Matrix Inversion Test ( Sum of Matrix series)

Question

Friends,I have a set of matrices of dimension $3\times3$ called $A_i$. ,

Following are the given conditions

a) each $A_i$ is non invertible except $A_0$ because their determinant is zero.

b) $\sum_{n=0}^\infty A_i$ is invertible and determinant is not zero

c)

1. This is the recursion available for $A_i$, $ A_{n}=\frac{1}{n} \{C_1* A_{n-1} +C_2 * A_{n-2}\} \tag 1$, where $A_0$ = Constant matrix ,$A_1$ =Constant matrix

2. $C_1,C_2 $ are constant matrices. $A_1$ and $A_0$ are initial values. $A_0,A_1,C_1,C_2,A_n $ have dimension $3\times 3$

3. $C_1,C_2,C_1+C_2 $ etc are skew symmetric matrices , not commutative, and also with diagonals as zeros

4. $A_n$ are converging series. Means last terms will be approaching to zero or very very small values

5. Determinant of $C_1*A_{n-1}$ and $C_2*A_{n-2}$ both are zero {Logic : det($C_1A_{n-1}$)=det($C_1$)det($A_{n-1}$),=0*det($A_{n-1}$),$=0 $ }

6. Given that SUM= $ \sum_{n=0}^{n= \infty} A_n \ne 0 $.

7. Let $S(x) = \sum_{n=0}^\infty A_nx^n$,$SUM=S(1)$.Given that $S(1)$ is invertible . Remember we still have not proved S(x) is invertible. What we only know is, S(1) is invertible from the given conditions

Question From the given condition can we say that $S(x)=\sum_{n=0}^\infty A_nx^n$ is invertible? If so. how do we prove that?. (x is not a matrix, it is just a variable)

Answer 1

The matrix $$ A(x):=\sum_{n=0}^\infty A_n x^n $$ is invertible at $x=1$. The set of invertible matrices is an open set in the space of all matrices. Thus, an idea to solve the problem is to show that $x\mapsto A(x)$ is continuous.

In order to prove that you need additional assumptions. One would be to assume that the power series $$ \sum_{n=0}^\infty \|A_n\|\cdot x^n $$ has convergence radius $r \ge1$. Here, $\|\cdot\|$ is a matrix norm. Then the series $$ \sum_{n=0}^\infty A_n x^n $$ would be convergent for all $x$ with $|x|<r$, hence $A(x)$ would be continuous on $(-r,+r) \cup \{1\}$ (assuming $x\in \mathbb R$). Then you obtain that $A_n(x)$ is invertible for $x$ near $1$.

Answer 2

Note: The question changed dramatically after this answer was written.

Since $\sum_n A_n$ exists, we see that $A_n \to 0$ and hence the radius of convergence of $\|A_n\|$ is at least one, so the function $f(x) = \sum_n A_i x^n$ is real analytic on $(-1,1)$. Furthermore, Abel's theorem (applied to each component) shows that $\lim_{x \uparrow 1} f(x) = \sum_n A_n$.

Then continuity of $\det$ shows that $f(x)$ must be invertible in some interval $(1-\epsilon,1]$ with $\epsilon>0$.

However, it is possible that $f$ is not defined for $x>1$.

First let $B_n=(-1)^n{1 \over n} I$ (each $B_n$ is invertible, but I will fix that in a moment). If $x>1$, we see that $\|B_n x^n\| \to \infty$, hence the sum does not converge. Now let $A_{3n-2}$ be the $(1,1)$ element of $B_n$, $A_{3n-1}$ be the $(2,2)$ element of $B_n$ and $A_{3n}$ be the $(3,3)$ element of $B_n$. Each $A_n$ is singular, and the sum does not converge for $x>1$.