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Let $T\in K(H)$ be a self-adjoint compact operator on Hilbertspace $H$.

I need to prove that: all eigenvalues of T are positive $\iff$ $T$ is positive ($\langle Tx,x\rangle \geq 0\thinspace \forall x\in H$)

So here is my attempt:


1.Let $T$ be positive and $x\in H$ an eigenvector. Then $0\leq\langle Tx,x\rangle=\lambda\lVert x\rVert^2\Rightarrow $ eigenvalues are positive

2. Let all eigenvalues be positive. We need to show $\langle Tx,x\rangle \geq 0\forall x\in H$. According to the spectral theorem I can write $x=y\thinspace +\sum\limits_{k}\thinspace \langle x,e_k\rangle e_k\thinspace\forall x\in H$ because $H = \ker(T) \oplus \overline{\operatorname{span}(\{e_1, e_2, \ldots \})}$. Therefore $Tx=Ty+\sum\limits_{k}\thinspace \langle x,e_k\rangle Te_k=\sum\limits_{k}\thinspace\lambda_k \langle x,e_k\rangle e_k$.

So we can say $\langle Tx,x\rangle=\langle Tx,y+\sum\limits_k\thinspace \langle x,e_k\rangle e_k\rangle=\langle Tx,y\rangle+\langle Tx,\sum\limits_k\thinspace \langle x,e_k\rangle e_k\rangle=\langle x, Ty\rangle+\langle Tx,\sum\limits_{k}\thinspace \langle x,e_k\rangle e_k\rangle =0+\langle Tx, \sum\limits_{k}\thinspace \langle x,e_k\rangle e_k\rangle=\langle \sum\limits_{k}\thinspace\lambda_k \langle x,e_k\rangle e_k, \sum\limits_k\thinspace \langle x,e_k\rangle e_k\rangle$

This is the part where I stuck. The sums are converging. Therefore I can't take them out of the scalar product.

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The series here converge in the norm of the Hilbert space. Since inner product is continuous for the norm topology we can write $\langle Tx,x \rangle =\lim_{N \to \infty} \langle \sum_{k=1}^{N},\lambda_k \langle x,e_k \rangle e_k,\sum_{k=1}^{N},\lambda_k \langle x,e_k \rangle e_k \rangle$ and you can expand the inner product. Finally you get $\langle Tx,x \rangle = \sum \lambda_k ^{2} \langle x,e_k \rangle^{2} \geq 0$.

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  • $\begingroup$ Great, thanks :) $\endgroup$ – ThomasMuller Apr 27 '18 at 11:09

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