Classification of Bounded Operators

Here we define the notion of the adjoint operator and using this notion, define self-adjoint, unitary and normal operators. We discuss simple algebraic properties of operators in these classes. We also discuss projections.

The Adjoint Operator

Theorem 2.1: Let $\mathcal{H}$ be a Hilbert space and $A \in \mathcal{B}(\mathcal{H})$. There exists a unique operator $A^* \in \mathcal{B}(\mathcal{H})$ such that:

Proof: Let $y \in \mathcal{H}$ and define the linear functional $f_y : \mathcal{H} \rightarrow \mathbb{C}$ as $f_y(x) \coloneqq (Ax,y)$. By the Cauchy-Schwarz inequality, for every $x \in \mathcal{H}$ we have:

So $f_y$ is a bounded linear functional on $\mathcal{H}$ and $\|f\| \leq \|A\|\|y\|$. According to the Riesz representation theorem there exists a unique $z_y \in \mathcal{H}$ such that for all $x \in \mathcal{H}$:

and $\|z_y\| = \|f\| \leq \|A\|\|y\|$. Define the operator $A^* : \mathcal{H} \rightarrow \mathcal{H}$ as $A^*x = z_y$. One can verify that $A^*$ is linear. We have already proved that for all $y \in \mathcal{H}$:

in other words $A^*$ is bounded and:

By construction 2.1 is satisfied and the uniqueness of $z_y$ guarantees that the constructed operator $A^*$ is the unique operator satisfying the inequality. $\blacksquare$

Definition 2.2: The operator $A^*$ from the previous theorem is called the adjoint of $A$.

The adjoint operator can also be defined in a more general situation when $B \in \mathcal{B}(X,Y)$ for Banach spaces $X,Y$; see Reed-Simon Section VI.2 for discussion. In this course we only consider the adjoint of operators on a Hilbert space.

Theorem 2.3: Let $A,A_1,A_2 \in \mathcal{B}(\mathcal{H})$, and $\alpha \in \mathbb{C}$. Then:

1. $(\alpha A_1 + A_2)^* \overline{\alpha}A_1^* + A_2^*$;
2. $(A_1A_2)^* = A_2^*A_1^*$;
3. $A^{**} = A$, where $A^{**} = (A^*)^*$;
4. $\|A^*\| = \|A\|$;
5. $\|A^*A\| = \|AA^*\| = \|A\|^2$;
6. If $A$ is invertible then $(A^{-1})^* = (A^*)^{-1}$.

Proof: In order to show that an operator is the adjoint of another, it is sufficient to show that it satisfies the adjoint equation (2.1) for all $x,y \in \mathcal{H}$. Then, the uniqueness of the adjoint tells us that it has to be the adjoint.

1. We compute for any $x,y \in \mathcal{H}$:
   $$((\alpha A_1 + A_2)x,y) = \alpha(A_1 x, y) + (A_2x,y)\\ = \alpha (x,A_1^* y) + (x,A_2^*y) \\ = (x,(\overline{\alpha}A_1^* + A_2^*)y).$$
2. We compute again, for $x,y \in \mathcal{H}$:
   $$(A_1A_2x,y)=(A_2x,A_1^*y)=(x,A_2^*A_1^*y).$$
3. For every $x,y \in \mathcal{H}$:
   $$(A^{**}x,y) = \overline{(y,A^{**}x)} \\ = \overline{(A^*y,x)} \\ = (x,A^*y) \\ = (Ax,y).$$
   Consequently for every $x,y \in \mathcal{H}$:
   $$((A-A^{**})x,y) = 0.$$
   Since this is in particular true for $y=(A-A^{**})x$ we obtain that for all $x \in \mathcal{H}$, $(A-A^{**})x = 0$.
4. This follows from 2.2 and (3):
   $$\|A\| = \|A^{**}\| \leq \|A^*\| \leq \|A\|.$$
5. By (4) and Theorem 1.5:
   $$\|A^*A\| \leq \|A^*\|\|A\| = \|A\|^2.$$

On the other hand:

Thus $\|A^*A\| = \|A\|^2$. Using this equality with $A^*$ instead of $A$ we derive from (3) and (4) that $\|AA^*\| = \|A\|^2$.

6. It is sufficient to take the adjoint operators in the equality:
   $$AA^{-1}= I = A^{-1}A.$$
   $\blacksquare$

Definition 2.4: An operator $A \in \mathcal{B}(\mathcal{H})$ is said to be:

1. normal if $AA^* = A^*A$,
2. self-adjoint if $A^* = A$, i.e. if for all $(x,y) \in \mathcal{H}$, $(Ax,y)=(x,Ay)$,
3. unitary if $A^*A=AA^*=I$ i.e. if $A^{-1}=A^*$.

Note that any self-adjoint operator is normal. Any unitary operator $U \in \mathcal{B}(\mathcal{H})$ is normal as well.

Unitary operators can be defined more generally as operators from one Hilbert space to another one: $U \in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)$ is called unitary if $U^*U=I_{\mathcal{H}_1}$, $UU^*=I_{\mathcal{H}_2}$.

Example 2.5: Consider the operator $T$ of multiplication by a function $t$ in $L^2(\mathbb{R})$. Then $T$ is self-adjoint if and only if $t$ is real-valued; $T$ is unitary if and only if $|t(x)| = 1$ for all $x$; $T$ is normal for any function $t$. Similar statements apply to the operator of multiplication by a sequence.

Example 2.6: An operator $T$ given by the matrix $\{t_{ij}\}_{i,j=1}^N$ in $\mathbb{C}^N$ is self-adjoint if and only if $\overline{t_{ij}} = t_{ji}$ for all $i,j$. Similarly an integral operator with the integral kernel $t(,y)$ is self-adjoint if $t(x,y) = \overline{t(y,x)}$ for all $x,y$. There are no known simple sufficient conditions for unitarity or normality of integral operators.

Theorem 2.7:

1. $A$ is self-adjoint, $\lambda \in \mathbb{R} \implies \lambda A$ is self-adjoint.
2. $A_1,A_2$ are self-adjoint $\implies A_1 + A_2$ is self-adjoint.
3. Let $A_1,A_2$ be self-adjoint. Then $A_1A_2$ is self-adjoint if and only if $A_1$ and $A_2$ commute.

Proof: Exercise. $\blacksquare$

Theorem 2.8: Let $A \ in \mathcal{B}(\mathcal{H})$; then $A$ is self-adjoint if and only if $(Ax,x)$ is real for all $x \in \mathcal{H}$.

Proof: If $A$ is self-adjoint then for all $x \in \mathcal{H}$:

in other words, $(Ax,x)$ is real.

Let us prove the converse. For any $x,y \in \mathcal{H}$, we have:

(this is the polarization identity for operators) and similarly:

For every $z \in \mathcal{H}$ we have:

Since this is true for $z \in \{x+y,x-y,x+iy,x-iy\}$, the right hand sides of the two above equations are equal. This implies that:

for all $x,y \in \mathcal{H}$, in other words, $A$ is self-adjoint. $\blacksquare$

Corollary 2.9: If $B_n, n \in \mathbb{N}$ are self-adjoint and $B_n \rightarrow B$ weakly, then $B$ is self-adjoint.

Proof: We have $(A_nx,x) \rightarrow (Ax,x)$ for all $x$. Since $(A_nx.x)$ are real, this implies that $(Ax,x)$ is real, hence $A$ is self-adjoint. $\blacksquare$