# Let $A = (a_{ij})$ be an an $n \times n$ matrix such that $\max\lvert a_{ij} \rvert < \cfrac{1}{n}$. Show that $I-A$ is invertible

Just wondering if the following solution I had works:

(Note here im using the norm $$\lVert A\rVert_{1} = \max_{1\leq j \leq n} \sum_{i=1}^n \rvert a_{ij} \lvert$$)

Suppose BWOC that $$I-A$$ is not-invertible. Then, $$I-A$$ has a non-trivial kernel $$\implies$$ $$1$$ is an eigenvalue of $$A \implies \exists \mathbf{v} \neq 0$$ such that $$A\mathbf{v} = \mathbf{v}$$. Now since $$\lVert A\mathbf{v} \rVert_1 \leq \lVert A \rVert_1 \lVert \mathbf{v} \rVert_1$$ (proved in previous part of the question) we have $$\lVert \mathbf{v} \rVert_1 \leq \lVert A \rVert_1 \lVert \mathbf{v} \rVert_1 \implies \lVert A \rVert_1 \geq 1$$. But, (using definition of norm above and that $$\max\lvert a_{ij} \rvert < \cfrac{1}{n}$$) $$\lVert A \rVert_1 < n\cfrac{1}{n} = 1$$. Contradiction.

• Yes, you are right. Commented Aug 13, 2021 at 1:20
• \cfrac should only be used on continued fractions. Please use \frac instead. Commented Aug 13, 2021 at 11:13

You can also approach this problem using the Gershgorin Circle theorem which says that any eigenvalue of $$I-A$$ has to be contained in some Gershgorin Disc $$D(1-a_{ii},R_i)$$ where $$R_i=\sum_{j\neq i}|a_{ij}|$$. Using the fact that $$\max_{i,j}|a_{ij}|<\frac{1}{n}$$ we can say $$R_i<1-\frac{1}{n}$$ and $$|1-a_{ii}|\geq |1-|a_{ii}||>1-\frac{1}{n}$$. So no Gershgorin Disc of $$I-A$$ will contain the origin implying $$I-A$$ cannot have a eigenvalue of $$0$$ and $$I-A$$ must be nonsingular.
The series $$\sum_{i=0}^{\infty} A^i$$ converges due to the norm condition. Let us take the limit to be $$B$$ it is clear that this is the inverse.
Another solution. $$I-A$$ has $$0$$ as an eigenvalue if and only if $$A$$ has 1 as an eigenvalue. Suppose 1 is an eigenvalue of $$A$$, let $$v$$ be an eigenvector for it. So $$Av=v$$ as a result $$A^nv=v$$ for all $$n$$. But as $$max \mid a_{ij} \mid <1/n$$ we see $$A^n$$ converges to 0. Giving us $$v=0$$ a contradiction.