**Proof of Lemma 2.** First, note that

$$|k\_d^{(n)} - k\_{\overline{d}}^{(n)}| \le \|d^{(n)}| - |\overline{d}^{(n)}| \le |d^{(n)} + \overline{d}^{(n)}| \le K\_d^{(n)} + K\_{\overline{d}}^{(n)}; \ \forall n (\ni n\_0) \in \mathbf{Z}\_+.$$

Thus,

$$\begin{split} \text{det}\mathcal{M}^{(n+1)} &= \text{det}\begin{bmatrix} M^{(n)} & 0 \\ m^{(n)\*} & d^{(n)} \end{bmatrix} + \text{det}\begin{bmatrix} M^{(n)} & m^{(n)} \\ m^{(n)\*} & \overline{d}^{(n)} \end{bmatrix} \\ = d^{(n)}\text{det}\mathcal{M}^{(n)} + \text{det}\left( \begin{bmatrix} M^{(n)} & 0 \\ 0 & \overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} I\_{n+1} + \begin{bmatrix} M^{(n)^{-1}} & 0 \\ 0 & 1/\overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} 0 & m^{(n)} \\ m^{(n)\*} & 0 \end{bmatrix} \right) \right) \\ = d^{(n)}\text{det}\mathcal{M}^{(n)} + \overline{d}^{(n)}\text{det}\mathcal{M}^{(n)}\text{det}\left(\overline{I}\_{n+1} + \begin{bmatrix} M^{(n)^{-1}} & 0 \\ 0 & 1/\overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} 0 & m^{(n)} \\ m^{(n)\*} & 0 \end{bmatrix} \right) \end{split} \tag{A1}$$
 
$$\forall n (\succeq n\_0) \in \mathbb{Z}\_+$$

if *M*(*n*)−<sup>1</sup> ∞ *n* = *n*0 exists so that det*M*(*n*) - 0, that is, if 0 < *k*(*n*) *M* = <sup>λ</sup>min*M*(*n*) ≤ *K*(*n*) *M* = <sup>λ</sup>max*M*(*n*) < +∞; <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+. Thus,

$$\begin{split} \left| \det \mathbf{M}^{(n+1)} \right| \leq \left| d^{(n)} \det \mathbf{M}^{(n)} \right| + \left| \overline{d}^{(n)} \det \mathbf{M}^{(n)} \right| \left| \det \left( I\_{n+1} + \begin{bmatrix} M^{(n)^{-1}} & 0 \\ 0 & 1/\sqrt{n} \end{bmatrix} \begin{bmatrix} 0 & m^{(n)} \\ m^{(n)\*} & 0 \end{bmatrix} \right| \right| \\ \leq & n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} + n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} \left( 1 + \frac{1}{\mathbf{K}\_{M}^{(n)} \mathbf{r}^{(n)}} \sqrt{m^{(n)}} m^{(n)} \right)^{n+1} \\ \leq n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} + n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} \left( 1 + \varepsilon^{(n)} \right)^{n+1} \leq n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} + n \mathbf{K}\_{d}^{(n)} K\_{M}^{(n)} \left( 1 + \left( 2^{n+1} - 1 \right) \varepsilon^{(n)} \right); \\ \forall n (\geq n\_{0}) \in \mathbf{Z}\_{+} \end{split} \tag{A2}$$

since <sup>λ</sup>max*m*(*n*)<sup>∗</sup>*m*(*n*) = *m*(*n*)<sup>∗</sup>*m*(*n*); <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+. Since it is assumed the existence of a real sequence <sup>ε</sup>(*n*)<sup>∞</sup>*n*=*n*0 ⊂ [0, 1) such that 1 *k*(*n*) *M k*(*n*) *d* √*m*(*n*)<sup>∗</sup>*m*(*n*) ≤ ε(*n*); <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ and since

$$\begin{aligned} \left(1+\varepsilon^{(n)}\right)^{n+1} = \sum\_{i=0}^{n+1} \binom{n+1}{i} \varepsilon^{(n)^i} &= 1 + \sum\_{i=1}^{n+1} \binom{n+1}{i} \varepsilon^{(n)^i} \le 1 + \binom{n+1}{2} \varepsilon^{(n)};\\ \forall n (\ge n\_0) &\in \mathbb{Z}\_+ \end{aligned} \tag{A3}$$

thus, it follows that det*M*(*n*+<sup>1</sup>) ≤ (*n* + 1)*K*(*n*+<sup>1</sup>) *M* ≤ *nK*(*n*) *M* holds for any given *n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ if

