In the case of n is a scalar, then

Proof:

The (i, j)^th block of A ⊗ (nB) is

$\begin{array}{c}\left[{\text{a}}_{\text{ij}}\left(\text{nB}\right)\right]=\text{n}\left[{\text{a}}_{\text{ij}}\text{B}\right]\\ =\text{n}\left[{\left(\text{i,j}\right)}^{\text{th}}\text{block}\text{of}\text{A}\otimes \text{B}\right]\end{array}$

In the case of matrix summation associated with Kronecker multiplication, the following are the results obtained (Shakeri et al., 2016; Mariet et al., 2016; Kepner et al., 2018):

Proof:

The product of a number of matrices is presented as (Shakeri et al., 2016; Mariet et al., 2016; Kepner et al., 2018):

The mixed product of a number of matrices is presented as (Shakeri et al., 2016; Mariet et al., 2016; Kepner et al., 2018):

Consider the two masteries A(n × n) and B(m × m), then the following (Shakeri et al., 2016; Mariet et al., 2016; Kepner et al., 2018), the Kronecker sum can be defined by the following expression:

where:

As stated earlier, the Kronecker multiplication is a powerful tool to build the transition rate matrix. Therefore, the Kronecker algebra can easily present any large system by combining all the sub-systems under it. Then, using Kronecker algebra in system building, a Matlab function found to use Kronecker algebra. Furthermore, this function is not within the Matlab program yet (Shakeri et al., 2016; Mariet et al., 2016; Kepner et al., 2018).

As mentioned earlier, the Kronecker sum is:

where:

In the system building function, the size of each sub-system is 2 × 2, this came from the equation

$\begin{array}{c}\frac{d{P}_1\left(t\right)}{dt}=-\lambda P_1\left(\text{t}\right)+\mu P_2\left(\text{t}\right)\\ \frac{d{P}_2\left(t\right)}{dt}=+\lambda P_1\left(\text{t}\right)-\mu P_2\left(\text{t}\right)\end{array}$

So each sub-system matrix will be as of:

$\begin{array}{c}{U}_n=\left[\begin{array}{cc}-{\lambda }_n& {\mu }_n\\ {\lambda }_n& -{\mu }_n\end{array}\right]\end{array}$

The size of the unit matrix will be 2 × 2. The overall size of the transition rate matrix can be calculated from the number of sub-systems (n) as follows:

Transition rate matrix = 2ⁿ.

Let a system contain two sub-systems A and B. Then:

Transition rate matrix = A ⊗ B = A ⊗ I_B + I_A ⊗ B

$\begin{array}{c}\begin{array}{cc}\text{A}=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\text{and}& \text{B}=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\end{array}\end{array}$

Then,

$\begin{array}{c}\text{A}\otimes {\text{I}}_{\text{B}}=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

and

$\begin{array}{c}{\text{I}}_{\text{A}}\otimes \text{B}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\\ \text{A}\otimes {\text{I}}_{\text{B}}=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{a}_{11}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& a_{12}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ a_{21}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& a_{22}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}\right]\\ {\text{I}}_{\text{A}}\otimes \text{B}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cc}1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\end{array}\right]\\ \left[\begin{array}{cc}{a}_{11}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& a_{12}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ a_{21}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& a_{22}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}\right]=\left[\begin{array}{cccc}{a}_{11}& 0& a_{12}& 0\\ 0& a_{11}& 0& a_{12}\\ a_{21}& 0& a_{22}& 0\\ 0& a_{21}& 0& a_{22}\end{array}\right]\\ \left[\begin{array}{cc}1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\end{array}\right]=\left[\begin{array}{cccc}{b}_{11}& b_{12}& 0& 0\\ b_{21}& b_{22}& 0& 0\\ 0& 0& b_{11}& b_{12}\\ 0& 0& b_{21}& b_{12}\end{array}\right]\end{array}$

Then

$\begin{array}{c}\text{A}\otimes \text{B}=\text{A}\otimes {\text{I}}_{\text{B}}+{\text{I}}_{\text{A}}\otimes \text{B}=\left[\begin{array}{cccc}{a}_{11}& 0& a_{12}& 0\\ 0& a_{11}& 0& a_{12}\\ a_{21}& 0& a_{22}& 0\\ 0& a_{21}& 0& a_{22}\end{array}\right]+\left[\begin{array}{cccc}{b}_{11}& b_{12}& 0& 0\\ b_{21}& b_{22}& 0& 0\\ 0& 0& b_{11}& b_{12}\\ 0& 0& b_{21}& b_{22}\end{array}\right]\end{array}$

Then the transition rate matrix for system containing two subsystems A and B are:

