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.