**2. Technical Results on Partitioned Hermitian Matrices, Cauchy's Interlacing Theorem and Stability**

The subsequent result relies on the conditions for the non-singularity of a partitioned Hermitian matrix of order (*n* + 1) which is built by aggregation from a principal Hemitian sub-matrix of order *n*. Mathematical proof is given in Appendix A.

**Lemma 1.** *Consider the partitioned matrix M* = *M m m*<sup>∗</sup> *d* + *d* ∈ **C**(*n*+<sup>1</sup>)×(*n*+<sup>1</sup>) *for any n* ∈ **Z**+*, where M*∈**C***<sup>n</sup>*×*<sup>n</sup> isHermitian,m*∈**C***<sup>n</sup> andd*,*d* ∈*R.Then,*

*(i) M is non-singular if and only if* det *M m m*<sup>∗</sup> *d* - −*d*det *M, equivalently, if and only if det M m m*<sup>∗</sup> 0 - −*det M* 0 *m*<sup>∗</sup> *d* + *d . M M*0

$$\begin{aligned} \text{(ii)} \quad & \text{Assume that } M > 0 \text{ and } d > 0. \text{ Then, } M\nu > 0 \text{ if and only if } \det\begin{bmatrix} M & m \\ m^\* & d \end{bmatrix} > -\det\begin{bmatrix} M & 0 \\ m^\* & d \end{bmatrix}. \text{ If } M > 0 \text{ and } d > 0, \\ & \text{and } d \ge 0 \text{ then } M\nu \le 0 \text{ and } d \le 0 \text{ then } M\nu \le 0 \text{ if } \det\begin{bmatrix} M & m \\ m^\* & d \end{bmatrix} \le 0. \end{aligned}$$

The subsequent result relies on some conditions which guarantee the boundedness of the determinant and eigenvalues of a recursive sequence of Hermitian matrices which were obtained and supported by Lemma 1 and Cauchy's interlacing theorem.

**Lemma 2.** *Consider the recursive sequence of Hermitian matrices* -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0 *for a given initial M*(*<sup>n</sup>*0) ∈ **C***<sup>n</sup>*0×*n*0 *for some given arbitrary n*0 ∈ **Z**+ *, where M*(*n*) ∈ *C<sup>n</sup>*<sup>×</sup>*n;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*, defined by M*(*n*+<sup>1</sup>) = *M*(*n*) *m*(*n*) *m*(*n*)<sup>∗</sup> *d*(*n*) + *d*(*n*) *;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *and assume that there is 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*) *, equivalently k*(*n*) *M* ≥ 1 *k*(*n*) *d* ε(*n*) √*m*(*n*)<sup>∗</sup>*m*(*n*)*;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *, where k*(*n*) *M* = <sup>λ</sup>*minM*(*n*) ≤ *K*(*n*) *M* = <sup>λ</sup>*maxM*(*n*)*; d*(*n*) ∈ *k*(*n*) *d* , *K*(*n*) *d ; d*(*n*) ∈ *k*(*n*) *d* , *K*(*n*) *d ;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+ *with K*(*n*) *d* ≥ *K*(*n*) *d and k*(*n*) *d* ≥ *k*(*n*) *d ;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*. Then, the following properties hold: (i)* lim sup *n*→∞ det*M*(*n*+<sup>1</sup>) − det*M*(*n*) ≤ 0*, detM*(*n*+<sup>1</sup>) ≤ *detM*(*n*) *for any given n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *if <sup>d</sup>*(*n*)*,* <sup>1</sup>−*K*(*n*)

*d* (*n*) *and m*(*n*) *satisfy the constraint K*(*n*) *d* ≤ *d* <sup>1</sup>+(<sup>2</sup>*n*+1−<sup>1</sup>)ε(*n*) *;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+ *with K*(*n*) *d* ≤ 1*, which becomes d*(*n*) ≤ <sup>1</sup>−*d*(*n*) <sup>1</sup>+(<sup>2</sup>*n*+1−<sup>1</sup>)ε(*n*) *;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+ *if d*(*n*) = *K*(*n*) *d* = *k*(*n*) *d* ≤ 1 *and d*(*n*) = *K*(*n*) *d* = *k*(*n*) *d ;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+*.*

