Show that a positive operator on a complex Hilbert space is self-adjoint

Question

Let $(\mathcal{H}, (\cdot, \cdot))$ be a complex Hilbert space, and $A : \mathcal{H} \to \mathcal{H}$ a positive, bounded operator ($A$ being positive means $(Ax,x) \ge 0$ for all $x \in \mathcal{H}$).

Prove that $A$ is self-adjoint. That is, prove that $(Ax,y) = (x, Ay)$ for all $x,y \in \mathcal{H}$.

Here's what I have so far. Because $A$ is positive we have $\mathbb{R} \ni (Ax,x) = \overline{(x,Ax)} = (x,Ax)$, all $x \in \mathcal{H}$.

Next, I have seen some hints that tell me to apply the polarization identity:

$$(x,y) = \frac{1}{4}((\lVert x+y \rVert^2 + \lVert x-y \rVert^2) - i(\lVert x + iy \rVert^2 - \lVert x - iy \rVert^2)),$$

where of course the norm is defined by $\lVert \cdot \rVert^2 = (\cdot, \cdot)$. So my guess is that I need to start with the expressions:

$$(Ax,y) = \frac{1}{4}((\lVert Ax+y \rVert^2 + \lVert Ax-y \rVert^2) - i(\lVert Ax + iy \rVert^2 - \lVert Ax - iy \rVert^2)),$$

$$(x,Ay) = \frac{1}{4}((\lVert x+Ay \rVert^2 + \lVert x-Ay \rVert^2) - i(\lVert x + iAy \rVert^2 - \lVert x - iAy \rVert^2)),$$

and somehow show they are equal. But here is where I have gotten stuck.

Hints or solutions are greatly appreciated.

Answer 1

You should apply the polarization identity in the form

$$4(Ax,y) = (A(x+y),x+y) - (A(x-y),x-y) -i(A(x+iy),x+iy) + i(A(x-iy),x-iy).$$

Since you already know $(Az,z) = (z,Az)$ for all $z \in \mathcal{H}$, it is not difficult to deduce $A^\ast = A$ from that.

Answer 2

Another solution is this one:

$T$ is self-adjoint iff $\langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Let $x\in \mathcal{H}$. Then $0=\langle Tx,x \rangle-\langle T^{*}x,x \rangle = \langle Tx,x \rangle-\overline{\langle Tx,x \rangle}=2i\operatorname{Im}\langle Tx,x\rangle \iff \langle Tx,x \rangle \in \mathbb{R}$. Thus $T=T^{*} \iff \langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Because $T$ is positive we have that $\langle Tx,x \rangle \in \mathbb{R}$.

Answer 3

I provide a proof motivated by one of the classical proofs of Cauchy–Schwarz inequality using a quadratic equation. It is quite elementary.

Claim Let $\mathcal{H}$ be a complex vector space with inner product such that $(Ax,x)\in \mathbb{R}$ for every $x$. Then $(Ax,y)=(x,Ay)$ for every $x,y\in \mathcal{H}$.

Proof. For every $t\in \mathbb{C}$ we have

$ (A(x+ty), x+ty)= (Ax,x) + |t|^2(Ay,y)+ \bar{t}(Ax,y)+ t(Ay,x) \in \mathbb{R}.$

In particular for $t=1$ we have $$(Ax,y)+ (Ay,x) \in \mathbb{R},$$

so $$Im \ (Ax,y) = - Im \ (Ay,x)= Im \ (x,Ay).$$

and for $t=i$ we have

$$i((Ay,x) -(Ax,y))= -i(Ax,y)+ i(Ay,x) \in \mathbb{R},$$

so $$Re \ (Ax,y) = Re \ (Ay,x)= Re \ (x,Ay).$$

We conclude that $(Ax,y)= (x,Ay).$

Answer 4

An alternative to polarization is to select arbitrary $v,w \in\mathcal H$ and use the following Gram matrix
$G:=\displaystyle \left[\begin{matrix}\langle Av, v\rangle & \langle Aw, v\rangle\\ \langle Av, w\rangle & \langle Aw, w\rangle\end{matrix}\right] = H +S$,
where $H$ is Hermitian and $S$ is skew-Hermitian. For any $\mathbf x\in \mathbb C^2$
$\mathbf x^* \big(G \mathbf x\big)=\mathbf x^*\displaystyle \left[\begin{matrix}\langle A(x_1\cdot v + x_2\cdot w), v\rangle\\ \langle A(x_1\cdot v+ x_2\cdot w), w\rangle\end{matrix}\right] $$=\big\langle A(x_1\cdot v + x_2\cdot w), x_1\cdot v+x_2\cdot w\big\rangle\in \mathbb R$
$\implies \mathbf x^* S\mathbf x =\mathbf x^* \big(G-H\big)\mathbf x\in \mathbb R$. All eigenvalues of $S$ are purely imaginary, hence must be zero and $\big\Vert S\big\Vert_F^2 =\text{trace}\big(S^*S\big)=-\text{trace}\big(S^2\big)=-(\lambda_1^2+\lambda_2^2) = 0$ $\implies S=\mathbf 0$

$\implies G=G^*\implies \langle Aw, v\rangle=\overline {\langle Av, w\rangle}={\langle w, Av\rangle} $.
Hence $A$ is self-adjoint since $w,v\in \mathcal H$ were arbitrary.

Answer 5

We can write A = R + iI where R, I are Hermitian (self adj.)

As is standard for self adjoint operators, ⟨x,Ry⟩, ⟨x, Iy⟩, ⟨Rx, y⟩, ⟨Ix, y⟩ are real.

Further, ⟨x,Ax⟩ = ⟨x,Rx⟩ + i ⟨x, Ix⟩ all real, by hypothesis.

So, ⟨x, Ix⟩ = 0 ∀ x ∈ V. By conjugacy ⟨Ix, x⟩ = 0 (2.6)

Now ⟨I(x + y), x + y⟩ = ⟨I(x), x⟩ + ⟨I(x), y⟩ + ⟨I(y), x⟩ + ⟨I(y), y⟩ ⇒

0 = ⟨I(x), y⟩ + ⟨I(y), x⟩ = ⟨I(x), y⟩ + ⟨y, Ix⟩ all real.

So, 0 = ⟨I(x), y⟩ + ⟨Ix, y⟩ by conjugacy.

So, 0 = 2 ⟨I(x), y⟩ . Therefore, ⟨I(x), y⟩ = 0 ∀x, y ∈ V. From this it follows Ix = 0 ∀x. That means A = R is hermitian.