On the Operator Equations $A^n=A^*A$

TL;DR

This paper investigates operator equations of the form $A^*A=A^n$ for $n eq 1$, characterizing when solutions are self-adjoint or normal, and introduces a new class of operators for $n extgreater 2$.

Contribution

It provides a detailed analysis of the operator equations $A^*A=A^n$ for $n eq 1$, identifying conditions for self-adjointness and normality, and introduces a new class of operators for $n extgreater 2$.

Findings

Operators satisfying $A^*A=A^n$ with $n eq 1$ are characterized.

For $n eq 1$, solutions are often self-adjoint or normal under certain conditions.

A new class of operators emerges for $n extgreater 2$, positioned between orthogonal projections and normal operators.

Abstract

Let $n\in\mathbb{N}$ and let $A$ be a closed linear operator (everywhere bounded or unbounded). In this paper, we study (among others) equations of the type $A^*A=A^n$ where $n\geq2$ and see when they yield $A=A^*$ (or a weaker class of operators). In case $n\geq3$, we have in fact a new class of operators which could placed right after orthogonal projections and just before normal operators.

Equations

$$A^{\ast }A=A^n,n\in \mathbb{N},n\ge 3.$$

$$A=e^{ie\pi }I$$

$$A^n=e^{ine\pi }I\ne I=A^{\ast }A,\forall n\in \mathbb{N},n\ge 2.$$

$$A^n=A^{\ast }A$$

$$A=\sum _{k=1}^n{e}^{\frac{2k\pi i}{n}}{P}_k$$

$$A^{n-1}=A^{\ast }.$$

$$\overline{\text{ran}{A}^n}=\overline{\text{ran}A}.$$

$$\overline{\text{ran}{A}^{n+1}}=\overline{A(\ker A^{\perp })}=\overline{\text{ran}A},$$

$$\ker A^{\ast }=(\text{ran}A{)}^{\perp }=(\text{ran}{A}^n{)}^{\perp }=(\text{ran}{A}^{\ast }A{)}^{\perp }=\ker A\subseteq \ker A^{n-1},$$

$$\phi (z)=z^{n-1}-\bar{z},z\in \mathbb{C}$$

$$\sigma (A)\subseteq \{0\}\cup \{e^{\frac{2k\pi i}{n}},k=1,\dots ,n\}.$$

$$A=\sum _{k=1}^n{e}^{\frac{2k\pi i}{n}}{P}_k.$$

$$A^{*}A=\left(\begin{array}[]{cc}C^{*}C&0\\ 0&B^{*}B\\ \end{array}\right)\text{ and }A^{2}=\left(\begin{array}[]{cc}BC&0\\ 0&CB\\ \end{array}\right).$$

$$A^{\ast }{A}^2=A^{\ast }A{A}^{\ast }⟹A^{\ast }A(A-A^{\ast })=0.$$

$$A^2=A{A}^{\ast }.$$

$$A^{\ast }A={A^{\ast }}^2{A}^2=A^3.$$

$$A=P_0+e^{\frac{2\pi i}{3}}{P}_1+e^{\frac{4\pi i}{3}}{P}_2.$$

$$B=\left(\begin{array}[]{cc}0&A\\ A^{2}&0\\ \end{array}\right).$$

$$\phi (z)=z-{\overline{z}}^2,z\in \mathbb{C}$$

$$\sigma (A)\subseteq \{0\}\cup \{e^{\frac{2k\pi i}{3}},k=0,1,2\}.$$

$$-i{A}^{\ast }=B^{\ast }=B=iA,$$

$$A^{\ast }A=q{A}^2⟹-A^2=q{A}^2⟹(1+q)A^2=0$$

$$D(A^2)=D(A^{\ast }A)=\{0\}$$

$$A^{\ast }A=A^2⟹A{A}^{\ast }A=A^3⟹A{A}^{\ast }A=A^{2}A⟹A{A}^{\ast }A=A^{\ast }AA,$$

