Self-Adjoint Operator Properties

Kreyszig Functional Analysis

Let $H$ be a complex Hilbert space. A bounded linear operator $T:H\to H$ is called self-adjoint (or Hermitian) if its Hilbert-adjoint operator is itself; i.e.

$$T=T^{\star },\forall x,y\in H:⟨Tx,y⟩=⟨x,Ty⟩$$

We can clearly see, by definition, that any linear combination of self-adjoint operators is again self-adjoint. The identity operator $I$ is also trivially self-adjoint. Thus, for any self-adjoint operator $T$ and complex number $\lambda$, $T_{\lambda }$ is self-adjoint. Furthermore, another elementary property of self-adjoint operators is that $⟨Tx,x⟩\in \mathbb{R}$ for all $x\in H$.

For the remainder of these notes, $H$ denotes a complex Hilbert space.

Theorem 1. Let $T:H\to H$ be self-adjoint. Then, $T$ has a real point spectrum. Proof) Assume $\lambda \in {\sigma }_p(T)$; there exists a nonzero $x\in H$ such that $T_{\lambda }x=0$. Then,

$$\lambda ⟨x,x⟩=⟨\lambda x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=⟨x,\lambda x⟩=\overline{\lambda }⟨x,x⟩$$

Thus, $\lambda \in \mathbb{R}$ and ${\sigma }_p(T)\subset \mathbb{R}$.

Theorem 2. Let $T:H\to H$ be self-adjoint. Eigenvectors corresponding to different eigenvalues of $T$ are necessarily orthogonal. Proof) Let ${\lambda }_1,{\lambda }_2\in {\sigma }_p(T)$ be different elements in the point spectrum and let $x_1$, $x_2$ be corresponding nonzero eigenvectors. Then, noting that the point spectrum is real,

$${\lambda }_1⟨x_1,x_2⟩=⟨{\lambda }_1{x}_1,x_2⟩=⟨T{x}_1,x_2⟩=⟨x_1,T{x}_2⟩=⟨x_1,{\lambda }_2{x}_2⟩={\lambda }_2⟨x_1,x_2⟩$$

Thus, $x_1\perp x_2$.

Theorem 3. Let $T:H\to H$ be self-adjoint. $T$ has an empty residual spectrum. Proof) Assume $\lambda \in {\sigma }_r(T)$. Then, $T_{\lambda }(H)$ is not a dense subset of $H$. As $\overline{T_{\lambda }(H)}\ne H$, this implies ${\overline{T_{\lambda }(T)}}^{\perp }$ is nonempty and there exists some nonzero $h\in H$ such that $T_{\lambda }x\perp h$ for all $x\in H$. Equivalently, for all $x\in H$,

$$0=⟨T_{\lambda }x,h⟩=⟨Tx,h⟩-\lambda ⟨x,h⟩=⟨x,Th-\overline{\lambda }h⟩\Rightarrow T_{\overline{\lambda }}h=0\Rightarrow \overline{\lambda }\in {\sigma }_p(T)$$

As $T$ has a real point spectrum, this implies $\lambda$ is real and is in the point spectrum of $T$. This is a contradiction as we assumed $\lambda$ is in the residual spectrum.

Theorem 4. Let $T:H\to H$ be self-adjoint. A complex number $\lambda$ is in the resolvent set $\rho (T)$ iff there exists a fixed $c>0$ such that for all $x\in H$,

$$\Vert T_{\lambda }x\Vert \ge c\Vert x\Vert$$

Proof) Assume that $\lambda \in \rho (T)$. Then, $T_{\lambda }$ has an inverse $R_{\lambda }(T)$ that is bounded and defined on a dense subset of $H$. For any $x\in H$,

$$\Vert x\Vert =\Vert R_{\lambda }(T)T_{\lambda }x\Vert \le \Vert R_{\lambda }(T)\Vert \Vert T_{\lambda }x\Vert$$

Clearly, the positive constant $c=1/\Vert R_{\lambda }(T)\Vert$ satisfies the desired constraint.

We show the converse. Assume $c>0$ satisfies $\Vert T_{\lambda }x\Vert \ge c\Vert x\Vert$ for all $x\in H$. Clearly, $\lambda \notin {\sigma }_p(T)$ as $T_{\lambda }x\ne 0$ for all $x\ne 0$. The only remaining possibility other than $\lambda \in \rho (T)$ is $\lambda$ being in the continuous spectrum of $T$; assume $\lambda \in {\sigma }_c(T)$. Let $R_{\lambda }(T)$ be the unbounded inverse of $T_{\lambda }$. For any $x\in H$,

$$\Vert T_{\lambda }x\Vert \ge c\Vert x\Vert =c\Vert R_{\lambda }(T)(T_{\lambda }x)\Vert \Rightarrow \Vert R_{\lambda }(T)(T_{\lambda }x)\Vert \le \Vert T_{\lambda }x\Vert /c$$

Note that $T_{\lambda }(H)$ is precisely the domain of $R_{\lambda }(T)$; hence $R_{\lambda }(T)$ is forced to be bounded and there is a contradiction.