$$n\mathbb{K}\_d^{(n)}\mathbb{K}\_M^{(n)} + n\mathbb{K}\_d^{(n)}\mathbb{K}\_M^{(n)} \left(1 + \left(2^{n+1} - 1\right)\varepsilon^{(n)}\right) \le n\mathbb{K}\_M^{(n)}$$

for any given *n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+, that is, if

$$n\kappa \mathsf{K}\_{\mathcal{M}}^{(n)} \left( \mathsf{K}\_{d}^{(n)} - 1 \right) + n\mathsf{K}\_{d}^{(n)} \mathsf{K}\_{\mathcal{M}}^{(n)} \left( 1 + \left( 2^{n+1} - 1 \right) \varepsilon^{(n)} \right) \le 0$$

for such *n* ∈ **Z**+, equivalently, if *nK*(*n*) *d* 1 + 2*n*+<sup>1</sup> − <sup>1</sup>ε(*n*) ≤ *n*1 − *K*(*n*) *d* , equivalently if *K*(*n*) *d* ≤ 1, and furthermore, and accordingly to the restriction on the sequence <sup>ε</sup>(*n*)<sup>∞</sup>*n*=*n*0 , if

$$K\_{\vec{d}}^{(n)} \le \frac{\left(1 - K\_{\vec{d}}^{(n)}\right)}{1 + (2^{n+1} - 1)\varepsilon^{(n)}} \le \frac{\left(1 - K\_{\vec{d}}^{(n)}\right) k\_{\vec{M}}^{(n)} k\_{\vec{d}}^{(n)}}{k\_{\vec{M}}^{(n)} k\_{\vec{d}}^{(n)} + (2^{n+1} - 1)\sqrt{\lambda\_{\text{max}} \left(m^{(n)\*} m^{(n)}\right)}}\tag{A4}$$

for such an *n* ∈ **Z**0+. In particular, if *d*(*n*) = *K*(*n*) *d* = *k*(*n*) *d* ≤ 1 and *d*(*n*) = *K*(*n*) *d* = *k*(*n*) *d* for some given *n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**0+, then (A4) holds if

$$\left| \overrightarrow{d}^{(n)} \right| \le \frac{\left( 1 - \left| d^{(n)} \right| \right)}{1 + \left( 2^{n+1} - 1 \right) \varepsilon^{(n)}} \le \frac{\left( 1 - \left| d^{(n)} \right| \right) k\_M^{(n)} \left| d^{(n)} \right|}{k\_M^{(n)} \left| d^{(n)} \right| + \left( 2^{n+1} - 1 \right) \sqrt{\lambda\_{\text{max}} \left( m^{(n)\*} m^{(n)} \right)}} \tag{A5}$$

and Property (i) has been proved. On the other hand, since Property (ii) assumes that the constraints of Property (i) hold the constraint (1) is a direct consequence of Property (i). Also, if *M*(*n*)0 and *M*(*n*+<sup>1</sup>)0 then the constraints (2) follow directly from (1) and Cauchy's interlacing theorem of the eigenvalues which are real non-negative. Property (ii) has been proved. Property (iii) follows since lim sup *n*→∞ det *M*(*n*+<sup>1</sup>) − det *M*(*n*) ≤ 0 directly from (A1) and the given assumptions. -

**Proof of Lemma 3.** By the inversion of a block partitioned matrix, one gets:

=

$$\boldsymbol{M}^{(n+1)^{-1}} = \begin{bmatrix} \boldsymbol{M}^{(n)} & \boldsymbol{m}^{(n)} \\ \boldsymbol{m}^{(n)\*} & \boldsymbol{d}^{(n)} + \overleftarrow{\boldsymbol{d}}^{(n)} \end{bmatrix}^{-1} = \begin{bmatrix} \overline{\boldsymbol{M}}^{(n)} & \overline{\boldsymbol{m}}^{(n)} \\ \overline{\boldsymbol{m}}^{(n)\*} & \overline{\boldsymbol{d}}^{(n)} + \overline{\boldsymbol{d}}^{(n)} \end{bmatrix} \tag{A6}$$

where

$$\overline{M}^{(n)} = M^{(n)} \left( l\_n + m^{(n)} \left( d^{(n)} + \widetilde{d}^{(n)} - m^{(n)\*} M^{(n)^{-1}} m^{(n)} \right)^{-1} m^{(n)\*} M^{(n)^{-1}} \right) \tag{A7}$$

