Let me first define the self-adjoint operator in a Hilbert space
$$A \in B(H)$$ be a self-adjoint operator on a Hilbert space $$H$$ if $$A=A^*$$ where $$A^*$$ is the adjoint of $$A$$.
and $$A \in B(H)$$ be a self-adjoint operator the $$A$$ is a positive operator if $$\geq0$$
My question let we have $$A$$ as a self-adjoint operator then I need to prove that $$A$$ is a positive operator if and only if every spectral value of $$A$$ is a non-negative real number.
Let me define the spectral value, let $$H$$ be a Hilbert space over a field $$K$$ and $$A\in B(H)$$ the set $$\sigma(A)$$={$$\lambda\in K$$ such that ($$A-\lambda I)$$is not invertible in $$B(H)$$} is called the spectrum of $$A$$ and the elements of $$\sigma(A)$$ are known as the spectral value of $$A$$

• Hint: If $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $\langle x, Ax\rangle = \lambda |x|^2$. Aug 2, 2022 at 12:43
• @eyeball That only gives you a complete proof for finite dimensional $H$. Aug 2, 2022 at 15:58
• @Exa Note that askers are expected to provide context for their questions, as is clarified here. Please edit your question to tell us, for instance, where you encountered this question, what you have tried so far, and any other thought you may have. Also, perhaps you could address the following: are you comfortable proving this in the case that $H$ is finite dimensional? Are you able to get either direction of the "if" and "only if"? Aug 2, 2022 at 16:02
• @BenGrossmann I believe it should work for separable infinite dimensional $H$ as well. I'll admit I'm less sure of inseparable $H$. Aug 2, 2022 at 16:07
• @eyeballfrog Even in separable spaces, not every element of the spectrum is an eigenvalue. For instance, $0$ is not an eigenvalue of the self-adjoint and compact operator $(x_1,x_2,x_3,\dots) \mapsto (x_1/1,x_2/2,x_3/3,\dots)$ over $\ell^2$, but it is an element of its spectrum. Aug 2, 2022 at 16:09

Theorem: Suppose that $$A$$ is a bounded self-adjoint linear operator on a complex Hilbert space $$X$$ such that $$\langle Ax,x \rangle \ge 0$$ for all $$x\in X$$. Then the spectrum of $$A$$ is a subset of the non-negative real axis.

Proof: By standard results, the spectrum $$\sigma(A)$$ is a subset of $$\mathbb{R}$$. Because of the positivity of $$A$$, it follows that, for any real $$\lambda > 0$$ and $$x\in X$$, one has $$\lambda\|x\|^2=\lambda\langle x,x\rangle \le \langle (A+\lambda I)x,x\rangle \le \|(A+\lambda I)x\|\|x\| \\ \lambda \|x\|\le \|(A+\lambda I)x\|. \tag{*}$$ From this it follows that $$\mathcal{N}(A+\lambda I)=\{0\}$$, and $$A+\lambda I$$ has a bounded inverse on its range. The range of $$A+\lambda I$$ is dense for all $$\lambda > 0$$ because $$\mathcal{R}(A+\lambda I)^{\perp}=\mathcal{N}(A+\lambda I)=\{0\}.$$ To see that the range $$\mathcal{R}(A+\lambda I)$$ is closed for $$\lambda > 0$$, suppose that $$\{ x_n \}$$ is a sequence such that $$(A+\lambda I)x_n$$ converges to some $$y$$ as $$n\rightarrow\infty$$. Then $$\{ (A+\lambda I)x_n \}$$ is a Cauchy sequence, which also forces $$\{ x_n \}$$ to be a Cauchy sequence by inequality $$(*)$$. So $$\lim_n x_n = x$$ exists and $$y=\lim_n (A+\lambda I)x_n = (A+\lambda I)x,$$ which proves that the range of $$A+\lambda I$$ is closed. Therefore, $$A+\lambda I$$ has a bounded inverse, which proves that $$-\lambda\in\rho(A)$$ for all $$\lambda > 0$$. Therefore, $$\sigma(A)\subseteq [0,\infty)$$. $$\;\;\blacksquare$$

I'll prove one direction. First I need to define the notion of a bounded below operator. A continuous linear operator $$T: H \rightarrow H$$ is called bounded below if there exists a positive number $$\delta>0$$ such that for any $$x\in H$$ we have $$||Tx||\geq||x|| \delta$$. You can easily see that a bounded below operator is one-one and has a closed range.

Now assume that for any $$h \in H, \geq 0$$. Let $$\alpha<0$$, then we show that $$A-\alpha I$$ is invertible therefore it is not the spectrum of $$A$$. For any $$h \in H$$,

$$$$\begin{split} ||(A-\alpha I)h||^2=<(A-\alpha I)h,(A-\alpha I)h> \\ &=||Ah||^2+|\alpha|^2||h||^2-2\alpha \\ &\geq |\alpha|^2||h||^2. \end{split}$$$$ Therefore $$||(A-\alpha I)h||\geq|\alpha|||h||$$, which means $$A-\alpha I$$ is bounded below, hence it is one-one ($$ker(A-\alpha I)=0$$) and has a closed range. To prove invertibility it remains to show that the range is $$H$$.

To prove the last part, I use the fact that $$T(H)=ker(T^*)^\perp$$ For any bounded operator $$T$$ with a closed range.

So, we have $$(A-\alpha I(H))=ker((A-\alpha I)^*)^\perp=ker(A^*-\bar\alpha I)^\perp=ker(A-\alpha I)^\perp=0^\perp=H$$, hence $$A-\alpha I$$ is on-to.

To prove the other direction, the only proof I know requires knowledge about the existence of a positive square root of a positive operator (Here by positive I mean an operator with nonnegative spectrum) which comes from the functional calculus theorem. So I will not discuss it.