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I am looking for a proof of the theorem that says:

A is a complex positive definite endomorphism and therefore is A self-adjoint. Does anybody of you know how to do this?

Julien
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1 Answers1

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In general, for $A:H\longrightarrow H$ bounded linear operator on a Hilbert space $H$, $A\geq 0$ implies $A^*=A$. Where $A\geq 0$ means $(Ax,x)\geq 0$ for every $x\in H$. Note that we could more generally simply assume that $(Ax,x)\in \mathbb{R}$ for every $x$ in $H$.

By assumption, $(Ax,x)\in\mathbb{R}$ whence $$ (Ax,x)=\overline{(Ax,x)}=(x,Ax)=(A^*x,x)\quad \Rightarrow \quad ((A-A^*)x,x)=0\quad \forall x\in H. $$

So it boils down to the following key property, which is false in the real case.

Fact: if $T$ is a (not necessarily bounded) linear operator on a complex Hilbert space $H$ such that $(Tx,x)=0$ for every $x\in H$, then $T=0$.

Proof: the usual polarization tricks, assuming semi-linearity in the first variable. With $x+y$, we get $$ 0=(T(x+y),x+y)=(Tx,y)+(Ty,x). $$ And with $x+iy$, $$ 0=(T(x+iy),x+iy)=i(Tx,y)-i(Ty,x). $$ It follows that $(Tx,y)=0$ for every $x,y$, whence $Tx=0$ for every $x$. $\Box$.

Julien
  • 44,791
  • In the finite dimensional case the fact can be proved by the existence of a nonzero eigenvector. Is there a way to extend this to the case of a (possibly continuous) operator on a general Hilbert space? – Elle Najt May 29 '15 at 17:11
  • (Though I think your proof is elegant.) – Elle Najt May 29 '15 at 17:14
  • How do you get $(Tx,y) = 0$ for all $x,y,$ implies $Tx = 0?$ for all $x$. What if $Tx$ is orthogonal to $y$? – Lemon Dec 22 '19 at 20:40
  • This is true if $A:D(A)\subset H \longrightarrow H$ and $(A(x),x)\geq 0$ for all $x \in D(A)$, with $\overline{D(A)}=H$? – Guilherme Apr 02 '20 at 00:33