$$
\overline{m}^{(n)} = -M^{(n)^{-1}}m^{(n)}\left(d^{(n)} + \overline{d}^{(n)} - m^{(n)}\mathcal{M}^{(n)^{-1}}m^{(n)}\right)^{-1} \tag{A8}
$$

$$
\overline{d}^{(n)} + \overline{d}^{(n)} = \left(d^{(n)} + \overline{d}^{(n)} - m^{(n)}\mathcal{M}^{(n)^{-1}}m^{(n)}\right)^{-1}
$$

since *M*(*n*) is non-singular and *d*(*n*) - *m*(*n*)<sup>∗</sup>*M*(*n*)−1*m*(*n*) − *d*(*n*) (i.e., *d*(*n*) + *d*(*n*) − *m*(*n*)<sup>∗</sup>*M*(*n*)−1*m*(*n*) is non-singular). The proof follows directly from Lemma 2 [(ii)-(iii)] by replacing *M*(*n*+<sup>1</sup>) → *M*(*n*+<sup>1</sup>) = *M*(*n*+<sup>1</sup>)−<sup>1</sup> ; <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+. -

−

**Proof of Lemma 4.** Since *M*(*<sup>n</sup>*0) 0 then its inverse is also positive definite and then both of them fulfil the positive semi-definiteness constraint of Lemma 2 [(ii),(iii)] and Lemma 3. The proof follows directly since sup *<sup>n</sup>*≥*n*0 maxdet*M*(*n*) , det*M*(*n*)−<sup>1</sup> < +∞ and lim sup *n*→∞ supmaxdet*M*(*n*) , det*M*(*n*)−<sup>1</sup> < +∞ then *M*(*n*) 0 for *n* ≥ *n*0 if *M*(*<sup>n</sup>*0) 0, lim sup *n*→∞ *M*(*n*) 0 and lim inf *n*→∞ *M*(*n*) 0. -

**Proof of Lemma 5.** Note that *M*(*<sup>n</sup>*0) = *A*(*<sup>n</sup>*0)∗*A*(*<sup>n</sup>*0) 0 since *A*(*<sup>n</sup>*0) is a stability matrix. From Lemma 2 and Lemma 3, the sequence -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0 consists of positive definite members. Thus, the singular values of the elements of the sequence -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=*n*0 are positive and bounded. Since *A*(*<sup>n</sup>*0) is stable and since the eigenvalues of any square matrix are continuous functions of its entries there is no zero eigenvalue in any member of the sequence -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=*n*0 . Any member of this sequence has no eigenvalues at the imaginary complex axis other than zero (i.e., any nonzero critically stable eigenvalues) since then the corresponding *M*(*n*) = *A*(*n*)∗*A*(*n*) is not positive definite contradicting the given assumption. As a result, no member of the sequence -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=*n*0 has a critically stable eigenvalue (i.e., located on the imaginary complex axis) or unstable eigenvalue (i.e., located on the complex open right half plane).

*M*(*<sup>n</sup>*0) 0 for some given arbitrary *n*0 ∈ **Z**+ and assume also that the conditions of Lemma 2 (iii) and Lemma 3 hold. Then, *M*(*n*+<sup>1</sup>) 0; <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+. -

**Proof of Lemma 7.** It is direct since lim *m*→∞-*A*(*n*)*m* ∞*<sup>n</sup>*=0 = 0 implies that lim*m*→∞-*A*(*n*+ξ)−*A*(*n*)*<sup>m</sup>* ∞*<sup>n</sup>*=0 = 0 for any given ξ ∈ **Z**+. -

**Proof of Lemma 8.** If for any given *n* ∈ **Z**0+, *A*(*n*)2 < 1 then lim*m*→∞*A*(*n*)*<sup>m</sup>* = 0 for any *n* ∈ **Z**0+, then lim *m*→∞-*A*(*n*)*m* ∞*<sup>n</sup>*=0 = 0, *M*(*n*)22 = *A*(*n*)∗*A*(*n*)22 ≤ *A*(*n*)42 < 1 and then *M*(*n*)2 < 1, lim*m*→∞*M*(*n*)*<sup>m</sup>* = 0 and lim *m*→∞-*M*(*n*)*m* ∞*<sup>n</sup>*=0 = 0 for any given *n* ∈ **Z**0+. Thus, if -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> is convergen<sup>t</sup> then -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> is convergent, hence Property (i) holds. On the other hand, since *M*(*n*) is semidefinite positive Hermitian by construction; ∀*n* ∈ **Z**0+ then the condition *M*(*n*)2 < 1; ∀*n* ∈ **Z**0+ leads to the convergence of - *M*(*n*) ∞*<sup>n</sup>*=1, and if *M*(*n*)2 < 1 for some *n* ∈ **Z**0+ then

$$\tau^2(M^{(n)}) = \lambda\_{\max}(M^{(n)\*}M^{(n)}) = \lambda\_{\max}^2\left(A^{(n)\*}A^{(n)}\right) = \lambda\_{\max}^2\left(M^{(n)}\right) = \left\|M^{(n)}\right\|\_2^2 < 1$$

