Existence of $(I-A)^{-1}$ implies convergence of $\sum A^k$?

Question

Suppose $A$ is a square matrix (over $\mathbb{R}$ or $\mathbb{C}$, take your pick) such that $(I-A)^{-1}$ exists. Then is it necessarily true that

$$I + A + A^2 + \dots + A^n + \dots = (I-A)^{-1}$$ ?

There is a well known theorem which claims that if $||A|| \lt 1$ then $(I-A)^{-1}$ exists and the above is true.

What I am asking is something like the converse of that theorem. My hunch is that this is false, but I am unable to come up with a counter-example.

@Jonas Thanks for the link! And looking at the answer, I feel stupid now... :-) — Hayley-Camilton, Aug 10 '14 at 14:29
$\sum A^k$ is convergent if $||A||<1$. This can be achieved using Jordan normal form. Thus the convergence of $\sum A^k$ is its corresponding function. — , Aug 10 '14 at 14:48
I find this may be not helpful to your question. sorry to misunderstand your question. — , Aug 10 '14 at 14:58

score 3 · Accepted Answer · answered Aug 10 '14 at 14:24

3

take $A=2I$ not working.........................

answered Aug 10 '14 at 14:24

Hamou

score 3 · Answer 2 · answered Aug 10 '14 at 14:24

3

Ask the question even for a $1\times 1$ matrix.

answered Aug 10 '14 at 14:24

Empy2

Thanks for the answer! I will accept the other answer as it is slightly more general. – Hayley-Camilton Aug 10 '14 at 14:30

2 Answers2