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.

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

2 Answers 2


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.


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