*(ii) Assume that sp*≤*M*(*n*) = λ(*n*) 1 , λ(*n*) 2 , ... , λ(*n*) *n and that K*(*n*) *d* ≤ <sup>1</sup>−*K*(*n*) *d* <sup>1</sup>+(<sup>2</sup>*n*+1−<sup>1</sup>)ε(*n*) *with K*(*n*) *d* ≤ 1*,* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+*, for some given n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*. Then, the following relations hold:*

$$\left| \lambda\_{n+1}^{(n+1)} \right| \le \left| \frac{\prod\_{i=1}^{n} \lambda\_i^{(n)}}{\prod\_{i=1}^{n} \lambda\_i^{(n+1)}} \right| = \frac{\left| \det \mathcal{M}^{(n)} \right|}{\left| \prod\_{i=1}^{n} \lambda\_i^{(n+1)} \right|} \tag{1}$$

*n*→∞

*(c)* *If, furthermore, M*(*n*)0 *and M*(*n*+<sup>1</sup>)0 *for some n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ then

$$\lambda\_n^{(n)} \le \lambda\_{n+1}^{(n+1)} \le \frac{\prod\_{i=1}^n \lambda\_i^{(n)}}{\prod\_{i=1}^n \lambda\_i^{(n+1)}};\\1 \le \frac{\lambda\_{n+1}^{(n+1)}}{\lambda\_n^{(n)}} \le \frac{\prod\_{i=1}^{n-1} \lambda\_i^{(n)}}{\prod\_{i=1}^n \lambda\_i^{(n+1)}};\\\frac{\prod\_{i=2}^{n+1} \lambda\_i^{(n+1)}}{\prod\_{i=2}^n \lambda\_i^{(n)}} \le \frac{\lambda\_1^{(n)}}{\lambda\_1^{(n+1)}} \le 1 \qquad \text{(2)}$$

*(iii) Assume that the constraints of Property (ii) hold with <sup>M</sup>*(*<sup>n</sup>*0)0*;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+ *and, furthermore,* lim sup *n*→∞ *d*(*n*) + *d*(*n*) ≤ 1 *and m*(*n*) = min*o M*(*n*) , *<sup>o</sup><sup>d</sup>*(*n*)*, which is guaranteed if m*(*n*) = *minoK*(*n*) *M* , *oK*(*n*) *d* . Then, *M*(*n*)0 and *<sup>M</sup>*(*n*+<sup>1</sup>)0; <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ *Z*+, det *<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0 is bounded and the sequence *spM*(*n*)<sup>∞</sup>*n*=*n*0 is bounded, if *det M*(*<sup>n</sup>*0) is finite, and then lim sup *n*→∞ det *M*(*n*+<sup>1</sup>) − det *M*(*n*) ≤ 0.

### **Remark 1.** *Concerning Lemma 2 (i), we can focus on the following particular cases of interest A:*

*(a) m*(*n*) = 0 *and* det*M*(*n*+<sup>1</sup>) ≤ det*M*(*n*) *fails for all n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *and some <sup>n</sup>*0(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*. Then,* |det*M*(*n*+<sup>1</sup>)| = |det*M*(*n*)||*d*(*n*) + *d*(*n*)| > |det*M*(*n*)| *so that K*(*n*) *d* + *K*(*n*) *d* ≥ *d*(*n*) + *d*(*n*) > 1 *and* <sup>1</sup>−*K*(*n*) *d* <sup>1</sup>+(<sup>2</sup>*n*+1−<sup>1</sup>)ε(*n*) ≥ *K*(*n*) *d* > 1 − *K*(*n*) *d so that* 1 ≤ 1 + 2*n*+<sup>1</sup> − <sup>1</sup>ε(*n*) < 1*, a contradiction. Thus, one has* det*M*(*n*+<sup>1</sup>) ≤ det*M*(*n*) *for any n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *such that m*(*n*) = 0 *and also lim sup detM*(*n*+<sup>1</sup>) − *detM*(*n*) ≤ 0*.*