$$A^{\ast }A=-A^2⟹B^{\ast }B=B^2.$$

$$A^{\ast }A=q{A}^2⟹A^2=q{A}^2⟹(1-q)A^2\subset 0$$

$$A^{\ast }A=A^n⟹A{A}^{\ast }A=A^{n+1}⟹A{A}^{\ast }A=A^{n}A⟹A{A}^{\ast }A=A^{\ast }AA,$$

$$D\left(A^2\right)\subseteq D\left(A^{\ast }A\right)=D\left(A^n\right)$$

$$D\left(A^2\right)=D\left(A^n\right).$$

$$D\left(A^3\right)\subseteq D\left(A^{\ast }AA\right)=D\left(A^{n+1}\right)$$

Full text

• Corresponding author.

1. Introduction

It was asked in [8] whether $A^{*}A=A^{2}$ entails the self-adjointness of $A\in B(H)$? This was first answered affirmatively in [18] on finite dimensional vector spaces. Then, the authors in [10] (who were probably not aware of [18]) too obtained positive results in both the finite and the infinite dimensional settings. It is noteworthy that the infinite dimensional case was only alluded superficially in [18] where the authors of that paper were informed by the referee of the possibility of obtaining the self-adjointness of $A$ by using the so-called "technique of sequences of local inverses". In this paper, we carry on this interesting investigation to deal with the unbounded case and we reprove some known results using simpler arguments. Some consequences are also given. Then, we treat the more general equations of the type

[TABLE]

Finally, we assume readers are familiar with notions and results in operator theory. Some general references are [4], [11], [13] and [14].

2. The equations $A^{*}A=A^{n}$ with $A\in B(H)$:

Definition**.**

Let $A\in B(H)$. If $A^{*}A=A^{n}$ for some $n\in\mathbb{N}$ such that $n\geq 3$, then $A$ is called a generalized projection.

Remarks*.*

1. (1)

First, we note that for a general $n\in\mathbb{N}$ (with $n\geq 3$), then $A^{*}A=A^{n}$ does not always gives the self-adjointness of $A\in B(H)$ even when $\dim H<\infty$ as we shall shortly see. Second, we notice that if $A$ is any orthogonal projection, then it does satisfy $A^{*}A=A^{n}$ for any $n\geq 2$. 2. (2)

In general, there are unitary operators which do not satisfy such equations even when $\dim H<\infty$. We need to find a unitary $A\in B(H)$ such that $A^{*}A\neq A^{n}$ for any $n$. Consider on a finite dimensional space $H$, the following:

[TABLE]

where $e$ is the usual transcendental number. Then $A^{*}A=I$ whilst

[TABLE]

The first major result of the paper is a complete characterization of this apparently new class of operators.

Theorem 2.1**.**

Let $H$ be a complex Hilbert space and let $A\in B(H)$ be a bounded operator and let $n\in\mathbb{N},n\geq 2$. Then $A$ is a solution of the equality

[TABLE]

if and only if

• •

$A=A^{*}$* (if $n=2$),*

• •

there is a family $P_{1},\ldots,P_{n}\in B(H)$ of orthogonal projections such that $P_{j}P_{k}=0,(j\neq k)$ such that

[TABLE]

(if $n\geq 3$). In this case, we also have $\|A\|=1$ (when $A\neq 0$).

Proof.

The ”if” part of the statement is clear. We show that the “only if” part is also true. First we are going to prove that

[TABLE]

It is clear from the hypothesis that $A^{*}=A^{n-1}$ on the range of $A$, hence $A^{*}=A^{n-1}$ also on $\overline{\text{\rm{ran}}A}$ because of continuity. It suffices therefore to prove that $A^{n-1}=0$ on $\ker A^{*}$, i.e., $\ker A^{*}\subseteq\ker A^{n-1}$. First we claim that

[TABLE]

Indeed, our assumption clearly implies that $\text{\rm{ran}}A^{n+1}=\text{\rm{ran}}AA^{*}A=A(\text{\rm{ran}}A^{*}A)$, hence by equality $\overline{\text{\rm{ran}}A^{*}A}=\ker A^{\perp}$ we conclude that

[TABLE]

that clearly gives (4). From this we conclude that

[TABLE]

which proves (3).