$\begin{array}{c}\text{A}\otimes \text{B}=\text{A}\otimes {\text{I}}_{\text{B}}+{\text{I}}_{\text{A}}\otimes \text{B}=\left[\begin{array}{cccc}{a}_{11}+b_{11}& b_{12}& a_{12}& 0\\ b_{21}& a_{11}+b_{22}& 0& a_{12}\\ a_{21}& 0& a_{22}+b_{11}& b_{21}\\ 0& a_{21}& b_{21}& a_{22}+b_{22}\end{array}\right]\end{array}$

It should be noted that the Kronecker multiplication of a two-unit matrix with the same size is equal to doubling the size of the matrix, i.e., let I₁ size 2 × 2; I₂ size 2 × 2 and I₃ size 2 × 2, then

Proof:

$\begin{array}{c}{\text{I}}_{\text{1}}\otimes {\text{I}}_{\text{2}}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\end{array}$

I₁ ⊗ I₂ ⊗ I₃ =

$\begin{array}{c}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cccc}1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}\right]\end{array}$

Then,

$\begin{array}{c}{\text{I}}_{\text{1}}\otimes {\text{I}}_{\text{2}}\otimes {\text{I}}_{\text{3}}=\left[\begin{array}{cccccccc}1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]\end{array}$

Three sub-systems A, B, and C then the transition rate matrix equal to:

$\begin{array}{c}\text{A}\otimes \text{B}\otimes \text{C=A}\otimes {\text{I}}_{\text{B}}\otimes {\text{I}}_{\text{C}}\otimes {\text{I}}_{\text{B}}\otimes \text{B}\otimes {\text{I}}_{\text{C}}{\text{+I}}_{\text{A}}\otimes {\text{I}}_{\text{B}}\otimes \text{C}\end{array}$

It is well known that the Kronecker Multiplication step:

A ⊗ I_B ⊗ I_C = A ⊗ I (where I size 4 × 4)

I_B ⊗ B ⊗ I_C =

$\begin{array}{c}\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\otimes \left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]=\left[\begin{array}{cc}{a}_{11}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]& a_{12}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ a_{21}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]& a_{22}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\end{array}\right]\\ =\left[\begin{array}{cccccccc}{a}_{11}& 0& 0& 0& a_{12}& 0& 0& 0\\ 0& a_{11}& 0& 0& 0& a_{12}& 0& 0\\ 0& 0& a_{11}& 0& 0& 0& a_{12}& 0\\ 0& 0& 0& a_{11}& 0& 0& 0& a_{12}\\ a_{21}& 0& 0& 0& a_{22}& 0& 0& 0\\ 0& a_{21}& 0& 0& 0& a_{22}& 0& 0\\ 0& 0& a_{21}& 0& 0& 0& a_{22}& 0\\ 0& 0& 0& a_{21}& 0& 0& 0& a_{22}\end{array}\right]\\ {\text{I}}_{\text{B}}\otimes \text{B}\otimes {\text{I}}_{\text{C}}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ =\left[\begin{array}{cc}1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]& 1\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cccc}{b}_{11}& b_{12}& 0& 0\\ b_{21}& b_{22}& 0& 0\\ 0& 0& b_{11}& b_{12}\\ 0& 0& b_{21}& b_{22}\end{array}\right]\otimes \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ =\left[\begin{array}{cccc}{b}_{11}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{12}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ b_{21}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{22}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{11}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{12}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& 0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{21}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& b_{22}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}\right]\\ =\left[\begin{array}{cccccccc}{b}_{11}& 0& b_{12}& 0& 0& 0& 0& 0\\ 0& b_{11}& 0& b_{12}& 0& 0& 0& 0\\ b_{21}& 0& b_{22}& 0& 0& 0& 0& 0\\ 0& b_{21}& 0& b_{22}& 0& 0& 0& 0\\ 0& 0& 0& 0& b_{11}& 0& b_{12}& 0\\ 0& 0& 0& 0& 0& b_{11}& 0& b_{12}\\ 0& 0& 0& 0& 0& 0& b_{22}& 0\\ 0& 0& 0& 0& 0& b_{21}& 0& b_{22}\end{array}\right]\\ {\text{I}}_{\text{A}}\otimes {\text{I}}_{\text{B}}\otimes \text{C}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\otimes \left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\\ =\left[\begin{array}{cccc}1\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 1\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 1\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\\ 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 0\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]& 1\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\end{array}\right]\\ =\left[\begin{array}{cccccccc}{c}_{11}& c_{12}& 0& 0& 0& 0& 0& 0\\ c_{21}& c_{22}& 0& 0& 0& 0& 0& 0\\ 0& 0& c_{11}& c_{12}& 0& 0& 0& 0\\ 0& 0& c_{21}& c_{22}& 0& 0& 0& 0\\ 0& 0& 0& 0& c_{11}& c_{12}& 0& 0\\ 0& 0& 0& 0& c_{21}& c_{22}& 0& 0\\ 0& 0& 0& 0& 0& 0& c_{11}& c_{12}\\ 0& 0& 0& 0& 0& 0& c_{21}& c_{22}\end{array}\right]\end{array}$