Proof Since $\Delta _{T}^{\mathbf {e}_i+\mathbf {e}_j}=0$ for all $i \neq j$ ,

$$ \begin{align*} D^2_{T_j}-T_i D^2_{T_j} T^*_i=D^2_{T_i}-T_j D^2_{T_i} T^*_j=0 \quad (i \neq j). \end{align*} $$

As the defect spaces are assumed to be finite dimensional, the above identities induce unitaries $W_j^{(i)}: \mathcal {D}_{T_j} \to \mathcal {D}_{T_j}$ for all $i\neq j$ , defined by

(3.1) $$ \begin{align} W_j^{(i)}(D_{T_j}h)=D_{T_j} T^*_i h\quad (h\in\mathcal{H}). \end{align} $$

Then observe that for $i, k=1,\ldots ,n$ and $i\neq k\neq j$ , $W_j^{(i)} W_j^{(k)}=W_j^{(k)} W_j^{(i)}$ . Indeed,

$$ \begin{align*} W_j^{(i)} W_j^{(k)} (D_{T_j} h) =W_j^{(i)} (D_{T_j} T^*_{k}h) = D_{T_j} T^*_i T^*_{k} h =W_j^{(k)} W_j^{(i)} (D_{T_j} h). \end{align*} $$

Now for $i=j$ , we set $W^{(j)}_j:=W_j^{ *(1)}\cdots W_j^{ *(j-1)} W_j^{* (j+1)}\cdots W_j^{* (n)}$ . Then it is clear from the definition that

$$\begin{align*}W_j^{(j)}(D_{T_j} P_T^* h)=D_{T_j} T^*_j h\quad (h\in\mathcal{H}),\end{align*}$$

and for each $j=1,\dots ,n$ , the tuple $(W_j^{(1)},\dots , W_j^{(n)})$ is a tuple of commuting unitaries and

(3.2) $$ \begin{align} W_j^{(1)}\cdots W_j^{(n)}=I_{\mathcal{D}_{T_j}}. \end{align} $$

We now have all the ingredients to define the triple corresponding to the dilating BCL n-tuple. We take $\mathcal {E}=\bigoplus ^n_{i=1} \mathcal {D}_{T_i}$ , and , where for $i=1,\dots ,n$ ,

$$ \begin{align*} U^*_i=\bigoplus^n_{j=1} W_j^{(i)} \!\quad \text{and} \!\quad P_i (x_1, \ldots, x_n)=(0,\ldots 0, x_i , 0, \ldots 0)\quad\! (x_l \in \mathcal{D}_{T_l}, l=1,\ldots,n). \end{align*} $$

By the construction of $U_i$ and $P_i$ and the identity 3.2, it easy to observe that for ${i,j=1,\ldots ,n}$ ,

$$\begin{align*}U_i U_j=U_j U_i, \quad U_i P_j=P_j U_i,\quad P_1+ \cdots+P_n=I_{\mathcal{E}}, \quad \text{and} \quad U_1 \cdots U_n=I_{\mathcal{E}}.\end{align*}$$

Thus, and satisfy all the conditions (1)–(4) as in Theorem 2.3, and by Proposition 2.7, the corresponding BCL n-tuple $(M_{\Phi _1},\ldots ,M_{\Phi _n})\in \mathfrak{B}^n_0(H^2_{\mathcal {E}}(\mathbb{D}))$ . In the rest of the proof, we construct the dilation map and show that $(M_{\Phi _1},\ldots ,M_{\Phi _n})$ is an isometric dilation of $(T_1,\dots , T_n)$ . The following identity,

$$ \begin{align*} D_{P_T}^2=D_{T_1}^2+T_1 D^2_{T_2}T^*_1+T_1 T_2 D^2_{T_3}T^*_1 T^*_2 +\cdots + T_1 \cdots T_{n-1} D^2_{T_n}T^*_1 \cdots T^*_{n-1} h \quad (h\in\mathcal{H}), \end{align*} $$

implies that the map $\iota :\mathcal {D}_{P_T} \to \mathcal {E}$ , defined by

(3.3) $$ \begin{align} D_{P_T} h \mapsto (D_{T_1} h, D_{T_2} T^*_1 h, D_{T_3}T^*_1 T^*_2 h, \ldots, D_{T_n}T^*_1 \cdots T^*_{n-1} h), \end{align} $$

is an isometry. Since $P_T$ is a pure contraction, the map $\pi :\mathcal {H} \to H^2_{\mathcal {D}_{P_T}}(\mathbb{D})$ , defined by $\pi h=\sum _{k \in \mathbb{Z}_+} z^k (D_{P_T} P_T^{* k}h) $ ( $z\in \mathbb{D}, h\in \mathcal {H}$ ), is an isometry. In fact, $\pi $ is the well-known dilation map for $P_T$ . Finally, the dilation map for us is the isometry $\Pi :=(I \otimes \iota )\pi : \mathcal {H}\to H^2_{\mathcal {E}}(\mathbb{D})$ . To complete the proof, we show that for all $i=1,\dots ,n$ ,

$$\begin{align*}\Pi T^*_i=M_{\Phi_i}^* \Pi.\end{align*}$$

Indeed, it is established by the rigorous computation below. For $ k\in \mathbb{Z}_+, \eta \in \mathcal {E}, h\in \mathcal {H}$ ,

$$ \begin{align*} &\langle M_{\Phi_i}^* \Pi h, z^k \eta\rangle\\ &\quad= \langle (I \otimes \iota)\sum_{m \in\mathbb{Z}_+} z^m (D_{P_T} P_T^{* m}h), (U_i P_i^\perp+z U_i P_i)z^k \eta \rangle \\ &\quad= \langle (P_i^\perp U_i^* \iota D_{P_T} P_T^{* k}h + P_i U_i^* \iota D_{P_T} P_T^{* k+1}h), \eta\rangle \\ &\quad= \langle P_i^\perp U_i^* (D_{T_1}P_T^{* k}h , D_{T_2} T^*_1 P_T^{* k}h,\ldots, D_{T_n}T^*_1 \cdots T^*_{n-1}P_T^{* k}h ), \eta\rangle \\ &\qquad\qquad\qquad\qquad+ \langle P_i U_i^* (D_{T_1} P_T^{* k+1}h, D_{T_2} T^*_1 P_T^{* k+1}h, \ldots, D_{T_n}T^*_1 \cdots T^*_{n-1} P_T^{* k+1}h ), \eta\rangle \\ &\quad= \langle U_i^* (D_{T_1}P_T^{* k}h,\ldots, D_{T_{i-1}} T^*_1 \cdots T^*_{i-2}P_T^{* k}h, 0, D_{T_{i+1}} T^*_1 \cdots T^*_{i}P_T^{* k}h, \ldots, D_{T_n}T^*_1 \cdots T^*_{n-1}P_T^{* k}h) , \eta\rangle \\ &\qquad\qquad\qquad\qquad+\langle U_i^* (0,\ldots, 0, D_{T_i} T^*_1 \cdots T^*_{i-1} P_T^{* k+1}h, 0, \ldots, 0), \eta\rangle \text{ (as } U_i P_i=P_i U_i)\\ &\quad=\langle (D_{T_1} T_i^* P_T^{* k}h,\ldots, D_{T_n}T^*_1 \cdots T^*_{n-1} T^*_i P_T^{* k}h ), \eta\rangle. \end{align*} $$