If $n=2$ then (3) expresses just that $A$ is self-adjoint. (Observe that up to this point we did not used that $H$ is complex). Suppose now that $n\geq 3$, then from (3) it follows that $A$ is normal and that the function

[TABLE]

vanishes on $\sigma(A)$. In particular, if $\lambda\in\sigma(A)$ then either $\lambda=0$ or $\lambda$ is a solution of $\lambda^{n}=1$, whence we conclude that

[TABLE]

Let us denote by $E$ the spectral measure of $A$ and set $P_{k}:=E(\{e^{\frac{2k\pi i}{n}}\})$ then it follows from the spectral theorem that $P_{k}P_{j}=0$ $(k\neq j)$ and that

[TABLE]

To show the last claim, just apply the spectral radius theorem to the normal operator $A$ to obtain $\|A\|=1$ when $A\neq 0$. This marks the end of the proof. ∎

Remark*.*

As alluded to above, any orthogonal projection satisfies the equations $A^{*}A=A^{n}$ with $n\geq 3$ ($n=2$ is also allowed). This new class of operators lies therefore just between orthogonal projections and normal operators.

Corollary 2.2**.**

Let $A\in B(H)$ be satisfying $AA^{*}=A^{2}$. Then $A$ is self-adjoint.

Remark*.*

From Theorem 2.1, it turns out that operators satisfying $A^{2}=A^{*}A$ are just the self-adjoint ones. However, a solution of $A^{n}=A^{*}A$, $n\geq 3$ need not be self-adjoint, as it can be seen immediately from the general form (2) of those operators.

As an immediate consequence, we have:

Proposition 2.3**.**

If $B,C\in B(H)$ are such that $C^{*}C=BC$ and $B^{*}B=CB$, then $B=C^{*}$.

Proof.

Let $A\in B(H\oplus H)$ be defined as A=\left(\begin{array}[]{cc}0&B\\ C&0\\ \end{array}\right). Then

[TABLE]

By hypothesis, we ought to have $A^{*}A=A^{2}$, whereby $A$ becomes self-adjoint, in which case, $B=C^{*}$, as wished. ∎

Corollary 2.4**.**

Let $B,C\in B(H)$ be self-adjoint and such that $C^{2}=BC$ and $B^{2}=CB$. Then $B=C$.

Another consequence is the following:

Proposition 2.5**.**

Let $A\in B(H)$ be such that $A^{*}A^{2}=A^{*}AA^{*}$. Then $A$ is self-adjoint.

Proof.

We may write

[TABLE]

Hence for any $x\in H$, we clearly have that $(A-A^{*})x\in\ker(A^{*}A)$. But it is well known that $\ker(A^{*}A)$ coincides with $\ker A$. Therefore, $(A-A^{*})x\in\ker A$ or simply

[TABLE]

A glance at Corollary 2.2 finally gives the self-adjointness of $A$, marking the end of the proof. ∎

The method of matrices of operators allows us to establish the following result:

Proposition 2.6**.**

Let $A\in B(H)$ be satisfying

[TABLE]

Then there exist three orthogonal projections, $P_{0}$, $P_{1}$, $P_{2}\in B(H)$ which are pairwise orthogonal such that

[TABLE]

Proof.