Furthermore, note that by Banach space properties $\lambda \in \rho (T)$ implies $R_{\lambda }(T)\in B(H,H)$.

Theorem 5. Let $T:H\to H$ be self-adjoint. The entire spectrum $\sigma (T)$ of $T$ is real. Proof) Let $\lambda =\alpha +\beta i$ be a non-real ($\beta \ne 0$) complex number. It suffices to show for all such numbers, $\lambda \in \rho (T)$. This is true as for any nonzero $x\in H$,

$$\Vert T_{\lambda }x\Vert \ge |⟨T_{\lambda }x,x⟩|/\Vert x\Vert =|⟨Tx,x⟩-\lambda ⟨x,x⟩|/\Vert x\Vert \ge |\beta |\Vert x\Vert$$

Therefore, $|\beta |>0$ satisfies $\Vert T_{\lambda }x\Vert \ge |\beta |\Vert x\Vert$ for all $x\in H$, and we are done.

Theorem 6. Let $M={\sup }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $m={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$. Then, $\sigma (T)\subset [m,M]$. Proof) We show that for any real $\lambda =M+c$ for some positive $c>0$, $\lambda \in \rho (T)$. This is true as for any $x\in H$,

$$\Vert T_{\lambda }x\Vert \ge |⟨T_{\lambda }x,x⟩|/\Vert x\Vert =|⟨Tx,x⟩-\lambda \Vert x{\Vert }^2\Vert /\Vert x\Vert \ge c\Vert x{\Vert }^2$$

by Theorem 4, we are done. The $\lambda <m$ case is identical.

Theorem 7. Let $T:H\to H$ be self-adjoint. Then,

$$\Vert T\Vert =\underset{\Vert x\Vert =1}{\sup }|⟨Tx,x⟩|=\max (|m|,|M|)$$

Proof) Let $K={\sup }_{\Vert x\Vert =1}|⟨Tx,x⟩|$. Clearly, $|⟨Tx,x⟩|\le \Vert Tx\Vert \Vert x\Vert \Rightarrow K\le \Vert T\Vert$. It suffices to show $K\ge \Vert T\Vert$. For all $x,y\in H$, we have the following identity:

$$2K(\Vert x{\Vert }^2+\Vert y{\Vert }^2)=K(\Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2)$$ $$\ge ⟨T(x+y),(x+y)⟩-⟨T(x-y),(x-y)⟩=2(⟨Tx,y⟩+⟨Ty,x⟩)$$ $$\therefore K(\Vert x{\Vert }^2+\Vert y{\Vert }^2)\ge ⟨Tx,y⟩+⟨Ty,x⟩$$

For any $x\in H$ such that $\Vert x\Vert =1$, take $y=Tx/\Vert Tx\Vert$. Substituting into the identity gives

$$2K\ge ⟨Tx,y⟩+⟨Ty,x⟩=2\Vert Tx\Vert \Rightarrow K\ge \Vert Tx\Vert$$

As this applies for any $x$ such that $\Vert x\Vert =1$,

$$K\ge \underset{\Vert x\Vert =1}{\sup }\Vert Tx\Vert =\Vert T\Vert$$

Therefore, $K=\Vert T\Vert$.

Theorem 8. Let $T:H\to H$ be self-adjoint, and define $m$ and $M$ as in Theorem 6. $m$ and $M$ are in the spectrum $\sigma (T)$.

Proof) Consider $T^{\prime }=T-mI$. By the spectral projection theorem, it suffices to show that $M^{\prime }=M-m$ is a spectral value of $T^{\prime }$. An analogous method works to show $m\in \sigma (T)$.

Note that $m^{\prime }=m-m=0\le M^{\prime }\Rightarrow M^{\prime }=\Vert T^{\prime }\Vert ={\sup }_{\Vert x\Vert =1}⟨T^{\prime }x,x⟩$. By the definition of supremum, there is a sequence $(x_n)\in H$ such that $\Vert x_n\Vert =1$ and $⟨T^{\prime }{x}_n,x_n⟩\to M^{\prime }$. As $⟨T^{\prime }{x}_n,x_n⟩\le \Vert T^{\prime }{x}_n\Vert \Vert x_n\Vert \le M^{\prime }$, $\Vert T^{\prime }{x}_n\Vert \to M^{\prime }$ as well. This sequence satisfies

$$\Vert T^{\prime }{x}_n-M^{\prime }{x}_n{\Vert }^2=⟨T^{\prime }{x}_n-M^{\prime }{x}_n,T^{\prime }{x}_n-M^{\prime }{x}_n⟩=\Vert T^{\prime }{x}_n{\Vert }^2+M^2\Vert x_n{\Vert }^2-2⟨T^{\prime }{x}_n,M^{\prime }{x}_n⟩$$

which converges to

$$\to M^{\prime 2}+M^{\prime 2}-2M^{\prime 2}\Rightarrow \Vert T^{\prime }{x}_n-M^{\prime }{x}_n{\Vert }^2=\Vert T_{M^{\prime }}^{\prime }{x}_n\Vert \to 0$$

By Theorem 4, this implies $M^{\prime }\in \sigma (T)$.