However, for $\eta \in \mathcal {E}, h\in \mathcal {H}, k\in \mathbb{Z}_+$ ,

$$ \begin{align*} \langle \Pi T^*_i h, z^k \eta\rangle &= \langle (I \otimes \iota)\sum_{m \in\mathbb{Z}_+} z^m (D_{P_T} P_T^{* m}T^*_ih), z^k \eta\rangle \\ &= \langle \iota D_{P_T} P_T^{* k}T^*_ih, \eta\rangle \\ &=\langle (D_{T_1} T_i^* P_T^{* k}h,\ldots, D_{T_n}T^*_1 \cdots T^*_{n-1} T^*_i P_T^{* k}h) , \eta\rangle. \end{align*} $$

Hence, $\Pi T^*_i=M_{\Phi _i}^* \Pi \,\, (i=1,\ldots ,n)$ , as asserted.

Proof Let the BCL n-tuple $(M_{\Phi _1}, \ldots , M_{\Phi _n})\in \mathfrak{B}^n_0(H^2_{\mathcal {E}}(\mathbb{D}))$ be an isometric dilation of T such that $\mathcal {E}$ is finite dimensional. The existence of such a dilation is shown in Theorem 3.1. Moreover, it also follows from Theorem 3.1 that

$$\begin{align*}T\cong (P_{\mathcal{Q}} M_{\Phi_{1}}|_{\mathcal{Q}},\ldots, P_{\mathcal{Q}} M_{\Phi_{n}}|_{\mathcal{Q}}),\end{align*}$$

where $\mathcal {Q}=\Pi (\mathcal {H})\subseteq H^2_{\mathcal {E}}(\mathbb{D})$ is a co-invariant subspace. Then, for $p\in \mathbb{C}[z_1, \ldots , z_n]$ ,

$$ \begin{align*} \|p(T)\|_{\mathcal{B}(\mathcal{H})}&=\|P_{\mathcal{Q}} p( M_{\Phi_{1}},\ldots, M_{\Phi_{n}})|_{\mathcal{Q}}\|_{\mathcal{B}(H^2_{\mathcal{E}}(\mathbb{D}))}\\ &\leq \| p( M_{\Phi_{1}},\ldots, M_{\Phi_{n}})\|_{\mathcal{B}(H^2_{\mathcal{E}}(\mathbb{D}))} \\ &=\|M_{p(\Phi_{1}(z),\ldots, \Phi_{n}(z))}\|_{\mathcal{B}(H^2_{\mathcal{E}}(\mathbb{D}))}. \end{align*} $$

Now, it is easy to observe that

$$\begin{align*}\|M_{p(\Phi_{1}(z),\ldots, \Phi_{n}(z))}\|_{\mathcal{B}(H^2_{\mathcal{E}}(\mathbb{D}))}= \|p(\Phi_{1}(z),\ldots, \Phi_{n}(z))\|_{H^\infty_{\mathcal{B}(\mathcal{E})}(\mathbb{D})}.\end{align*}$$

Clearly, the right-hand side is equal to

$$\begin{align*}\sup_{\theta\in [0,2\pi]}\|p(\Phi_1(e^{i\theta}),\ldots, \Phi_n(e^{i\theta}))\|_{\mathcal{B}(\mathcal{E})}.\end{align*}$$

Thus, we have

$$\begin{align*}\|p(T)\|_{\mathcal{B}(\mathcal{H})}\leq\sup_{\theta\in [0,2\pi]}\|p(\Phi_1(e^{i\theta}),\ldots, \Phi_n(e^{i\theta}))\|_{\mathcal{B}(\mathcal{E})}.\end{align*}$$

Since $\text {dim}\,(\mathcal {E})<\infty $ , for $z\in \mathbb{T}$ , $(\Phi _1(z),\ldots , \Phi _n(z))$ is an n-tuple of commuting unitary matrices. Also since $\Phi _1(z)\cdots \Phi _n(z)=zI_{\mathcal {E}}$ for all $z\in \overline {\mathbb{D}}$ , it follows that for any $(z_1,\dots ,z_n)\in \sigma (\Phi _1(z),\dots ,\Phi _n(z))$ ,

(3.5) $$ \begin{align} \underline{\mathbf{z}}:=z_1\cdots z_n=z. \end{align} $$

Therefore,

$$ \begin{align*} &\sup_{z\in\mathbb T}\|p(\Phi_1(z),\ldots , \Phi_n(z))\|_{\mathcal{B}(\mathcal{E})} \\&\quad=\sup \{|p(z_1,\ldots,z_n)|: (z_1,\ldots,z_n)\in \sigma(\Phi_1(z),\ldots, \Phi_n(z)), z\in \mathbb T \}\\ &\quad=\sup \{|p(\mathbf{z})|: \mathbf{z}=(z_1,\ldots,z_n)\in \mathbb T^n\cap \sigma(\Phi_1(\underline{\mathbf{z}}),\ldots, \Phi_n(\underline{\mathbf{z}}))\} \, \, (\text{by }(3.5))\\ &\quad= \sup_{\mathbf{z}\in W\cap {\mathbb T}^n} |p(\mathbf{z})|\\ &\quad \leq \sup_{\mathbf{z}\in W\cap \overline{\mathbb{D}}^n} |p(\mathbf{z})|, \end{align*} $$

where W is the one-dimensional symmetric algebraic variety as in (3.4). The above computation implies that $\sup _{\theta \in [0,2\pi ]}\|p(\Phi _1(e^{i\theta }),\ldots , \Phi _n(e^{i\theta }))\|_{\mathcal {B}(\mathcal {E})}\leq \sup _{\mathbf {z}\in W\cap \overline {\mathbb{D}}^n} |p(\mathbf {z})|$ . This completes the proof.