Let $A\in B(H)$ and define $B\in B(H\oplus H)$ by:

[TABLE]

Then B^{2}=\left(\begin{array}[]{cc}A^{3}&0\\ 0&A^{3}\\ \end{array}\right). Since B^{*}B=\left(\begin{array}[]{cc}{A^{*}}^{2}A^{2}&0\\ 0&A^{*}A\\ \end{array}\right), by hypothesis we must therefore have $B^{*}B=B^{2}$. Hence, $B$ is self-adjoint by Theorem 2.1. This just means that $A={A^{*}}^{2}$. Consequently, $A$ is obviously normal and

[TABLE]

vanishes on $\sigma(A)$. From that it is readily seen that if $\lambda\in\sigma(A)$ then either $\lambda=0$ or $\lambda$ is a solution of $\lambda^{3}=1$. Whence, we conclude that

[TABLE]

From the spectral theorem it follows that $A$ can be written as (5) for some orthogonal projections $P_{0}$, $P_{1}$, $P_{2}$ with pairwise orthogonal ranges. The proof is complete. ∎

We can also treat the "skew-adjointness" case. First, we give a result which might already be known to some readers and so it is preferable to include a proof. Recall that a bounded hyponormal operator having a real spectrum is self-adjoint (see e.g. [15]).

Lemma 2.7**.**

Let $A\in B(H)$ be hyponormal and having purely imaginary spectrum. Then, $A$ is skew-adjoint (that is, $A^{*}=-A$).

Proof.

Set $B=iA$ and so $B$ too is hyponormal. Hence $\sigma(B)\subset\mathbb{R}$ for by assumption $\sigma(A)\subset i\mathbb{R}$. Hence $B$ is self-adjoint, i.e.

[TABLE]

i.e. $A$ is clearly skew-adjoint. ∎

Mutatis mutandis, the following result is then easily obtained:

Proposition 2.8**.**

Let $A\in B(H)$ be satisfying $A^{*}A=-A^{2}$. Then $A$ is skew-adjoint.

The following sharp result is also of interest.

Proposition 2.9**.**

If $A\in B(H)$ is such that $A\neq 0$ and $A^{*}A=qA^{2}$ where $q\in\mathbb{R}^{*}$, then either $q=1$ or $q=-1$.

Proof.

By considering the cases $q<0$ and $q>0$ separately, we may as above establish the skew-adjointness of $A$ and self-adjointness of $A$ respectively. Now, in case $A$ is skew-adjoint (when $q<0$), we may write

[TABLE]

which gives $q=-1$. In the event of the self-adjointness of $A$, we may just reason similarly to get $q=1$, and this finishes the proof. ∎

3. The equations $A^{*}A=A^{n}$ with a closed and densely defined operator $A$:

First, we stop by some examples.

Examples 3.1**.**

1. (1)

If $A$ is a linear operator, then $A^{*}A=A^{2}$ does not necessarily give $A=A^{*}$. The most trivial example is to consider a densely defined and unclosed operator $A$ (hence such $A$ cannot be self-adjoint) such that

[TABLE]

as in [12], say. Then $A^{*}A=A^{2}$ is trivially satisfied. 2. (2)

$A^{*}A\subset A^{2}\not\Rightarrow A=A^{*}$* even when $A$ is closed*: Indeed, consider any closed, densely defined and symmetric operator $A$ which is not self-adjoint. Then $A\subset A^{*}$ and so $A^{*}A\subset A^{2}$. 3. (3)

$A^{2}\subset A^{*}A\not\Rightarrow A=A^{*}$* even when $A$ is closed*: In this case, consider any closed, densely defined and symmetric operator $A^{*}$ which is not self-adjoint. A similar observation as just above then yields $A^{2}\subset A^{*}A$. We may even consider a closed, symmetric and semi-bounded such that $D(A^{2})=\{0\}$ (see [1], cf. [3]). Then trivially $A^{2}\subset A^{*}A$ and $A$ is not self-adjoint.

Now, we deal with the equation $A^{*}A=A^{2}$ for a closed and densely defined $A$.

