On the geometric series convergence for the expression: $(I-\frac{1}{z}A)^{-1}$ where $A$ is a matrix

Question

I was reading up about the rigour behind the Cauchy Integral Formula for matrices, and stumbled across this answer on the site, from years ago.

It states that, if $|z|\gt\|A\|$ for some matrix norm $\|\cdot\|$ and $z\in\Bbb C$:

$$(I-\frac{1}{z}A)^{-1}=\sum_{k=0}^\infty\frac{A^k}{z^k}$$

Now, this is very much like the standard result: $A/z$ appears to be lesser in magnitude than $1$, and the inverse is a "division", $I$ appears to be like the number $1$... but in the interests of rigour, I find this result needs a little work.

The standard result for geometric series is proven as follows:

$$S_n(z)=\sum_{k=0}^n z^k\implies S_n(z)-zS_n(z)=1-z^{k+1}\implies S_n(z)=\frac{1-z^{k+1}}{1-z}\\\lim_{n\to\infty}S_n(z)=\lim_{n\to\infty}\frac{1-z^{n+1}}{1-z}=\frac{1}{1-z}-\frac{1}{1-z}\lim_{n\to\infty}z^{n+1}$$

And if $|z|\lt1$, successive powers of $z$ will be lower and lower in absolute value, and will approach $0$ in the limit, giving $(1-z)^{-1}$ as the final result. We crucially need $z^n\to0$. To readapt this proof for our inverse here, we must assert that $A^n/z^n\to0$ as $n\to\infty$. So clearly $\|A^n/z^n\|\to0$ by definition of $\|\cdot\|$ and $z$, if you take $\|AB\|\le\|A\|\cdot\|B\|$ to be a property of the matrix norm - I have asked another very related question, just now, about this. If that is the case, then by definition of $\|\cdot\|$, $A^n/z^n\to0$ also.

Is this correct? And am I right in thinking that this is only correct for a slightly stricter definition of matrix norm?

Answer 1

Let me see if I can help clarify things for you. We need to prove two statements:

1. That the infinite sum $\sum_{k=0}^\infty\frac{A^k}{z^k}$ of matrices is convergent, and
2. The resulting matrix is the inverse of $I - \frac{1}{z}A$.

Your instinct to mirror the proof of summability of $\sum_{k=0}^\infty \omega^k$, where $\omega \in \Bbb{C}$ such that $|\omega| < 1$ is helpful more to part 2 than to part 1. The reason is, when it comes to the division by $1 - \omega$ step, the analogue for matrices would be multiplying by $\left(I - \frac{1}{z}A\right)^{-1}$, which we don't necessarily know exists yet! (Or at least, we shouldn't have to assume it). But, what it does tell us is this: $$\left(I - \frac{1}{z}A\right)\left(\sum_{k=0}^N\frac{A^k}{z^k}\right) = \left(\sum_{k=0}^N\frac{A^k}{z^k}\right)\left(I - \frac{1}{z}A\right) = I - \frac{A^{N+1}}{z^{N+1}} \to I,$$ as $N \to \infty$ (and yes, it is important to show that $\frac{A^{N+1}}{z^{N+1}} \to 0$, which you show precisely as you did, with the submultiplicativity assumption). So, if 1. holds, then taking the limits yields $$\left(I - \frac{1}{z}A\right)\left(\sum_{k=0}^\infty\frac{A^k}{z^k}\right) = \left(\sum_{k=0}^\infty\frac{A^k}{z^k}\right)\left(I - \frac{1}{z}A\right) = I.$$ Actually, submultiplicativity is handy in this step too, as it proves multiplication is continuous.

To show part 1, you should show that the partial sums are Cauchy. This reduces to showing that sums of the form $$\sum_{k=M}^N \frac{A^k}{z^k}$$ can be made arbitrarily small for sufficiently large $M < N$. By triangle inequality and submultiplicativity, $$\left\|\sum_{k=M}^N \frac{A^k}{z^k}\right\| \le \sum_{k=M}^N \frac{\|A^k\|}{|z|^k} \le \sum_{k=M}^N \frac{\|A\|^k}{|z|^k},$$ which can be made arbitrarily small as $\sum_{k=0}^\infty \frac{\|A\|^k}{|z|^k}$ is a convergent geometric series, and has Cauchy partial sums. Using completeness, this means that $\sum_{k=0}^\infty \frac{A^k}{z^k}$ exists, proving 1.