*(b) d*(*n*) + *d*(*n*) = 0 *and* det*M*(*n*+<sup>1</sup>) ≤ det*M*(*n*) *fails for all n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *and some <sup>n</sup>*0(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*. Note from the definition of the recursive sequence* -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0*that*

$$\begin{split} \det \mathbf{M}^{(n+1)} &= \det \begin{bmatrix} M^{(n)} & 0 \\ m^{(n)\*} & d^{(n)} \end{bmatrix} + \det \begin{bmatrix} M^{(n)} & m^{(n)} \\ m^{(n)\*} & \widetilde{d}^{(n)} \end{bmatrix} \\ = \mathbf{d}^{(n)} \det \mathbf{M}^{(n)} + \det \begin{bmatrix} M^{(n)} & 0 \\ 0 & \overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} I\_{n+1} + \begin{bmatrix} M^{(n)^{-1}} & 0 \\ 0 & 1/\overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} 0 & m^{(n)} \\ m^{(n)\*} & 0 \end{bmatrix} \end{bmatrix} \\ = -\widetilde{d}^{(n)} \det \mathbf{M}^{(n)} \begin{bmatrix} 1 - \det \begin{bmatrix} I\_{n+1} + \begin{bmatrix} M^{(n)^{-1}} & 0 \\ 0 & -1/\overline{d}^{(n)} \end{bmatrix} \begin{bmatrix} 0 & m^{(n)} \\ m^{(n)\*} & 0 \end{bmatrix} \end{bmatrix} \end{split} \tag{3}$$

$$\begin{split} \text{Sinc } & \frac{1}{k\_M^{(n)}} \sqrt{m^{(n)} m^{(n)}} \le \varepsilon^{(n)} \text{ and } \left\{ \varepsilon^{(n)} \right\}\_{n=n\_0}^{\infty} \to 0 \text{ then} \\ & \limsup\_{n \to \infty} \left( \left| \det M^{(n+1)} / \det M^{(n)} \right| - \left| \overleftarrow{d}^{(n)} \right| \left| 1 - \det \left( I\_{n+1} + \begin{bmatrix} 0 & M^{(n)^{-1}} m^{(n)} \\ - \left( 1/d^{(n)} \right) m^{(n)\*} & 0 \end{bmatrix} \right| \right| \right) \\ & = \limsup\_{n \to \infty} \left| \det M^{(n+1)} / \det M^{(n)} \right| = 0 \end{split}$$

*and* lim sup *n*→∞ det*M*(*n*+<sup>1</sup>) − det*M*(*n*) ≤ 0 *since, otherwise, if* lim sup *n*→∞ det*M*(*n*+<sup>1</sup>) − det*M*(*n*) > 0 *then* lim sup *n*→∞ det*M*(*n*+<sup>1</sup>)/det*M*(*n*) ∈ (−∞,<sup>−</sup><sup>1</sup>)<sup>∪</sup> (1,+∞)*, a contradiction to* lim sup *n*→∞ det*M*(*n*+<sup>1</sup>)/det*M*(*n*) = 0*. Note that if d*(*n*) = 0 *then d*(*n*) ≤ 1 *under the given constraints so that if m*(*n*) ∞*n*=*n*0 → 0*;* <sup>∀</sup>*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *and some n*0 ≥ *n*0 *then lim sup n*→∞ *detM*(*n*+<sup>1</sup>) − *detM*(*n*) ≤ 0*.*

Now, one gets from Lemma 2 [(ii), (iii)] the subsequent dual result concerning the recursion obtained from the inverse of *M*(*n*). The use of this result will make it possible to give sufficiency-type

conditions regarding the non-singularity of the recursive calculation for any positive integer *n*, and also as *n* tends to infinity.

**Lemma 3.** *For some given arbitrary n*0 ∈ **Z**+ *and all n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+*, define:*

$$\overline{\mathcal{M}}^{(n)} = \mathcal{M}^{(n)^{-1}} \left( I\_n + m^{(n)} \left( d^{(n)} + \widetilde{d}^{(n)} - m^{(n)\*} \mathcal{M}^{(n)^{-1}} m^{(n)} \right)^{-1} m^{(n)\*} \mathcal{M}^{(n)^{-1}} \right) \tag{4}$$

$$\overline{m}^{(n)} = -M^{(n)^{-1}}m^{(n)}\left(d^{(n)} + \widetilde{d}^{(n)} - m^{(n)\*}M^{(n)^{-1}}m^{(n)}\right)^{-1} \tag{5}$$

−

$$\overline{d}^{(n)} + \overline{d}^{(n)} = \left( d^{(n)} + \widetilde{d}^{(n)} - m^{(n)\*} \mathcal{M}^{(n)^{-1}} m^{(n)} \right)^{-1} \tag{6}$$