Then, *A*(*n*)2 = λ 1/2 max*A*(*n*)∗*A*(*n*) = *M*(*n*) 1/2 2 < 1. Thus, if the above holds for any *n* ∈ **Z**0+, one concludes that -*A*(*n*) ∞*<sup>n</sup>*=0 is convergen<sup>t</sup> if -*M*(*n*) ∞*<sup>n</sup>*=0 is convergent. Hence, Property (ii) follows. Property (iii) is a combination of the other two properties since for any *n* ∈ **Z**0+, *M*(*n*)2 < 1, *A*(*n*)2 < 1 implies that *M*(*n*)2 < 1 and *M*(*n*)2 < 1 implies that *A*(*n*)2 < 1. -

**Proof of Theorem 1.** Properties (i),(ii) follows from Lemma 7and Lemma 8 by taking into account the factorization (28). Property (iii) follows from Property (i) and (20) since

$$\|\|\overline{\mathbf{x}}^{(n+m)}\|\| \le \left(\prod\_{i=0}^{m-1} \|\overline{A}^{(n-i)}\|\|\|\overline{\mathbf{x}}^{(n)}\|\| \le \max\_{0 \le i \le m-1} \|\overline{A}^{(n-i)}\|\|\|\overline{\mathbf{x}}^{(n)}\|\|; \ \forall n \in \mathbb{Z}\_{0+\tau} \forall m \in \mathbb{Z}\_{+}\tag{A9}$$

and then *x*(*n*+*m*) → 0 as *m* → ∞ for any *n* ∈ **Z**0+. Property (iv) is proved in the same way as Property (iii) via Property (ii). The proof of Property (v) is made by comparing (28) with (10)–(12) by replacing λ(*n*)*M*(*n*) → *A*(*n*)*<sup>T</sup> B*(*n*) and Δ(*n*) → *B*(*n*)*<sup>T</sup> B*(*n*) . One gets, via complete induction, that if *M*(0) is convergen<sup>t</sup> and, furthermore,

$$\begin{split} \lambda \lambda\_{\max} \Big( \mathcal{B}^{(n)^{\top}} A^{(n)} A^{(n)^{\top}} B^{(n)} \Big) &< \left( 1 - \varepsilon^{(n+1)} - \max \{ 1 - \varepsilon^{(n)}, \left. \left| \lambda\_{\max} \left( B^{(n)^{\top}} B^{(n)} \right) \right| \right| \right)^{2}; \\ &\forall n \in \mathbb{Z}\_{0+} \end{split} \tag{A10}$$

Then, -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> is convergent, since *M*(0) is convergent, provided that ε(*n*+<sup>1</sup>) < <sup>ε</sup>(*n*), that is <sup>ε</sup>(*n*)<sup>∞</sup>*n*=<sup>0</sup> ⊂ [0, 1) is strictly decreasing, and

$$\left| \lambda\_{\max} \left( B^{(n)^T} B^{(n)} \right) \right| < 1 - \varepsilon^{(n+1)}; \ \forall n \in \mathbf{Z}\_{0+} \tag{A11}$$

The constrains (A10), (A11) are jointly fulfilled if <sup>λ</sup>max*B*(*n*)*<sup>T</sup> B*(*n*) < min1 − ε(*n*+<sup>1</sup>) , ε(*n*) − ε(*n*+<sup>1</sup>); ∀*n* ∈ **Z**0+.-

**Proof of Lemma 9.** Since *Am* = *EA* = *IEA* then *vec* (*EA*) = *vec*(*IEA*) = *I* ⊗ *A<sup>T</sup> vec*(*E*), and (32) becomes for I being the identity matrix of the same order as *E* and *A*:

$$\text{rect}\left(B \otimes \mathbb{C}^{\text{T}}\right) \text{vec}\left(K\right) = \text{vec}\left((E - I)A\right) = \left(I \otimes A^{\text{T}}\right) \text{vec}\left(E\right) - \text{vec}\left(A\right) \tag{A12}$$

whose solutions are given by (34) if (33) holds and the whole set of solutions reduces to (35) if the solution is unique. The corresponding Equations (36)–(38) are go<sup>t</sup> for the particular case when *Am* = ρ*A* with ρ < 1/*A*2. Property (i) has been proved. Property (ii) follows since the least-squares best approximation to the corresponding incompatible algebraic system (31), or (32), is (40), that is, (34) for *kw* = 0, [13,14]. -