Proof For $X\in \{M\}^\prime $ , since $X M_z=M_z X$ , then there exists $\Theta \in H_{\mathcal {B}(\mathcal {E})}^\infty (\mathbb{D})$ such that $X=M_{\Theta }$ . Let $\Theta (z)=\sum _{ k \geq 0}A_k z^k$ , where $A_k \in \mathcal {B}(\mathcal {E})$ ( $k \geq 0$ ). We fix $i\in \{1,\dots ,n\}$ . Since $X M_{\Phi _i}=M_{\Phi _i} X$ , then the identity $ \Phi _i(z) \Theta (z)=\Theta (z) \Phi _i(z)$ implies that for all $z\in \mathbb{D}$ ,

$$ \begin{align*} U_i P_i^\perp A_0 +\sum_{ k \geq 1}( U_i P_i^\perp A_k+ U_i P_i A_{k-1}) z^{k} = A_0 U_i P_i^\perp +\sum_{ k \geq 1}(A_k U_i P_i^\perp +A_{k-1} U_i P_i) z^{k}. \end{align*} $$

Comparing the coefficients, we obtain

(4.2) $$ \begin{align} U_i P_i^\perp A_0=A_0 U_i P^\perp_i\,\, \text{and}\,\, U_i P_i^\perp A_k+ U_i P_i A_{k-1}=A_k U_i P_i^\perp +A_{k-1} U_i P_i \,\, (k\geq 1). \end{align} $$

Note that as $M\in \mathfrak{B}^n_0(H^2_{\mathcal {E}}(\mathbb{D}))$ , we have $U_i P_j=P_j U_i$ for all $i,j=1,\dots ,n$ . Then the identity $U_i P_i^\perp A_0=A_0 U_i P^\perp _i$ together with $A^*_0 P_i^\perp U_i^* = P^\perp _iU_i^*A_0^*$ , obtained by taking adjoint, imply that $\mathcal {E}_i^\perp $ is a reducing subspace for $A_0$ . That is, $A_0 P_i^{\perp }=P_i^{\perp }A_0$ . Letting $k=1$ in 4.2, we get

$$\begin{align*}U_i P_i^\perp A_1+ U_i P_i A_0=A_1 U_i P_i^\perp +A_0 U_i P_i.\end{align*}$$

Consequently,

$$ \begin{align*} U_i P^\perp_i A_1P^\perp_i= (U_i P_i^\perp A_1+ U_i P_i A_0)P_i^\perp=(A_1 U_i P_i^\perp +A_0 U_i P_i)P_i^\perp =A_1 U_i P^\perp_i. \end{align*} $$

By the same argument as before, the above identity together with $P_i^\perp A^*_1 P^\perp _i U^*_i=P^\perp _i U^*_i A_1^*$ imply that $\mathcal {E}_i^\perp $ is a reducing subspace for $A_1$ . Proceeding in the same way, we can show that $\mathcal {E}_i^\perp $ is a reducing subspace for $A_k$ for all $k\ge 2$ . We set $A^{(i)}_k:=A_k|_{\mathcal {E}_i}$ for all $k\ge 0$ , and $\Theta _i(z):=\sum _{k\geq 0}A^{(i)}_k z^k$ . Then $\Theta _i\in H_{\mathcal {B}(\mathcal {E}_i)}^\infty (\mathbb{D})$ . Since i was arbitrarily chosen, we have $\Theta _i\in H_{\mathcal {B}(\mathcal {E}_i)}^\infty (\mathbb{D})$ for all $i=1,\dots ,n$ , and it is now easy to see that

$$\begin{align*}\Theta(z)= \oplus_{i=1}^n \Theta_i(z), \text{ and therefore } M_{\Theta}=\oplus_{i=1}^n M_{\Theta_i}. \end{align*}$$

This completes the proof.

Proof Since X is a contraction in $\{M\}^\prime $ , using Proposition 4.1, X has the block matrix from as in 4.1 for some $\Theta _i \in H_{\mathcal {B}(\mathcal {E}_i)}^\infty (\mathbb{D}) (i=1,\dots ,n)$ . We consider

$$\begin{align*}G_i=\begin{bmatrix} 0 & \dots &0&\dots&0\\ \vdots &\ddots &\vdots &\ddots &\vdots \\ 0 &\dots & I_{H^2_{\mathcal{E}_i}(\mathbb{D})}- M_{\Theta_i}M^*_{\Theta_i}&\dots &0\\ \vdots &\ddots & \vdots &\ddots &\vdots\\ 0 &\dots &0 &\dots & 0 \end{bmatrix},\quad (i=1,\dots,n)\end{align*}$$

where the block matrix representation is taken with respect to the decomposition $H^2_{\mathcal {E}}(\mathbb{D})=H^2_{\mathcal {E}_1}(\mathbb{D})\oplus \cdots \oplus H^2_{\mathcal {E}_n}(\mathbb{D})$ . Since $M_{\Theta _i}$ is a contraction, it follows that $G_i\ge 0$ and $I-XX^*=G_1+\dots +G_n$ . Now for $i\neq j$ , using the block matrix form 2.3 of $M_{\Phi _j}$ and $M_{\Phi _j}X=XM_{\Phi _j}$ , it can be easily verified that $M_{\Phi _j}G_iM_{\Phi _j}^*=G_i$ . The proof now follows.

Proof Using Agler’s hereditary functional calculus ([Reference Agler1]) to the following identity, for $(z_1, \ldots , z_n)$ and $(w_1, \ldots , w_n) \in \mathbb{D}^n$ ,

$$ \begin{align*} &(1-z_1\bar{w}_1 \cdots z_n \bar{w}_n)\\& \quad =\sum_{i=1}^n (1-z_i \bar{w}_i)-\sum^n_{\underset{i< j}{i,j=1}}(1-z_i\bar{w}_i)(1-z_j\bar{w}_j) + \cdots+ (-1)^{n-1} (1-z_1\bar{w}_1)\cdots (1-z_n \bar{w}_n), \end{align*} $$

we get

$$ \begin{align*} I -P_T P_T^*=\sum_{i=1}^n D^2_{T_i}-\sum^n_{\underset{i< j}{i,j=1}} \Delta^{\mathbf{e}_i+\mathbf{e}_j}_{T}+ \sum_{\underset{i< j< k}{i,j,k =1}}^n \Delta^{\mathbf{e}_i+\mathbf{e}_j+\mathbf{e}_k}_{T}- \cdots +(-1)^{n-1} \Delta^{\mathbf{e}_i+\cdots+\mathbf{e}_n}_{T}. \end{align*} $$

Hence, by Lemma 2.1, $I-P_T P_T^* =\sum _{j=1}^n D^2_{T_j}$ .

Proof First, we define

$$\begin{align*}G_j=\text{SOT}-\lim_{\alpha=(\alpha_1,\dots,\alpha_{n-1})\to \infty} \tilde{T}_j^\alpha (I-XX^*)\tilde{T}_j^{*\alpha}\quad (j=1,\ldots, n).\end{align*}$$

Clearly, the above SOT limit exists as $T_i$ ’s are contractions. It then follows from the definition that $(I-C_{T_i})(G_j)=G_j -T_i G_j T^*_i = 0$ for all $i\neq j$ . Now,

$$ \begin{align*} \sum_{j=1}^n G_j=\text{SOT}-\lim_{\alpha\to \infty} \sum_{j=1}^n \tilde{T}_j^\alpha (I-XX^*)\tilde{T}_j^{*\alpha} =\text{SOT}-\lim_{\alpha\to \infty} \big( \sum_{j=1}^n \tilde{T}_j^\alpha \tilde{T}_j^{*\alpha}-X(\sum_{j=1}^n \tilde{T}_j^\alpha\tilde{T}_j^{*\alpha})X^*\big). \end{align*} $$