*and assume that:*

*(1) there is a real sequence* <sup>ε</sup>(*n*)<sup>∞</sup>*n*=*n*0 ⊂ [0, 1) *such that* 1 *k*(*n*) *M* (*n*)*k*(*n*) *d <sup>m</sup>*(*n*)<sup>∗</sup>*m*(*n*) ≤ ε(*n*)*;*∀*n*(<sup>≥</sup> *<sup>n</sup>*0) ∈ **Z**+ *, where*

$$
\overline{\mathcal{K}}\_{M}^{(n)} = \left| \lambda\_{\text{min}} \left( \overline{\mathcal{M}}^{(n)} \right) \right| \le \overline{\mathcal{K}}\_{M}^{(n)} = \left| \lambda\_{\text{max}} \left( \overline{\mathcal{M}}^{(n)} \right) \right|; \ \overline{d}^{(n)} \in \left[ \overline{k}\_{d}^{(n)}, \overline{\mathcal{K}}\_{d}^{(n)} \right];
$$

$$
\overline{\mathcal{d}}^{(n)} \in \left[ \overline{k}\_{\overline{d}^{-}}^{(n)}, \overline{\mathcal{K}}\_{\overline{d}}^{(n)} \right]; \ \forall n (\ge n\_{0}) \in \mathbb{Z}\_{+},
$$

$$
\overline{d}^{(n)} \ne m^{(n)\*} \mathcal{M}^{(n)^{-1}} m^{(n)} - d^{(n)}
$$

$$\begin{split} (2) \quad & \overline{\mathcal{M}}^{(n)} \succeq 0, \quad \limsup\_{n \to \infty} \Big( \overline{d}^{(n)} + \overline{d}^{(n)} \Big) \le 1 \quad \text{and} \overline{\mathcal{m}}^{(n)} = \min \Big( o \big( \|\overline{\mathcal{M}}^{(n)}\| \big), o \Big( \overline{d}^{(n)} \Big) \Big) \quad \text{which is guaranteed if } \overline{\mathcal{M}}^{(n)} \le \min \, \overline{\mathcal{M}}^{(n+1)}, \\ & \overline{\mathcal{m}}^{(n)} = \min \Big( o \big( \overline{\mathcal{K}}^{(n)}\_{\mathcal{M}} \big), o \Big( \overline{\mathcal{K}}^{(n)}\_{\mathcal{\overline{d}}} \big) \Big). \text{Then, } \overline{\mathcal{M}}^{(n)} \succeq 0 \text{ and } \overline{\mathcal{M}}^{(n+1)} \succeq 0, \forall n \ge n\_0 \in \mathbb{Z}\_+, \left\{ \det \overline{\mathcal{M}}^{(n)} \right\}\_{n=n\_0}^{\text{op}} \text{ is bounded,} \\ & \text{the sequence} \{ \operatorname{sp} \{ \overline{\mathcal{M}}^{(n)} \big) \Big\}\_{n=n\_0}^{\infty} \text{ is bounded,} \newline \text{if } \det \overline{\mathcal{M}}^{(n\_0)} \text{ is finite, and then } \limsup\_{n \to \infty} \{ \operatorname{det} \Big| \overline{\mathcal{M}}^{(n+1)} \big| - \det \left| \overline{\mathcal{M}}^{(n)} \right| \} \le 0. \end{split}$$

One gets by combining Lemma 2 and Lemma 3 the two subsequent direct results:

**Lemma 4.** *Assume that 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*+*.*

**Lemma 5.** *Assume that, for some finite n*0 ∈ *Z*+*, A*(*<sup>n</sup>*0) ∈ **C***<sup>n</sup>*0×*n*0 *is a stability matrix and construct a sequence* -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0*according to the recursive rule:*

$$\mathcal{M}^{(n+1)} = \mathcal{A}^{(n+1)\*} \mathcal{A}^{(n+1)} = \begin{bmatrix} M^{(n)} & m^{(n)} \\ m^{(n)\*} & d^{(n)} + \widetilde{d}^{(n)} \end{bmatrix} = \begin{bmatrix} A^{(n)\*} A^{(n)} & m^{(n)} \\ m^{(n)\*} & d^{(n)} + \widetilde{d}^{(n)} \end{bmatrix} ; \forall n \ge n\_0$$

*with initial condition M*(*<sup>n</sup>*0) = *A*(*<sup>n</sup>*0)∗*A*(*<sup>n</sup>*0) 0*. Assume also that* -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=*n*0 *and the sequence of its inverses satisfy the constraints of Lemma 2 [(ii),(iii)] and Lemma 3.*

*Then,* -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=*n*0*is a sequence of stability matrices.*

The above result can be directly extended for the case when *A*(*<sup>n</sup>*0) ∈ **C***<sup>n</sup>*0×*n*0 is antistable, that is, when all its eigenvalues have positive real parts and *M*(*<sup>n</sup>*0) 0. Then, by using similar arguments, as in the proof of Lemma 5 based on the continuity of the matrix eigenvalues with respect to its entries and supported by Lemmas 2,3, according to Cauchy's interlacing theorem, one concludes that -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=*n*0 consists of antistable members.

