# Matrix Inversion Test ( Sum of Matrix series)

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)

• Whoa buddy, that's a pretty huge change to the question. – copper.hat Jul 25 '14 at 8:01
• I am guessing that with the recursion above that you can show that $S$ is defined for all $x$. Continuity of $\det$ does the rest, but gives no estimate of the interval over which the $\det$ is non zero. – copper.hat Jul 25 '14 at 8:16
• Are you talking about det(S(x))? – Nirvana Jul 25 '14 at 8:19
• by x=1, we can say it has a determinant . I need to make sure whether it is correct for all x.. bit messed up coz of that – Nirvana Jul 25 '14 at 8:23

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 $x>1$.

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 $1$.
• You have to show that the power series $\sum_{n=0}^\infty \|A_n\| x^n$ has convergence radius larger than $1$. Here, $\|\cdot\|$ is any matrix norm. The question does not mention in which sense the limit $\lim_{n\to\infty} \sum_{k=0}^n A_k$ exists. You need to add some assumptions. – daw Jul 25 '14 at 7:24
• $A_0$ is invertible rest are not.. I will update that..Typo error..Sorry for that – Nirvana Jul 25 '14 at 8:13