Showing that a discrete-time system has one equilibrium

Question

I have the following discrete-time system:

$$x(k + 1) = Bx(k) + c$$

where $B \in \mathbb{R}^{n \times n}$ and $c \in \mathbb{R}^n$. Assuming $B$ is convergent, how can I show the following?

1. The matrix $(I_n - B)$ is invertible

2. The only equilibrium point of the system is $(I_n - B)^{-1}c$

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Thus far, I have:

$$I_n - B = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right] - \left[ {\begin{array}{*{20}c} b_{11} & b_{12} \\ 0 & b_{22} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 1-b_{11} & b_{12} \\ 0 & 1-b_{22} \\ \end{array} } \right]$$

Since the determinant of this resulting matrix is never zero (assuming $b_{11} \neq 1$ and $b_{22} \neq 1$), does this show that $(I_n - B)$ is invertible?

Answer 1

We can verify by multiplication that $(I-B)(I+B+B^{2}+....+B^{n})=I-B^{n+1}$.

Since $B$ is convergent we have that lim$\left\|B^{n} \right\|$=$0$ $\,\,(1)$

we can show that the series $I+B+B^{2}+.....$ converges, since by Corollary p.151 Kantorovich: Functional Analysis a necessary and

sufficient for the series to converge is that for some $k$ we have

$\left\|B^{k} \right\|<1$. But we already have that, since

lim$\left\|B^{n} \right\|$=$0$ .

So the series $I+B+B^{2}+.....=V$ and we have $(I-B)V=I$ and likewise that $V(I-B)=I$.

So $I-B$ is invertible and hence the equilibrium point is $x=(I-B)^{-1}c$.