**Lemma 6.** *Lemma 5 holds "mutatis-mutandis" if A*(*<sup>n</sup>*0) ∈ *C<sup>n</sup>*0×*n*0 *is antistable.*

### **3. Some Extended Results Related to Sequences of Convergent Matrices**

In order to be able to adapt the above results to discrete dynamic systems, the well-known result that that the stability domain of a convergen<sup>t</sup> matrix (i.e., a "stable" discrete matrix) is the open unit circle of the complex plane centered at zero has to be taken into account. Note that, in particular, *A* ∈ **C***<sup>n</sup>*×*<sup>n</sup>* is convergen<sup>t</sup> if *sp*(*A*) ⊂ *C*1 = {*z* ∈ **C** : |*z*| < 1} so that {*Am*}∞0 → 0 as *m* → ∞. It turns out that convergen<sup>t</sup> matrices describe the stability property in the discrete sense. In other words, the solution of the discrete difference vector equation *zm*+1 = *Azm*, where *A* ∈ **C***<sup>n</sup>*×*n*, converges to 0 ∈ **R***n* for any given *z*0 ∈ **C***<sup>n</sup>* if and only if *A* ∈ **C***<sup>n</sup>*×*<sup>n</sup>* is convergent. The relevant results of Section 2 can be extended to this situation as follows, provided that *A* ∈ **C***<sup>n</sup>*×*<sup>n</sup>* is also Hermitian. Consider the following cases:

*Case a*: *sp*(*A*) ⊂ *C*1+ = {*z* ∈ **C** : 0 < *z* < 1} ⊂ *C*1 so that *<sup>A</sup>*(∈ **C***<sup>n</sup>*×*n*) 0. Then, it is convergen<sup>t</sup> if and only if (*In* − *A*) 0. The proof is direct since if (*In* − *A*) 0 then *x*<sup>∗</sup>*Ax* < *x*<sup>∗</sup>*x* for any *x*(- 0) ∈ **C***<sup>n</sup>*. Thus, by taking any <sup>λ</sup>(- 0) ∈ *sp*(*A*) of eigenvector *x* ∈ **C***<sup>n</sup>*, one determines that λ < 1 if λ - 0 and λ = 0 directly fulfills the constraint. This proves the sufficiency part. The "only if part" follows, since if (*In* − *A*) 0 fails, there is <sup>λ</sup>(- 0) ∈ *sp*(*A*) of eigenvector *x* ∈ **C***<sup>n</sup>* such that *x*<sup>∗</sup>*Ax* ≥ *x*<sup>∗</sup>*x* then λ ≥ 1 and *A* is not convergent.

*Case b*: *sp*(*A*) ⊂ *C*1− = {*z* ∈ **C** : −1 < *z* < 0} ⊂ *C*1 so that *A* (∈ **C***<sup>n</sup>*×*n*) ≺ 0. Then, it is convergen<sup>t</sup> if and only if (*In* + *A*) 0. The proof is direct, since if (*In* + *A*) 0, then *x*<sup>∗</sup>*Ax* > −*x*<sup>∗</sup>*<sup>x</sup>* for any *x*(- 0) ∈ **C***<sup>n</sup>*. Thus, by taking any <sup>λ</sup>(- 0) ∈ *sp*(*A*) of eigenvector *x* ∈ **C***<sup>n</sup>*, one determines that 0 >λ> −1. The remainder of the proof follows Case a closely.

*Case c*: *sp*(*A*) ⊂ *C*1 so that *<sup>A</sup>*<sup>2</sup>(∈ **C***<sup>n</sup>*×*n*) 0 so that *spA*<sup>2</sup>⊂ *C*1+. Then, it is convergen<sup>t</sup> if and only if *In* − *A*2 0 according to *Case a* by replacing *A* → *A*<sup>2</sup> . Note that *Case c* is included *Case a* and *Case b*.