To complete the proof, it is enough to show that $\text {SOT}-\lim _{\alpha \to \infty } \sum _{j=1}^n \tilde {T}_j^\alpha \tilde {T}_j^{*\alpha }=I_{\mathcal {H}}$ . Since $\Delta ^{\mathbf {e}_i+\mathbf {e}_j}_{T}=0$ (that is, $D_{T_j}^2=T_i D^2_{T_j}T^*_i$ ( $i\neq j$ )), then

$$\begin{align*}D_{T_j}^2=\tilde{T}_j^\alpha D^2_{T_j} \tilde{T}_j^{*\alpha}\quad (j=1,\ldots,n, \alpha=(\alpha_1, \ldots, \alpha_{n-1})\in \mathbb{Z}^{n-1}_+). \end{align*}$$

Taking sum over j in both sides to the above identity, we get for all $\alpha \in \mathbb{Z}^{n-1}_+$ ,

$$ \begin{align*} \sum_{j=1}^n \tilde{T}_j^\alpha\tilde{T}_j^{*\alpha}& =\sum_{j=1}^n D^2_{T_j}+ \sum_{j=1}^n \tilde{T}_j^\alpha T_j T^*_j \tilde{T}_j^{*\alpha}\\ &=I_{\mathcal{H}}-P_T P_T^* + \sum_{j=1}^n \tilde{T}_j^\alpha T_j T^*_j \tilde{T}_j^{*\alpha}\quad(\text{by Lemma } 4.3). \end{align*} $$

Consequently, for $\alpha \ge \mathbf e=(1,\dots , 1)$ ,

$$ \begin{align*} I_{\mathcal{H}}-\sum_{j=1}^n \tilde{T}_j^\alpha\tilde{T}_j^{*\alpha}= P_T(I_{\mathcal{H}} - \sum_{j=1}^n \tilde{T}_j^{\alpha-e} \tilde{T}_j^{*{\alpha-e}}) P_T^*. \end{align*} $$

Set $Y_{\alpha }:= I_{\mathcal {H}}-\sum _{j=1}^n \tilde {T}_j^\alpha \tilde {T}_j^{*\alpha }, (\alpha \in \mathbb{Z}^{n-1}_+)$ . By the above identity, for $\alpha \ge \mathbf e$ , if $k=\text {min}\{\alpha _1,\dots ,\alpha _{n-1}\}$ , then

$$ \begin{align*} Y_{\alpha}= P_T Y_{\alpha-\mathbf e}P_T^*=\cdots=P^k_{T} Y_{\alpha-k\mathbf e}P_T^{* k}. \end{align*} $$

Thus, for $h\in \mathcal {H}$ ,

$$ \begin{align*} \|Y_\alpha h\|=\|P_T^k Y_{\alpha-ke} P_T^{*k} h\|\leq \|Y_{\alpha-ke}\| \|P_T^{*k} h\|\leq (n+1)\|P_T^{*k} h\|. \end{align*} $$

Since $P_T$ is pure, it then follows that $\text {SOT}-\lim _{\alpha \to \infty } Y_\alpha =0$ . This completes the proof.

Proof Using the power series expansion of $\tau _{W}(z)$ , it is easy to see that the identity in $(1)$ is equivalent to

(4.4) $$ \begin{align} AU=UA \text{ and } (B D^{k-1} C)U=U(B D^{k-1} C)\quad (k\geq 1). \end{align} $$

Now we assume that $(2)$ holds. Then by a routine block matrix multiplication, $AU=U A$ , $BY=UB$ , $CU=YC$ , and $DY=YD$ . Thus, $AU=U A$ and

$$ \begin{align*} B D^{k-1} CU= B D^{k-1} YC=B Y D^{k-1} C=U B D^{k-1} C \quad (k\geq 1). \end{align*} $$

Hence, by the observation we made earlier, $M_{\tau _W} (I_{H^2(\mathbb{D})}\otimes U)=(I_{H^2(\mathbb{D})}\otimes U) M_{\tau _W}$ . This proves $(2) \Longrightarrow (1)$ .

For the proof of the moreover part, note that the block matrix representation of W with respect to the decomposition $\mathcal {E}\oplus \mathcal {F}_1\oplus \mathcal {F}_2$ is

$$\begin{align*}W=\begin{bmatrix} A & B|_{\mathcal{F}_1}& 0\\ P_{\mathcal{F}_1}C & D_1 & 0\\ 0& 0 & D_2 \end{bmatrix}. \end{align*}$$

Since W is a unitary,

$$ \begin{align*} A^* A+ C^* P_{\mathcal{F}_1} C=I_{\mathcal{E}}, \quad P_{\mathcal{F}_1} B^* B|_{\mathcal{F}_1} +D_1^* D_1=I_{\mathcal{F}_1}, \end{align*} $$

$$\begin{align*}P_{\mathcal{F}_1} B^* A +D_1^* P_{\mathcal{F}_1} C=0, \text{ and } P_{\mathcal{F}_1} C C^*|_{\mathcal{F}_1}+ D_1 D_1^*=I_{\mathcal{F}_1}.\end{align*}$$

Claim:

$$\begin{align*}\bigvee_{k\geq 0} D_1^k P_{\mathcal{F}_1} C (\mathcal{E})= \mathcal{F}_1. \end{align*}$$

The proof of the claim follows from [Reference Ball and Cohen8, Theorem 4.2]. However, we supply a proof for the completeness. To this end, it is enough to show that $\mathcal {S}:=(\mathcal {F}_1 \ominus \bigvee _{k\geq 0} D_1^k P_{\mathcal {F}_1} C (\mathcal {E}))=\{0\}$ . Let $h\in \mathcal {S}$ ; that is, $h\in \mathcal {F}_1$ and $h\perp \bigvee _{k\geq 0} D_1^k P_{\mathcal {F}_1} C (\mathcal {E})$ . Therefore, $h \perp D_1^k P_{\mathcal {F}_1} C \eta $ ( $k\geq 0, \eta \in \mathcal {E}$ ). In particular, $h \perp P_{\mathcal {F}_1} C \eta $ ( $\eta \in \mathcal {E}$ ); that is,

$$\begin{align*}\langle h, P_{\mathcal{F}_1} C \eta \rangle=0 \quad (\eta \in \mathcal{E}).\end{align*}$$

This implies $h\in \text {ker }C^*$ . In other words, $\mathcal {S}\subseteq \text {ker }C^*$ . Now it is easy to see that $\mathcal {S}$ is a $D^*_1$ -invariant subspace and the identity $P_{\mathcal {F}_1} C C^*+ D_1 D_1^*=I_{\mathcal {F}}$ restricted on $\mathcal {S}$ becomes $D_1 D_1^*|_{\mathcal {S}}=I_{\mathcal {S}}$ , which contradicts that $D_1$ is completely non-coisometric. Hence, the claim is proved.

