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 $\|A_1^* A_1 + A_2^* A_2\| = 2 \neq \|A_1 A_1^* + A_2 A_2^*\| =1$.

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  • $\begingroup$ the operators $A_k$ are acting on the same Hilbert space $E$ $\endgroup$ – Student Jan 27 '19 at 14:54
  • $\begingroup$ on the Hilbert space $H_1 \oplus H_2 \oplus H_3$ $\endgroup$ – hänsel Jan 27 '19 at 14:55
  • $\begingroup$ 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 '19 at 15:02

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 $\|A_1^* A_1 + A_2^* A_2\|=2 \neq \|A_1 A_1^* + A_2 A_2^*\|=1$.

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  • 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 '19 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 '19 at 12:02
  • $\begingroup$ Please what is explicitely the expresion of $A_1$? Thanks $\endgroup$ – Student Mar 25 '19 at 11:15
  • $\begingroup$ $A_1(x_1,x_2,x_3,⋯⊕y_2,y_3,…)=(0,x_1,x_2,x_3,...⊕0,0,0,....)$. $\endgroup$ – hänsel Apr 25 '19 at 13:59
  • $\begingroup$ Thank you very much but please why in the expressions of $A_1$ and $A_2$, you start with $y_2$ and not $y_1$? $\endgroup$ – Student Apr 25 '19 at 19:42

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