**Proof of Theorem 2.** The solvability of (41) in the form (42) follows from the Rouché-Froebenius rank conditions from the algebraic compatibility under Assumptions 1,2. By defining

$$A^{0(n+1)} = \begin{bmatrix} A^{(n)} \ 0\_{(\Sigma\_{i=0}^{n+1}{}\_{n}) \times n\_{n+1}} \end{bmatrix}; \; B^{0(n+1)} = B^{(n)}$$

in order to complete a square "a priori" matrix of dynamics of the (*n* + 1)-the aggregated system obtained after the aggregation of the (*n* + 1)-th subsystem, note that

$$A^{0(n+1)} = \begin{bmatrix} A^{0(n)} + \mathcal{B}\_0^{(n)} \mathcal{K}\_0^{(n)} & A^{0(n)} & 0\\ \hat{A}^{(n+1)} + \mathcal{B}\_0^{(n)} \mathcal{K}\_0^{a(n)} \mathcal{C}^{(n)} & \mathcal{D}^{(n)} + \widetilde{\mathcal{D}}^{(n)} + \mathcal{B}\_0^{(n)} \mathcal{K}\_0^{(n)} & 0 \end{bmatrix} \in \mathbb{R}^{(\sum\_{i=0}^{n+1} n\_i) \times (\sum\_{i=0}^{n+1} n\_i)} \tag{A13}$$

$$\boldsymbol{B}^{\boldsymbol{\theta}(n+1)} = \begin{bmatrix} \left(\boldsymbol{\mathcal{B}}\_{1}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{1}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-1)} & \boldsymbol{0} & \left(\boldsymbol{\mathcal{B}}\_{j}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{r}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-r)} & \boldsymbol{0} \\ \left(\boldsymbol{\mathcal{B}}\_{1}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{1}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-1)} & \left(\boldsymbol{\mathcal{B}}\_{1}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{1}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-1)} & \left(\boldsymbol{\mathcal{B}}\_{r}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{r}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-r)} & \left(\boldsymbol{\mathcal{B}}\_{j}^{(n)} + \boldsymbol{\mathcal{B}}\_{0}^{(n)}\boldsymbol{\mathcal{K}}\_{r}^{(n)}\right)\boldsymbol{\mathsf{C}}^{(n-r)} \end{bmatrix} \tag{A14}$$

∀*n* ∈ **Z**0+. From (41), with corresponding associated controller explicit solutions (42), one gets that the (*n* + 1)-th aggregated delay-free dynamics is described by the matrix:

$$A^{\mathbf{0}(n+1)} = \begin{bmatrix} A^{(n)} \ 0\_{(\Sigma\_{i=0}^{n+1}n)\times n\_{n+1}} \end{bmatrix} = \begin{bmatrix} A\_f^{(n)} & A^{\mathbf{0}(n)} & \mathbf{0} \\ A\_f^{\mathbf{a}(n+1)} & A\_f^{(n)} & \mathbf{0} \end{bmatrix} \in \mathbf{R}^{(\Sigma\_{i=0}^{n+1}n\_i)\times(\Sigma\_{i=0}^{n+1}n\_i)}; \ \forall n \in \mathbf{Z}\_{0+} \tag{A15}$$

