# $\sum_{k=1}^dA_k^*A_k$ and $\sum_{k=1}^dA_kA_k^*$ have the same norms if $A_k$ are commuting

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

• the operators $A_k$ are acting on the same Hilbert space $E$ Jan 27, 2019 at 14:54
• on the Hilbert space $H_1 \oplus H_2 \oplus H_3$ Jan 27, 2019 at 14:55
• 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 $\mathbb{C}^3$. Jan 27, 2019 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 $$H_1 \rightarrow H_2 \rightarrow H_3 \rightarrow ...$$

and

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

• 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 $A_2$? Jan 27, 2019 at 18:57
• $A_1$ yes. but you have double-size Hilbert space $\ell^2(\mathbb{N}) \oplus \ell^2(\mathbb{N})$. $A_2(x_1,x_2,x_3, \dots \oplus y_2,y_3,\ldots)=(0,\ldots \oplus x_1,y_2,y_3,y_4,\ldots)$. Jan 28, 2019 at 12:02
• Please what is explicitely the expresion of $A_1$? Thanks Mar 25, 2019 at 11:15
• $A_1(x_1,x_2,x_3,⋯⊕y_2,y_3,…)=(0,x_1,x_2,x_3,...⊕0,0,0,....)$. Apr 25, 2019 at 13:59
• Thank you very much but please why in the expressions of $A_1$ and $A_2$, you start with $y_2$ and not $y_1$? Apr 25, 2019 at 19:42