We now proceed to construct a contraction Y on $\mathcal {F}$ such that $W\tilde {U}=\tilde {U}W$ ; that is,

$$\begin{align*}BY= UB, CU=YC, \text{ and } D Y=YD. \end{align*}$$

Since $\mathcal {F}_1=\bigvee _{k\geq 0} D_1^k P_{\mathcal {F}_1} C (\mathcal {E})$ , we define

$$\begin{align*}\!Y:\mathcal{F}_1 \oplus \mathcal{F}_2 \to \mathcal{F}_1 \oplus \mathcal{F}_2,\text{ by } (D_1^k P_{\mathcal{F}_1} C \eta, f) \mapsto (D_1^k P_{\mathcal{F}_1} C U \eta, 0)\quad\! (\eta\in \mathcal{E}, f\in \mathcal{F}_2, k\geq 0).\end{align*}$$

It remains to show that Y is a well-defined contraction on $\mathcal {F}$ and $BY=UB$ as the identities $CU=YC$ and $D Y=YD$ are immediate from the definition of Y. In fact, the reader must have noticed that it is the only way one could define Y with required properties, and therefore, Y is unique in this case. We show below that $Y|_{\mathcal {F}_1}$ is a unitary. To this end, it is enough to show that

$$\begin{align*}\langle D_1^k P_{\mathcal{F}_1} C \eta, D_1^l P_{\mathcal{F}_1} C \eta^\prime\rangle =\langle D_1^k P_{\mathcal{F}_1} C U \eta, D_1^l P_{\mathcal{F}_1} C U \eta^\prime \rangle\end{align*}$$

for all $k, l\in \mathbb{Z}_+$ and $\eta , \eta ^\prime \in \mathcal {E}$ . When $l=k=0$ , note that

$$ \begin{align*} \langle P_{\mathcal{F}_1} C \eta, P_{\mathcal{F}_1} C \eta^\prime \rangle &= \langle \eta,(I_{\mathcal{E}}-A^* A) \eta^\prime \rangle \quad(\text{as }A^* A+ C^* P_{\mathcal{F}_1} C=I_{\mathcal{E}})\\ &=\langle U\eta,U\eta^\prime \rangle- \langle UA \eta, UA \eta^\prime \rangle\quad (\text{as } U \text{ is an isometry})\\ &=\langle U \eta, U \eta^\prime \rangle- \langle AU \eta, A U \eta^\prime \rangle \quad (\text{as } AU =UA)\\ &= \langle P_{\mathcal{F}_1} CU \eta, P_{\mathcal{F}_1} CU \eta^\prime\rangle. \end{align*} $$

For the general case, we use the following identity repeatedly:

(4.5) $$ \begin{align} \langle D_1^m P_{\mathcal{F}_1} C \eta, D_1^n P_{\mathcal{F}_1} C \eta^\prime \rangle&=\langle D_1^{m-1}P_{\mathcal{F}_1} C \eta, D_1^*D_1D_1^{n-1} P_{\mathcal{F}_1} C \eta^\prime \rangle\nonumber\\ &= \langle D_{1}^{m-1} P_{\mathcal{F}_1} C\eta, D_{1} ^{n-1} P_{\mathcal{F}_1} C \eta^\prime \rangle- \langle B D_{1} ^{m-1} P_{\mathcal{F}_1} C\eta, B D_{1}^{n-1} P_{\mathcal{F}_1} C \eta^\prime \rangle, \end{align} $$

as $P_{\mathcal {F}_1} B^* B|_{\mathcal {F}_1}+ D_1^* D_1=I_{\mathcal {F}_1}$ , where $m,n\ge 1$ . Now consider the case when at least one of l and k is nonzero. We assume, without any loss of generality, that $k\geq l$ . Then for any $\eta , \eta ^\prime \in \mathcal {E}$ , by applying (4.5) l times, we have

$$ \begin{align*} \langle D_1^k P_{\mathcal{F}_1} C \eta, D_1^l P_{\mathcal{F}_1} C \eta^\prime \rangle &=\langle D_1^{k-l} P_{\mathcal{F}_1} C\eta, P_{\mathcal{F}_1} C \eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} C\eta, B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle \nonumber\\ &=\langle D_1^{k-l-1} P_{\mathcal{F}_1} C \eta, D_1^* P_{\mathcal{F}_1} C \eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} C\eta, B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle\nonumber\\ &=-\langle D_1^{k-l-1} P_{\mathcal{F}_1} C\eta, B^* A \eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} C\eta, B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle \nonumber\\ &=-\langle B D_1^{k-l-1} P_{\mathcal{F}_1} C\eta, A \eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} C\eta, B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle. \end{align*} $$

In the above identity, replacing $\eta $ and $\eta ^\prime $ by $U\eta $ and $U\eta ^\prime $ , respectively, and using (4.4), we get

$$ \begin{align*} &\langle D_1^k P_{\mathcal{F}_1} C U\eta, D_1^l P_{\mathcal{F}_1} C U\eta^\prime \rangle\\ &\quad=-\langle B D_1^{k-l-1} P_{\mathcal{F}_1} CU\eta, A U\eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} CU\eta, B D_1^{l-j} P_{\mathcal{F}_1} C U\eta^\prime \rangle\\ &\quad=-\langle U B D_1^{k-l-1} P_{\mathcal{F}_1} C\eta, UA \eta^\prime \rangle- \sum^{l}_{j=1} \langle U B D_1^{k-j} P_{\mathcal{F}_1} C\eta, U B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle\\ &\quad=-\langle B D_1^{k-l-1} P_{\mathcal{F}_1} C \eta, A \eta^\prime \rangle- \sum^{l}_{j=1} \langle B D_1^{k-j} P_{\mathcal{F}_1} C \eta, B D_1^{l-j} P_{\mathcal{F}_1} C \eta^\prime \rangle\\ &\quad= \langle D_1^k P_{\mathcal{F}_1} C \eta, D_1^l P_{\mathcal{F}_1} C \eta^\prime \rangle. \end{align*} $$

Thus, Y is a unitary on $\mathcal {F}_1$ . From the definition of Y, it follows that Y is a contraction. Now for $g=\sum _{k=0}^lD_1^k P_{\mathcal {F}_1} C \eta _k$ ( $\eta _k\in \mathcal {E}$ ) and $f\in \mathcal {F}_2$ ,

$$ \begin{align*} B Y (g,f)=B Y \big(\sum_{k=0}^lD_1^k P_{\mathcal{F}_1} C \eta_k, f \big)&= \big(\sum_{k=0}^l B D_1^k P_{\mathcal{F}_1} C U \eta_k, 0\big)\\ & = UB \big (\sum_{k=0}^l D_1^k P_{\mathcal{F}_1} C \eta_k, f\big )\\ &= U B(g,f). \end{align*} $$

Therefore, $BY =UB$ , and this completes the proof.