Now, for *Case a*, replace *<sup>M</sup>*(*n*+<sup>1</sup>), defined in Lemma 2, by *In*+<sup>1</sup> − *M*(*n*+<sup>1</sup>) = ⎡⎢⎢⎢⎢⎣ *In* − *M*(*n*) −*<sup>m</sup>*(*n*) −*<sup>m</sup>*(*n*)<sup>∗</sup> 1 − *d*(*n*) + *d*(*n*) ⎤⎥⎥⎥⎥⎦ and it has to be guaranteed that if *M*(*n*) is Hermitian, then *In* − *M*(*n*) is also Hermitan, and *In* − *<sup>M</sup>*(*<sup>n</sup>*0) 0 for some *n*0 ∈ **Z**+ then *In*+<sup>1</sup> − *M*(*n*+<sup>1</sup>) 0; ∀*n* ≥ *n*0.

For *Case b*, replace *M*(*n*+<sup>1</sup>) → *In*+<sup>1</sup> + *M*(*n*+<sup>1</sup>) = *In* + *M*(*n*) *m*(*n*) *m*(*n*)<sup>∗</sup> 1 + *d*(*n*) + *d*(*n*) and it has to be guaranteed that if *M*(*n*) is Hermitian, then *In* + *M*(*n*) is also Hermitan, and *In* + *<sup>M</sup>*(*<sup>n</sup>*0) 0 for some *n*0 ∈ **Z**+ then *In*+<sup>1</sup> + *M*(*n*+<sup>1</sup>) 0; ∀*n* ≥ *n*0.

For *Case c*, note that *In* − *M*(*n*)<sup>2</sup> = *In* + *M*(*n*)*In* − *M*(*n*) so that

$$\left(I\_{\mathrm{Il}} - M^{(n)}\right)^2 = \begin{bmatrix} I\_n + M^{(n)} & m^{(n)} \\ m^{(n)\*} & 1 + d^{(n)} + \widetilde{d}^{(n)} \end{bmatrix} \begin{bmatrix} I\_{\mathrm{Il}} - M^{(n)} & -m^{(n)} \\ -m^{(n)\*} & 1 - \left(d^{(n)} + \widetilde{d}^{(n)}\right) \end{bmatrix}$$

then, replace

$$M^{(n+1)} \rightarrow \left(I\_{n+1} - M^{(n+1)}\right)^2 = \begin{bmatrix} I\_n - M^{(n)2} & -\left(\left(d^{(n)} + \widetilde{d}^{(n)}\right)I\_n + M^{(n)}\right)m^{(n)} \\ -m^{(n)\*}\left(\left(d^{(n)} + \widetilde{d}^{(n)}\right)I\_n + M^{(n)\*}\right) & 1 - \left(d^{(n)} + \widetilde{d}^{(n)}\right)^2 - m^{(n)\*}m^{(n)} \end{bmatrix}$$

and it has to be guaranteed that if *M*(*n*)<sup>2</sup> is Hermitian and *In* − *<sup>M</sup>*(*<sup>n</sup>*0)<sup>2</sup> 0 for some *n*0 ∈ **Z**+ then *In*+<sup>1</sup> − *M*(*n*+<sup>1</sup>)<sup>2</sup> 0; ∀*n* ≥ *n*0. Since *A* 0 is Hermitian, it is of the form *A* = *E*∗*E* for some full rank *n*-matrix *E*. Then, *A*<sup>2</sup> = (*E*∗*E*)<sup>2</sup> = *E*∗*EE*∗*E* = *A*∗*A*. If

*A* ≺ 0 then (−*<sup>A</sup>*) 0 and (−*<sup>A</sup>*)<sup>2</sup> = (*F*∗*F*)<sup>2</sup> = *F*∗*FF*∗*F* = (−*A*∗)(−*<sup>A</sup>*) = *A*∗*A* for some full rank *n*-matrix *F*. Then, Cases a and b can be dealt with using *Case c* by replacing *A*<sup>2</sup> → *A*∗*A*.

