If $M \in [0,1)^{n \times n}$ has zero diagonal, is $I - M$ invertible?

Question

As the title states, if $M$ is a $n$ by $n$ matrix such that $M_{ii} = 0$ and $0 \leq M_{ij} < 1$ for $i \neq j$, is it always true that $I - M$ is invertible? I've tried some small examples, and they would suggest for this to be the case. Any hints?

Answer 1

This is helpful:

https://en.wikipedia.org/wiki/Gershgorin_circle_theorem

You must find a matrix where the sum of the absolute values in the rows are big enough. In particular $2\times 2$ matrices will not provide a counterexample (OK, this is trivial anyway), and if you are looking for a $3\times 3$ counterexample, you need to have the sum of absolute values in some row outside the diagonal to be at least 1.

This is what I found:

$$M= \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix}$$

Can you see that $I-M$ is not invertible?