# Let $T,S$ be unilateral shifts in $H,K$ and $A\in B(H,K)$ a contraction. If $S^*A=AT^*$, then why is $A$ a transposed infinite Toeplitz matrix?

Let $$H, K$$ be Hilbert spaces.

As the Toeplitz Matrix, I define an operator $$P_n$$ in the form:

$$P_n = \begin{pmatrix} Q_0 & 0 & 0 & \ldots & 0 \\ Q_1 & Q_0 & 0 & \ldots & 0\\ Q_2 & Q_1 & Q_0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ Q_{n-1} & Q_{n-2} & Q_{n-3} & \ldots & Q_0 \end{pmatrix}$$

where $$Q_n \in B(H, K)$$.

Here, an infinite Toeplitz matrix would be simply $$P_{\infty}$$.

As unilateral shifts, I define:

$$T = \begin{pmatrix} 0 & 0 & 0 & 0 & \ldots\\ I & 0 & 0 & 0 & \ldots\\ 0 & I & 0 & 0 & \ldots\\ 0 & 0 & I & 0 & \ldots\\ 0 & 0 & 0 & I & \ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ \end{pmatrix}$$

$$S = \begin{pmatrix} 0 & 0 & 0 & 0 & \ldots\\ I & 0 & 0 & 0 & \ldots\\ 0 & I & 0 & 0 & \ldots\\ 0 & 0 & I & 0 & \ldots\\ 0 & 0 & 0 & I & \ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ \end{pmatrix}$$

for $$T \in B(H)$$, $$S \in B(K)$$.

So if we now have an intertwining operator $$A \in B(H, K)$$ for $$T^*$$ and $$S^*$$ i.e. $$S^* A = A T^*$$ then why is the form of this operator (if a contraction) the one of the transposed infinite Toeplitz matrix (what I mean is: Why is then $$A$$ a transposed infinite dimensional lower triangular matrix with operators as their elements)?

For more literature about this if needed, you can find more in "The Commutant Lifting Approach to Interpolation Problems" by Foias and Frazho, more precisely in the chapter about the Nevanlinna-Pick interpolation problem.

I think that $$A,T,S$$ are not operators from and to $$H$$ or $$K$$ but from and to $$\bigoplus^\bot_{n\in\Bbb N}H$$ or $$\bigoplus^\bot_{n\in\Bbb N}K.$$
$$T_{i,j}=I$$ if $$i=j+1$$ and $$0$$ else, hence $$(T^*)_{i,j}=I$$ if $$j=i+1$$ and $$0$$ else. Same for $$S.$$ Hence (using that $$(UV)_{i,j}=\sum_kU_{i,k}V_{k,j}$$) $$\forall i,j\ge1\quad (S^*A)_{i,j}=(AT^*)_{i,j}$$ boils down to: $$\forall i,j\ge1\quad A_{i+1,1}=0\quad\text{and}\quad A_{i+1,j+1}=A_{i,j}$$ i.e. the (infinite) matrix $$A$$ is the transpose of Toeplitz matrix.
• Ah, Ok, I think I understand. Because if you have $$S^* A = A T^*$$ then you can rewrite this as $$AT = SA$$ and if you write it out, you get for $AT$ and $SA$ each that $A_{i+1, 1} = 0$ and $A_{i+1, j+1} = A_{i,j}$
• No, you cannot "rewrite this as $AT = SA$". You write out directly $(S^* A)_{i,j}=(A T^*)_{i,j}.$ Commented Mar 20, 2023 at 14:52
• $(S^*)_{i,j}=(S_{j,i})^*.$ And $(S^*A)_{i,j}=\sum_k(S^*)_{i,k}A_{k,j}.$ Commented Mar 20, 2023 at 14:55
• You get $A_{i+1,j} =A_{i,j-1}$ only when $j-1\ge1.$ When $j=1,$ what you get is $A_{i+1,j} =0.$ Commented Mar 20, 2023 at 16:23
• $(AT^*)_{i,1}=\sum_kA_{i,k}(T^*)_{k,1}=\sum_kA_{i,k}0=0.$ Commented Mar 20, 2023 at 16:29