrong).

Here's my reasoning: if $A$ is positive ($⟨Av,v⟩≥0$ for all $v∈V$), then the inner product is necessarily real, which means that it is equal to its conjugate, so we can say:

$⟨Av,v⟩ = ⟨Av,v⟩^* = ⟨v,Av⟩$

By definition of Hermitian conjugate:

$⟨v,Av⟩ = ⟨A^\dagger v,v⟩$

Therefore:

$⟨Av,v⟩ = ⟨A^\dagger v,v⟩$ for all $v∈V$

Is this enough to conclude that $A = A^\dagger$, proving that $A$ is Hermitian?

What am I doing wrong?

Thanks!

J.

(My first post on math.stackexchange, so be gentle :))

Community
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Jacques de Molay
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1 Answers1

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Yes. The polarization identity shows that if $⟨Bv,v⟩ = 0$ for all $v$ then $⟨Bv,w⟩ = 0$ for all $v,w$, and thus $B=0$.

Robert Israel
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  • Hello Robert, and thank you for your prompt answer. Do you mind if I ask you to elaborate a bit further? I am just cutting my teeth with linear algebra and I am not sure I see how the way you use the polarization identity confirms my deduction about how $A$ is Hermitian. Thanks! – Jacques de Molay Jun 26 '15 at 19:27
  • Nevermind. I got it now. Thanks again! :) – Jacques de Molay Jun 27 '15 at 11:52