By taking advantage from the fact that a complex square matrix *A* is convergen<sup>t</sup> (i.e., stable in the discrete sense) if and only if the Hermitian matrix *A*∗*A* is convergent, we now build a sequence - *<sup>M</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> of Hermitian matrices as follows, in order to discuss the convergence of its members, provided that *M*(*<sup>n</sup>*0) is convergen<sup>t</sup> for some given *n*0 ∈ **Z**0+ or, with no loss in generality, provided that *M*(0) is convergent. Then,

$$M^{(n+1)} = \begin{bmatrix} M^{(n)} & \lambda^{(n)} m^{(n)} \\ \lambda^{(n)} m^{(n)\*} & \delta^{(n)} \end{bmatrix} = \begin{bmatrix} M^{(n)} & 0 \\ 0 & \delta^{(n)} \end{bmatrix} + \begin{bmatrix} 0 & \lambda^{(n)} m^{(n)} \\ \lambda^{(n)} m^{(n)\*} & 0 \end{bmatrix}; \forall n \in \mathbb{Z}\_{0+} \tag{7}$$

with <sup>λ</sup>(*n*)<sup>∞</sup>*n*=<sup>0</sup> ⊂ [0, 1).Then,

$$\begin{split} \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} \mathcal{M}^{(n)} & \lambda^{(n+1)}m^{(n)} \\ \lambda^{(n)}m^{(n)\*} & \delta^{(n)} \end{array} \right] \right| & \leq \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} \mathcal{M}^{(n)} & 0 \\ 0 & \delta^{(n)} \end{array} \right] \right| + \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} 0 & \lambda^{(n)}m^{(n)} \\ \lambda^{(n)}m^{(n)\*} & 0 \end{array} \right] \right| \\ & = \max\left( \left| \lambda\_{\text{max}} (\mathcal{M}^{(n)}) \right|, \left| \delta^{(n)} \right| \right) + \lambda^{(n)} \sqrt{m^{(n)\*}m^{(n)}} \\ & \leq \max\left( 1 - \varepsilon^{(n)}, \left| \delta^{(n)} \right| \right) + \lambda^{(n)} \sqrt{m^{(n)\*}m^{(n)}} < 1 - \varepsilon^{(n)}; \ \forall n \in \mathbb{Z}\_{0+} \end{split} \tag{8} \\ \end{split} \tag{9}$$

which holds if

$$m^{(n)\*}m^{(n)} < \frac{1}{\lambda^{(n)2}} \left(1 - \varepsilon^{(n)} - \max\left(1 - \varepsilon^{(n)}, \left|\delta^{(n)}\right|\right)\right)^2; \forall n \in \mathbf{Z}\_{0+} \tag{9}$$

or, *m*(*n*)2 < 1 λ(*n*)<sup>2</sup> 1 − ε(*n*) − max<sup>1</sup> − ε(*n*) , δ(*n*); ∀*n* ∈ **Z**0+, provided that ε(*n*+<sup>1</sup>) < ε(*n*); ∀*n* ∈ **Z**0+, that is - ε(*n*) ∞*<sup>n</sup>*=0 ⊂ [0, 1) is strictly decreasing, so - ε(*n*) ∞*<sup>n</sup>*=0 → 0, and δ(*n*) < 1 − ε(*n*+<sup>1</sup>); ∀*n* ∈ **Z**0+. 

Now, assume that the iterations to build -*<sup>M</sup>*(*n*+<sup>1</sup>)<sup>∞</sup>*n*=<sup>0</sup> do not add a new row and column to obtain *M*(*n*+<sup>1</sup>) from *M*(*n*) via the contribution of the members of an updating sequence *<sup>M</sup>*(*n*)<sup>∞</sup>*n*=0; ∀*n* ∈ **Z**0+ but a set of the, in general. Then, one may ge<sup>t</sup> that:

$$M^{(n+1)} = \begin{bmatrix} M^{(n)} & \lambda^{(n)} \overline{M}^{(n)} \\ \lambda^{(n)} \overline{M}^{(n)\*} & \Delta^{(n)} \end{bmatrix} = \begin{bmatrix} M^{(n)} & 0 \\ 0 & \Delta^{(n)} \end{bmatrix} + \begin{bmatrix} 0 & \lambda^{(n)} \overline{M}^{(n)} \\ \lambda^{(n)} \overline{M}^{(n)\*} & 0 \end{bmatrix}; \forall n \in \mathbb{Z}\_{0+} \tag{10}$$