Proof Since

$$ \begin{align*} \begin{bmatrix} A& B\\ C& D \end{bmatrix}\begin{bmatrix} Qh\\ G T^* h\end{bmatrix}=\begin{bmatrix} QX^* h\\ G h \end{bmatrix}, \end{align*} $$

we have

(4.6) $$ \begin{align} A Q h+ B G T^* h=QX^* h, \end{align} $$

and

(4.7) $$ \begin{align} C Qh+ D G T^* h=G h. \end{align} $$

Now, replacing h by $T^* h$ in 4.7, we have

$$ \begin{align*} G T^* h=C Q T^*h+ D G T^{* 2} h. \end{align*} $$

Then, substituting $G T^*h$ by $C Q T^*h+ D G T^{* 2} h$ in 4.6, we obtain

$$ \begin{align*} QX^* h =A Qh+ BC Q T^*h+ B D G T^{* 2} h. \end{align*} $$

By repeating the procedure m times, we get

$$ \begin{align*} Q X^* h=A Qh+\sum^{m-1}_{k=0} B D^{ k} C Q T^{* k+1}h+ B D^m G T^{* m+1} h. \end{align*} $$

Since T is pure, for all $h\in \mathcal {H}$ , $\|B D^m G T^{* m+1} h\|\leq \|T^{* m+1} h\|\to 0$ as $m\to \infty $ . Therefore,

(4.8) $$ \begin{align} Q X^*=A Q+\sum^{\infty}_{k=0} B D^{ k} C Q T^{* k+1}. \end{align} $$

Now, for $ k\in \mathbb{Z}_+, \eta \in \mathcal {E}, h\in \mathcal {H}$ ,

$$ \begin{align*} \langle M_{\Theta}^* \Pi h, z^k \eta\rangle &= \langle \sum_{m\ge 0}z^m Q S^* T^{* m}h, (A^*+C^*z (I-z D^*)^{-1} B^*)z^k \eta\rangle\\ &= \langle \sum_{m\ge 0}z^m Q S^* T^{* m}h, (A^*+ \sum_{m\ge 0}z^{m+1 }C^* D^{* m} B^*)z^k \eta\rangle\\ &= \langle A Q S^* T^{* k}h, \eta \rangle + \sum_{m\ge 0} \langle B D^ m C Q S^* T^{* m+k+1}h, \eta\rangle\\ &= \langle \big(A Q + \sum_{m\ge 0} B D^ m C Q T^{* m+1}\big) S^* T^{* k}h, \eta\rangle\\ &= \langle Q X^* S^* T^{* k}h, \eta\rangle \quad (\text{by the identity}\ 4.8), \end{align*} $$

and on the other hand,

$$ \begin{align*} \langle \Pi X^* h, z^k \eta\rangle = \langle \sum_{m\ge 0}z^m Q S^* T^{* m} X^* h, z^k \eta\rangle=\langle Q X^* S^* T^{* k } h, \eta\rangle. \end{align*} $$

Hence, $\Pi X^* = M^*_{\Theta } \Pi $ . This completes the proof.

Proof Since the proof uses the explicit isometric dilation of $T\in \mathfrak{B}^n_0(\mathcal {H})$ as constructed in Theorem 3.2, we first recall some of the necessary ingredients from the proof of Theorem 3.2. By Theorem 3.2, the Hilbert space $\mathcal {E}$ has the decomposition $\mathcal {E}=\oplus ^n_{j=1}\mathcal {K}_j$ , where $\mathcal {D}_{T_j}\subseteq \mathcal {K}_j$ for all $j=1,\dots ,n$ . The dilation map $\Pi : \mathcal {H} \to H^2_{\mathcal {E}}(\mathbb{D})$ is an isometry given by

$$\begin{align*}\Pi h=\sum^\infty_{k=0} z^k (D_{T_1} P_T^{* k} h, D_{T_2} T^*_1 P_T^{* k} h, \ldots, D_{T_n}T^*_1 \cdots T^*_{n-1} P_T^{* k} h)\quad (h\in\mathcal{H}). \end{align*}$$

The dilating BCL n-tuple M corresponds to the triple , where and are defined as

(4.9) $$ \begin{align} U^*_i=\bigoplus^n_{j=1} W_j^{(i)}\,\, \text{and } P_i (x_1, \ldots, x_n)=(0,\ldots 0, x_i , 0, \ldots 0)\quad (x_l \in \mathcal{K}_l, l=1,\ldots,n), \end{align} $$

for some unitaries $W_j^{(i)}: \mathcal {K}_j \to \mathcal {K}_j$ satisfying

(4.10) $$ \begin{align} W_j^{(i)}(D_{T_j}h)=D_{T_j} T^*_i h\quad ( i\neq j) \text{ and } W_j^{(j)}(D_{T_j} P_T^* h)=D_{T_j} T^*_j h \end{align} $$

for all $h\in \mathcal {H}$ . Moreover, for all $j=1,\dots ,n$ ,

(4.11) $$ \begin{align} W_j^{(j)}=W_j^{(1)*} \cdots W_j^{(j-1) *} W_j^{(j+1) *} \cdots W_j^{(n)*}. \end{align} $$

For all $j=1,\ldots ,n$ , we define

$$\begin{align*}\Pi_j:\mathcal{H} \to H^2_{\mathcal{K}_j}(\mathbb{D}) \text{ by } h\mapsto \sum^\infty_{k=0} z^k D_{T_j}T^*_1 \cdots T^*_{j-1} P_T^{* k}h. \end{align*}$$

It is now clear that for all $h\in \mathcal {H}$ , $\Pi h=\oplus _{j=1}^n \Pi _jh $ . Suppose any contractive multiplier $\Theta \in H^\infty _{\mathcal {B}(\mathcal {E})}(\mathbb{D})$ is in $\{M\}^\prime $ . Then, by Proposition 4.1, $M_\Theta $ has the form $M_{\Theta }=\oplus _{j=1}^n M_{\Theta _j}$ , where $\Theta _j\in H_{\mathcal {B}(\mathcal {K}_j)}^\infty (\mathbb{D})$ for all $j=1,\dots ,n$ . By the explicit form of $M_{\Phi _i}$ , as in 2.3, $M_{\Theta }$ commutes with $M_{\Phi _i}$ for all $i=1,\dots ,n$ if and only if

(4.12) $$ \begin{align} \Theta_j(z)W_j^{(i)}=W_j^{(i)}\Theta_j(z)\quad (z\in\mathbb{D}, i,j=1,\dots,n). \end{align} $$

Moreover, $\Pi X^*= M^*_{\Theta } \Pi $ if and only if for all $h\in \mathcal {H}$ ,

(4.13) $$ \begin{align} \bigoplus_{j=1}^n \Pi_j X^*h=\bigoplus_{j=1}^n M^*_{\Theta_j} \Pi_jh, \text{ equivalently } \Pi_j X^*=M^*_{\Theta_j} \Pi_j\quad (j=1,\dots,n). \end{align} $$

Thus, by the above discussion, to complete the proof, we need to construct ${\Theta _j\in H_{\mathcal {B}(\mathcal {K}_j)}^\infty (\mathbb{D})}$ for all $j=1,\dots ,n$ such that (4.13) and (4.12) holds. To this end, for the rest of the proof, we fix j and find an auxiliary Hilbert space $\mathcal {G}$ and a contraction R on $\mathcal {K}_j\oplus \mathcal {G}$ such that $\Theta =\tau _{R^*}$ , the transfer function corresponding to $R^*$ . Also note that for $\tau _{R^*}$ to satisfy equation (4.12), by Theorem 4.5, it is sufficient to find contractions $Y_i$ on $\mathcal {G}$ such that

