Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all operators on $E$.

Let $A_1,\cdots,A_d$ be pairwise commuting operators on $E$. Is the equality $$\left\|\displaystyle\sum_{k=1}^dA_k^*A_k \right\|=\left\|\displaystyle\sum_{k=1}^dA_kA_k^* \right\|,$$ need not hold?

Recall that an operator $T\in \mathcal{L}(E)$ is said to be isometry if $T^*T=I$.

I try to find two commuting isometries $A_1$ and $A_2$ such that $A_1A_1^*+A_2A_2^*=I$. However, this is not possible.

Indeed, if $A_1A_1^*+A_2A_2^*=I$, multiplying on the LHS by $A_1^*$ you get $$A_1^*A_1A_1^*+A_1^*A_2A_2^*=A_1^* \\ A_1^*+A_1^*A_2A_2^*=A_1^* \\ A_1^*A_2A_2^*=0 \\ $$ Multiply on the RHA by $A_2$ to get $$A_1^*A_2=0 \\ A_1^*A_2A_1 =0$$

Now, since $A_1,A_2$ commute you get $$A_1^*A_1A_2=0 \\ A_2=0$$

Also the same idea shows that it is not possible for three commuting isometies $A_1,A_2,A_3$ to get $A_1A_1^*+A_2A_2^*+A_3A_3^*=I$.


No. 3 copies of Hilbert spaces $H_1,H_2,H_3$. $A_1$ a partial isomtry copying $H_1$ to $H_2$, and $A_2$ a partial isometry copying $H_1$ to $H_3$. Then $A_1 A_2 =A_2A_1 =0$. But $\|A_1^* A_1 + A_2^* A_2\| = 2 \neq \|A_1 A_1^* + A_2 A_2^*\| =1$.

  • $\begingroup$ the operators $A_k$ are acting on the same Hilbert space $E$ $\endgroup$ – Student Jan 27 at 14:54
  • $\begingroup$ on the Hilbert space $H_1 \oplus H_2 \oplus H_3$ $\endgroup$ – hänsel Jan 27 at 14:55
  • $\begingroup$ I hope that you can explain me your answer. What you mean by partial isometry? Thanks. $\endgroup$ – Student Jan 27 at 14:57
  • $\begingroup$ partial isometry is $A$ with $A A^* A=A$, and copies isometrically the source Hilbert space $A^* A (H)$ to $AA^*(H)$, see lietrature. you may take $H_1=H_2=H_2 =\mathbb{C}$, and $A_1,A_2$ 3x3 matrices, where $A_1(e_1)=e_2$, otherwise 0, and $A_2(e_1)=e_3$, otherwise 0, where $e_1,e_2,e_3$ canoncal basis of $\mathbb{C}^3$. $\endgroup$ – hänsel Jan 27 at 15:02
  • $\begingroup$ Thanks a lot. The example is really good. The equality holds if the $A_k$ are normal. Do you think that it holds also if the operators $A_k$ are hyponormal? $\endgroup$ – Student Jan 27 at 15:22

Alternative answer: Even No, if operators $A_i$ are hyponormal.

We modify above answeer: Take copies of Hilbert spaces: $H_1,H_2,H_3,...$ and $K_2,K_3,K_4,...$.

$A_1$ partial isometry which is shift $H_1 \rightarrow H_2 \rightarrow H_3 \rightarrow ...$


$A_2$ partial isometry which is shift $H_1 \rightarrow K_2 \rightarrow K_3 \rightarrow ...$

Again, $A_1 A_2=A_2 A_1=0$, but $\|A_1^* A_1 + A_2^* A_2\|=2 \neq \|A_1 A_1^* + A_2 A_2^*\|=1$.

  • 1
    $\begingroup$ Thanks for the second answer. $A_1: \ell_{\mathbb{N}^*}^2(\mathbb{C})\rightarrow \ell_{\mathbb{N}^*}^2(\mathbb{C})$ be defined by $$A_1(x_1,x_2,\cdots)=(0,x_1,x_2,\cdots),$$ Please what is the difference between $A_1$ and $A_2$? $\endgroup$ – Student Jan 27 at 18:57
  • $\begingroup$ $A_1$ yes. but you have double-size Hilbert space $\ell^2(\mathbb{N}) \oplus \ell^2(\mathbb{N}) $. $A_2(x_1,x_2,x_3, \dots \oplus y_2,y_3,\ldots)=(0,\ldots \oplus x_1,y_2,y_3,y_4,\ldots)$. $\endgroup$ – hänsel Jan 28 at 12:02
  • $\begingroup$ Please the definition of $A_2$ is not clear to me. I hope that you provide me a bit explanation. Thanks a lot. I understand that a sequence $(Z_n)\in \ell^2(\mathbb{N})\oplus \ell^2(\mathbb{N})$ should be written as $$Z_n=x_n+y_n,$$ with $(x_n,y_n)\in \ell^2(\mathbb{N})\times \ell^2(\mathbb{N})$. So a shift operator on $\ell^2(\mathbb{N})\oplus \ell^2(\mathbb{N})$ should be written as $$A_2(x_1+y_1,x_2+y_2,\cdots)=(0,x_1+y_1,x_2+y_2,\cdots).$$ Please correct me. $\endgroup$ – Student Jan 28 at 13:24

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.