Theorem 3.2**.**

Let $H$ be a complex Hilbert space and let $A$ be a closed and densely defined (unbounded) operator verifying $A^{*}A=A^{2}$. Then $A$ is self-adjoint on its domain $D(A)\subset H$.

Proof.

Plainly,

[TABLE]

showing the quasinormality of $A$ (as defined in [5], say). By consulting [6] and [9], we know that quasinormal operators are hyponormal. That is, $A$ is hyponormal.

According to the proof of Theorem 8 in [2], closed hyponormal operators having a real spectrum are automatically self-adjoint. Once that’s known and in order that $A$ be self-adjoint, it suffices therefore to show the realness of its spectrum given that $A$ is already closed.

So, let $\lambda\in\sigma(A)$. Since $A$ is closed, we have by invoking a spectral mapping theorem (e.g. Theorem 2.15 in [7]) that $\lambda^{2}\geq 0$ for $A^{*}A$ is self-adjoint and positive. Now, this forces $\lambda$ to be real. Accordingly, $\sigma(A)\subset\mathbb{R}$, as needed. ∎

Remark*.*

Notice that the previous proof may well be applied to the first claim of Theorem 2.1 when $H$ is a Hilbert space over $\mathbb{C}$.

As in the bounded case, we have:

Proposition 3.3**.**

Let $B,C$ be two densely defined and closed operators obeying $C^{*}C=BC$ and $B^{*}B=CB$. Then $B=C^{*}$.

By adopting a very similar idea to the bounded case (by observing that Lemma 2.7 holds for unbounded and closed operators as well), we may easily establish the following result. We include, however, a somewhat different proof which could have been used above anyway.

Proposition 3.4**.**

Let $A$ be a closed and densely defined (unbounded) operator such that $A^{*}A=-A^{2}$. Then $A$ is skew-adjoint.

Proof.

Set $B=iA$. Then

[TABLE]

Since $B$ is closed, Theorem 3.2 applies and gives the self-adjointness of $B$ or the skew-adjointness of $A$, as required. ∎

An unbounded version of Proposition 2.9 is also available.

Proposition 3.5**.**

If $A$ is a closed, unbounded and densely defined operator such that $A^{*}A=qA^{2}$ where $q\in\mathbb{R}^{*}$, then either $q=1$ or $q=-1$.

Proof.

The proof is similar to the one in the bounded case. For instance, when $q>0$, we obtain the self-adjointness of $A$. Hence

[TABLE]

which forces $q=1$ (remember that $A^{2}$ is unbounded). ∎

Finally, we treat the unbounded case. Somehow expectedly, we show the impossibility of the equations $A^{*}A=A^{n}$ (with $n\geq 3$) for unbounded closed operators.

Theorem 3.6**.**

Let $A$ be a closed and densely defined operator with a domain $D(A)\subset H$ and let $n\in\mathbb{N}$ be such that $n\geq 3$. If $A^{*}A=A^{n}$, then $A\in B(H)$ (and so $A$ can be written in the form (2)).

Proof.

Let $A$ be a closed and densely defined operator which obeys $A^{*}A=A^{n}$ where $n\geq 3$. Then (as in the bounded case)

[TABLE]

showing the quasinormality of $A$. It then follows that $A$ is hyponormal and so $D(A)\subset D(A^{*})$. Hence

[TABLE]

or merely

[TABLE]

Also

[TABLE]

so that

[TABLE]

Now, since $A$ is closed, it follows that $A^{2}$ is closed as it is already quasinormal (see e.g. Proposition 5.2 in [16]). Also, the quasinormality of $A$ yields that of $A^{2}$ (by Corollary 3.8 in [5], say) and so $A^{2}$ is hyponormal. Therefore,

[TABLE]

and

[TABLE]

In the end, according to Corollary 2.2 in [17], it follows that $A^{2}$ is everywhere bounded on $H$. Hence $D(A)=H$ and so the Closed Graph Theorem intervenes now to make $A\in B(H)$, as coveted. ∎