(4.14) $$ \begin{align} R^* \begin{bmatrix} W_j^{(i)}& 0\\ 0&Y_i \end{bmatrix}= \begin{bmatrix} W_j^{(i)}& 0\\ 0&Y_i \end{bmatrix} R^* \quad (i=1,\dots,n). \end{align} $$

This is the point of departure to the construction of the space $\mathcal {G}$ and contractions R and $Y_i$ with required properties. Since X is a contraction in $\{T\}^\prime $ , by Theorem 4.4, there exist positive operators $G_1, \ldots , G_n$ on $\mathcal {H}$ such that

(4.15) $$ \begin{align} I-X X^*=G_1 +\cdots + G_n \text{ and } (I-C_{T_i})(G_m)=G_m -T_i G_m T^*_i = 0 \quad (i\neq m). \end{align} $$

Repeated application of the second identity above yields $T_jG_jT_j^*=P_T G_j P_T^*$ . Now,

$$ \begin{align*} D^2_{T_j}-X D^2_{T_j} X^*&=I-T_j T_j^* -X(I-T_j T^*_j)X^*\\ &=(I-C_{T_j})(I-X X^*)\\ &=(I-C_{T_j})(G_1+\cdots+ G_n)\\ &=G_j-T_j G_j T^*_j \quad (\text{by}\ 4.15). \end{align*} $$

A rearrangement of the above identity is

(4.16) $$ \begin{align} D^2_{T_j}+P_T G_j P_T^*=X D^2_{T_j} X^*+G_j. \end{align} $$

We set $\tilde {\mathcal {G}}:=\overline {\text {ran}}\, G_j^{1/2}$ ,

$$\begin{align*}\mathcal{M}:=\{(D_{T_j}h, G_j^{1/2}P_T^*h):h\in\mathcal{H}\}\subseteq \mathcal{K}_j\oplus \tilde{\mathcal{G}}, \text{ and } \mathcal{N}:=\{(D_{T_j}X^*h, G_j^{1/2}h):h\in\mathcal{H}\}\subseteq \mathcal{K}_j\oplus \tilde{\mathcal{G}}.\end{align*}$$

Therefore, the identity 4.16 induces an isometry $\tilde {R}:\mathcal {M} \to \mathcal {N}$ , defined by

(4.17) $$ \begin{align} \tilde{R}(D_{T_j}h, G^{1/2}_j P_T^* h)=(D_{T_j}X^* h, G^{1/2}_j h)\quad (h\in \mathcal{H}). \end{align} $$

Again, using the identity $G_j=T_i G_j T^*_i$ ( $i \neq j$ ), we define an isometry for each $i\neq j$ ,

(4.18) $$ \begin{align} \tilde{Y_i}:\tilde{\mathcal{G}} \to \tilde{\mathcal{G}}, \text{ by } G_j^{1/2}h\mapsto G_j^{1/2} T^*_i h\quad (h\in \mathcal{H}). \end{align} $$

Since $(\tilde {Y}_1, \ldots , \tilde {Y}_{j-1}, \tilde {Y}_{j+1}, \ldots , \tilde {Y}_n)$ is an $(n-1)$ -tuple of commuting isometries on $\tilde {\mathcal {G}}$ , it has minimal unitary extension $(Y_1, \ldots , Y_{j-1}, Y_{j+1}, \ldots , Y_n)$ on $\mathcal {G}$ , say. We set

(4.19) $$ \begin{align} Y_j:=Y_1^* \cdots Y^*_{j-1} Y^*_{j+1} \cdots Y^*_n. \end{align} $$

For the simplicity of notations, we set

$$\begin{align*}V_i:=\begin{bmatrix} W_j^{(i)}& 0\\ 0&Y_i \end{bmatrix}\quad (i=1,\dots,n). \end{align*}$$

The auxiliary space $\mathcal {G}$ and the required contractions $Y_i$ ’s on $\mathcal {G}$ are already constructed. To construct the contraction R, we take extension of $\tilde {R}$ , as in (4.17), in certain way so that (4.14) holds. It follows from (4.10) and (4.18) that $\mathcal {M}$ and $\mathcal {N}$ are invariant subspaces of each of the unitary $V_i$ on $\mathcal {K}_j\oplus \mathcal {G}$ with $i\neq j$ . Also note that for $h\in \mathcal {H}$ and $i\neq j$ ,

$$ \begin{align*} \tilde{R}V_i(D_{T_j}h, G_j^{1/2}P_T^*h)=\tilde{R} (D_{T_j}T^*_i h, G_j^{1/2}T^*_i P_T^*h)= (D_{T_j}T^*_i X^*h, G_j^{1/2}T^*_ih), \end{align*} $$

and on the other hand,

$$ \begin{align*} V_i \tilde{R} (D_{T_j}h, G_j^{1/2}P_T^*h)= V_i (D_{T_j}X^* h, G_j^{1/2}h) =(D_{T_j}T^*_i X^*h, G_j^{1/2}T^*_ih). \end{align*} $$

Thus, $ \tilde {R} V_i|_{\mathcal {M}}=V_i \tilde {R}|_{\mathcal {M}}$ for all $i\neq j$ . We consider

$$\begin{align*}\tilde{\mathcal{M}}:=\bigvee_{\alpha\in \mathbb{Z}^{n-1}_+}\tilde{V}^{* \alpha}(\mathcal{M}) \text{ and } \tilde{\mathcal{N}}:=\bigvee_{\alpha\in \mathbb{Z}^{n-1}_+}\tilde{V}^{* \alpha}(\mathcal{N}),\end{align*}$$

where $\tilde {V}:=(V_1, \ldots , V_{j-1}, V_{j+1}, \ldots , V_n)$ , and extend $\tilde {R}$ to get an operator from $\tilde {\mathcal {M}}$ onto $\tilde {\mathcal {N}}$ by the following prescription:

$$\begin{align*}\sum_{0\leq \alpha\leq \gamma}\tilde{V}^{* \alpha} x_\alpha\mapsto \sum_{0\le \alpha\leq \gamma}\tilde{V}^{* \alpha} \tilde{R} x_\alpha\quad (x_\alpha\in \mathcal{M}). \end{align*}$$