so that

$$\begin{split} \left| \lambda\_{\text{max}} M^{(n+1)} \right| &= \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} M^{(n)} & \lambda^{(n)} \overline{M}^{(n)} \\ \lambda^{(n)} \overline{M}^{(n)\*} & \Delta^{(n)} \end{array} \right] \right| \leq \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} M^{(n)} & 0 \\ 0 & \Delta^{(n)} \end{array} \right] \right| + \left| \lambda\_{\text{max}} \left[ \begin{array}{cc} 0 & \lambda^{(n)} \overline{M}^{(n)} \\ \lambda^{(n)} \overline{M}^{(n)\*} & 0 \end{array} \right] \right| \\ &= \max \left| \left| \lambda\_{\text{max}} (M^{(n)}) \right|, \left| \lambda\_{\text{max}} (\Delta^{(n)}) \right| \right| + \lambda^{(n)} \lambda\_{\text{max}}^{1/2} (\overline{M}^{(n)\*} \overline{M}^{(n)}) \\ &\leq \max \left( 1 - \varepsilon^{(n)}, \left| \lambda\_{\text{max}} (\Delta^{(n)}) \right| \right) + \lambda^{(n)} \lambda\_{\text{max}}^{1/2} (\overline{M}^{(n)\*} \overline{M}^{(n)}) < 1 - \varepsilon^{(n+1)}; \forall n \in \mathbb{Z}\_{0+} \end{split} \tag{11}$$

which holds by complete induction if *M*(0) is convergen<sup>t</sup> and

$$\|\overline{M}^{(n)}\|\_{2}^{2} = \lambda\_{\max} \Big( \overline{M}^{(n)} \overline{M}^{(n)} \Big) < \frac{1}{\lambda^{(n)2}} \Big( 1 - \varepsilon^{(n+1)} - \max \{ 1 - \varepsilon^{(n)}, \left| \lambda\_{\max} \{ \Delta^{(n)} \} \right| \Big) \Big)^{2};\tag{12}$$
  $\forall n \in \mathbb{Z}\_{0+}$ 

or, *M*(*n*)2 < 1λ(*n*) 1 − ε(*n*+<sup>1</sup>) − max<sup>1</sup> − ε(*n*) , <sup>λ</sup>maxΔ(*n*); ∀*n* ∈ **Z**0+, provided that ε(*n*+<sup>1</sup>) < <sup>ε</sup>(*n*), that is ε(*n*) ∞*<sup>n</sup>*=0 ⊂ [0, 1) is strictly decreasing, so <sup>ε</sup>(*n*)<sup>∞</sup>*n*=<sup>0</sup> → 0, and <sup>λ</sup>maxΔ(*n*) < 1 − ε(*n*+<sup>1</sup>); ∀*n* ∈ **Z**0+. This implies that -*<sup>M</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup>is convergent.

In the particular case that for some λ ∈ [0, <sup>1</sup>), λ(*n*) = λ*<sup>n</sup>*; ∀*n* ∈ **Z**0+, such a λ is a forgetting factor of the iteration.

We now consider the matrix factorization *M*(*n*) = *<sup>A</sup>*(*n*)∗*A*(*n*); ∀*n* ∈ **Z**0+. By construction, *M*(*n*) is Hemitian (then square), even if *A*(*n*) is not square; ∀*n* ∈ **Z**0+. In the case when *A*(*n*) is not square, and since its order strictly increases as *n* increases, it is possible to consider the convergence of the sequence -*A*(*n*) ∞*<sup>n</sup>*=0 (without invoking the values of its eigenvalues) as the following property -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> is asymptotically convergen<sup>t</sup> if lim*m*→∞- *A*(*n*+ξ)−*A*(*n*) *m* ∞*<sup>n</sup>*=0 = 0 for any given ξ ∈ **Z**+. -*<sup>A</sup>*(*n*)<sup>∞</sup>*n*=<sup>0</sup> is convergen<sup>t</sup> if lim*m*→∞-*A*(*n*)*<sup>m</sup>* ∞*<sup>n</sup>*=0 = 0. The following related results are direct of simple proofs given in Appendix A:

**Lemma 7.** *If* -*A*(*n*) ∞*<sup>n</sup>*=0 *is convergent then it is asymptotically convergent. The inverse is, in general, not true.*

**Lemma 8.** *Assume that M*(*n*) = *<sup>A</sup>*(*n*)∗*A*(*n*)*;* ∀*n* ∈ *Z*0+ *is a complex square matrix of any arbitrary order. Then:*


*(iii) For any n* ∈ **Z**0+*, A*(*n*)2 < 1 *if and only if M*(*n*)2 < 1*.* -*A*(*n*) ∞*<sup>n</sup>*=0 *is convergent if and only if* -*M*(*n*) ∞*<sup>n</sup>*=0*is convergent.*