Having in mind (27), construct

*M*(*n*+<sup>1</sup>) = *A*(*n*) *A*(*n*)*<sup>T</sup>* = ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*ˆ0(*n*) *A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *f A*<sup>ˆ</sup>(*n*) *f* 0 0 *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) 2 *i B* <sup>0</sup>(*n*) *I* 0 ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*ˆ0(*n*) *A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *f A*<sup>ˆ</sup>(*n*) *f* 0 0 *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) 2 *i B* <sup>0</sup>(*n*) *I* 0 ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ *T* = ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*(*n*) *T f* + *A*ˆ0(*n*)*A*ˆ0(*n*)*<sup>T</sup>* + *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) *T* 1 *A*(*n*) *f A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *T f* + *A*ˆ0(*n*)*A*<sup>ˆ</sup>(*n*) *T f* + *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) *T* 2 ⎡ ⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*ˆ0(*n*) *A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *f A*<sup>ˆ</sup>(*n*) *f* ⎤ ⎥⎥⎥⎥⎥⎥⎦ ⎡ ⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*ˆ0(*n*) *A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *f A*<sup>ˆ</sup>(*n*) *f* ⎤ ⎥⎥⎥⎥⎥⎥⎦ *T i B* <sup>0</sup>(*n*) *I* ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ = ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*(*n*) *T f* + *A*ˆ0(*n*)*A*ˆ0(*n*)*<sup>T</sup>* + *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) *T* 1 *A*(*n*) *f A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *T f* + *A*ˆ0(*n*)*A*<sup>ˆ</sup>(*n*) *T f* + *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) *T* 2 ⎡ ⎢⎢⎢⎢⎢⎢⎣ *A*(*n*) *f A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *f* ⎤ ⎥⎥⎥⎥⎥⎥⎦ ⎡ ⎢⎢⎢⎢⎢⎣ *A*ˆ0(*n*) *A*<sup>ˆ</sup>(*n*) *f* ⎤ ⎥⎥⎥⎥⎥⎦ *A*(*n*) *T f A*<sup>ˆ</sup>*<sup>a</sup>*(*n*+<sup>1</sup>) *T f A*ˆ0(*n*) *T f A*<sup>ˆ</sup>(*n*) *T f i B* <sup>0</sup>(*n*) δ(*n*)*I* ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ +*i B* <sup>0</sup>(*n*) ⎡ ⎢⎢⎢⎢⎣ 0 0 0 1 − δ(*n*) *I* ⎤ ⎥⎥⎥⎥⎦

where

*B* <sup>0</sup>(*n*) = ⎡ ⎢⎢⎢⎢⎢⎣ *B* <sup>0</sup>(*n*) 1 *B* <sup>0</sup>(*n*) 2 ⎤ ⎥⎥⎥⎥⎥⎦ = ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎣ *B*(*n*) 1 + *B*(*n*) 0 *K*(*n*) 1 *C*(*<sup>n</sup>*−<sup>1</sup>) 0 *B*ˆ *a*(*n*) 1 + *B*<sup>ˆ</sup>(*n*) 0 *K*ˆ *a*(*n*) 1 *C*(*<sup>n</sup>*−<sup>1</sup>) *B*<sup>ˆ</sup>(*n*) 1 + *B*<sup>ˆ</sup>(*n*) 0 *K*<sup>ˆ</sup>(*n*) 1 *C*<sup>ˆ</sup>(*<sup>n</sup>*−<sup>1</sup>) ··· *B*(*n*) *r* + *B*(*n*) 0 *K*(*n*) *r C*(*<sup>n</sup>*−*<sup>r</sup>*) 0 *B*ˆ *a*(*n*) *r* + *B*<sup>ˆ</sup>(*n*) 0 *K*ˆ *a*(*n*) *r C*(*<sup>n</sup>*−*<sup>r</sup>*) *B*<sup>ˆ</sup>(*n*) *r* + *B*<sup>ˆ</sup>(*n*) 0 *K*<sup>ˆ</sup>(*n*) *r C*<sup>ˆ</sup>(*<sup>n</sup>*−*<sup>r</sup>*) ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎦ (A16)

and

(a) *i B* <sup>0</sup>(*n*) is a binary indicator function defined as *i B* <sup>0</sup>(*n*) = 1 if *B* <sup>0</sup>(*n*) - 0 and *i B* <sup>0</sup>(*n*) = 0 if *B* <sup>0</sup>(*n*) = 0. The reason of the use of this indicator is that, in fact, if the delayed dynamics is zero then the dimension of the extended state, so that of *<sup>A</sup>*(*n*), decreases since the resulting block identity matrices are removed, 

$$
\begin{aligned}
\text{(b)} \quad \left\| \delta^{(n)} \right\|\_{n=0}^{\infty} &\subset \left[ 0, 1 \right] \text{ is a design sequence which satisfies } \left\{ \delta^{(n)} \right\}\_{n=0}^{\infty} \to 0. \\
\text{Note that the fact that } I = i \left( \overline{\mathcal{B}}^{0(n)} \right) I = i \left( \overline{\mathcal{B}}^{0(n)} \right) \delta^{(n)} I + i \left( \overline{\mathcal{B}}^{0(n)} \right) \left( 1 - \delta^{(n)} \right) I \text{ justifies (A16).}
\end{aligned}
$$