The extension is well-defined as

$$ \begin{align*} \| \sum_{0\le \alpha\leq \gamma}\tilde{V}^{* \alpha} \tilde{R} x_\alpha\|^2 &=\sum_{0\le \alpha, \beta\le \gamma}\langle \tilde{V}^{* \alpha} \tilde{R} x_\alpha, \tilde{V}^{* \beta} \tilde{R} x_\beta \rangle\\ &=\sum_{0\le \alpha, \beta\le \gamma} \langle \tilde{V}^{ (\beta -\alpha)_+} \tilde{R} x_\alpha, \tilde{V}^{ (\beta -\alpha)_-} \tilde{R} x_\beta \rangle\\ &=\sum_{0\le \alpha,\beta\le \gamma}\langle \tilde{R}\tilde{V}^{ (\beta -\alpha)_+} x_\alpha, \tilde{R}\tilde{V}^{ (\beta-\alpha)_-} x_\beta \rangle\quad (\text{as } \tilde{R} V_i|_{\mathcal{M}}=V_i \tilde{R}|_{\mathcal{M}}, i\neq j)\\ &=\sum_{0\le \alpha,\beta\le \gamma}\langle \tilde{V}^{ (\beta -\alpha)_+} x_\alpha, \tilde{V}^{ (\beta-\alpha)_-} x_\beta \rangle\\ &= \| \sum_{0\le \alpha\leq \gamma}\tilde{V}^{* \alpha} x_\alpha\|^2. \end{align*} $$

We denoted the extended operator again by $\tilde {R}: \tilde {\mathcal {M}}\to \tilde {\mathcal {N}}$ , and it follows from the definition that

$$\begin{align*}\tilde{R} V_i|_{\tilde{\mathcal{M}}}= V_i\tilde{R}|_{\tilde{\mathcal{M}}}\quad (i\neq j).\end{align*}$$

Finally, we define the contraction R on $\mathcal {K}_j\oplus \mathcal {G}$ by setting $R= \tilde {R}$ on $\tilde {\mathcal {M}}$ and $R\equiv 0$ on $\tilde {\mathcal {M}}^{\perp }$ . Since $\tilde {\mathcal {M}}$ is a reducing subspace for $V_i (i\neq j)$ , we have $RV_i=V_iR$ for all $i\neq j$ . Also since $V_j=V_1^*\cdots V_{j-1}^*V_{j+1}^*\cdots V_n^*$ , which can be read off from (4.11) and (4.19), it follows that $RV_j=V_jR$ . Then by Fuglede-Putnam, we have (4.14). It only remains to prove that

$$\begin{align*}\Pi_j X^*=M^*_{\Theta_j}\Pi_j,\end{align*}$$

where $\Theta _j(z)=\tau _{R^*}(z)=A^*+C^*z (I-z D^*)^{-1} B^*$ , transfer function of $R^*$ and

$$\begin{align*}R=\begin{bmatrix} A& B\\ C& D \end{bmatrix}: \mathcal{K}_j \oplus \mathcal{G}\to \mathcal{K}_j \oplus \mathcal{G}.\end{align*}$$

This is obtained by applying Lemma 4.6 for $S:=T_1\cdots T_{j-1}$ , $T:= P_T$ , $G:=G_j^{1/2}$ , $Q:=D_{T_j}$ , $\Pi :=\Pi _j$ , $W:=R$ , and using 4.17. Hence, the theorem is proved.

Proof Since X is a contraction, suppose Y on $\mathcal {K}$ is the minimal isometric dilation of X. That is,

$$\begin{align*}X^k=P_{\mathcal{H}} Y^k|_{\mathcal{H}} \, (k\in \mathbb{Z}_+) \text{ and } \mathcal{K}=\bigvee_{k\in \mathbb{Z}_+} Y^k(\mathcal{H}).\end{align*}$$

We define, for each $i=1,\ldots ,n$ , $\tilde {V_i}:\mathcal {K} \to \mathcal {K}$ by

$$\begin{align*}\tilde{V}_i(\sum^m_{k=0} Y^k h_k)=\sum^m_{k=0} Y^k V_i h_k \quad (h_k\in \mathcal{H}).\end{align*}$$

It is now enough to prove that each $\tilde {V_i}$ is well-defined, which readily follows from the following computation. For any $h, h^\prime \in \mathcal {H}$ and $k, l\in \mathbb{Z}_+$ with $k\geq l$ ,

$$ \begin{align*} \langle Y^k V_i h, Y^l V_i h^\prime \rangle= \langle Y^{k-l} V_i h, V_i h^\prime \rangle=\langle X^{k-l} V_i h, V_i h^\prime \rangle=\langle V_i X^{k-l} h, V_i h^\prime \rangle=\langle X^{k-l} h, h^\prime \rangle. \end{align*} $$

This completes the proof.

Proof Without of loss generality, let $T_1$ be a pure contraction. Since $(T_1,\ldots , T_n)\in \mathfrak{B}^n_0(\mathcal {H})$ , then, for $i=2, \ldots , n$ , $\Delta ^{(1,1)}_{(T_1,T_i)}=D^2_{T_i}-T_1 D^2_{T_i} T^*_1=0$ , where $D^2_{T_i}=I-T_i T^*_i$ . Iterating the identity k-times, we have

$$\begin{align*}D^2_{T_i}=T^k_1 D^2_{T_i} T^{*k}_1\quad (k\in \mathbb{Z}_+). \end{align*}$$

Using pureness of $T_1$ , we obtain $D^2_{T_i}=0$ , and therefore, $T_i$ is a co-isometry. Conversely, let $T_2, \ldots , T_n $ be co-isometries. Since $P_T$ is a pure contraction, for $h\in \mathcal {H}$ ,

$$\begin{align*}\|T^{* k}_1 h\|=\|T^{k}_2 \cdots T^{k}_n T^{* k}_n \cdots T^{* k}_2 T^{*k}_1 h\|\leq \| P^{* k}_T h\| \to 0\text{ as } n\to \infty.\end{align*}$$

This completes the proof.

Proof Since A and B are commuting matrices, then the pair $(A^*,B^*)$ can be simultaneously upper tringularizable. Suppose that

$$\begin{align*}(A^*,B^*)\cong \left( \begin{bmatrix} a_{11}& \cdots & \cdots& a_{1m}\\ 0& a_{22}& \cdots & a_{2 m}\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0 & \cdots & a_{mm} \end{bmatrix}, \begin{bmatrix} b_{11}& \cdots & \cdots& b_{1m}\\ 0& b_{22}& \cdots & b_{2 m}\\ \vdots & \vdots & \ddots & \vdots\\ 0 &0 & \cdots & b_{mm} \end{bmatrix} \right). \end{align*}$$

Then the joint eigenvalues of $A^*$ and $B^*$ are given by $\{(a_{ii}, b_{ii}): i=1,\dots ,m\}$ . Let $v_i$ be a joint eigenvector of norm one corresponding to the joint eigenvalues $(a_{ii}, b_{ii})$ . Then

$$ \begin{align*} 0=\langle \Delta^{(1,1)}_{(A,B)}v_i, v_i\rangle =\langle (I-AA^*-BB^*+ABA^*B^*)v_i, v_i\rangle=(1-|a_{ii}|^2)(1-|b_{ii}|^2). \end{align*} $$

This shows that either $|a_{ii}|=1$ or $|b_{ii}|=1$ . Also note that both $a_{ii}$ and $b_{ii}$ cannot have modulus one, as $A^{*n}B^{* n}\to 0$ in SOT as $n\to \infty $ . Now the result follows by rearranging all the unimodular eigenvalues of $A^*$ and $B^*$ .