Lisantiom

2022-10-08

Let $A=({a}_{ij})$ be an an $n\times n$ matrix such that $max|{a}_{ij}|<\frac{{\textstyle 1}}{{\textstyle n}}$. Show that I−A is invertible

Jaylyn George

Beginner2022-10-09Added 6 answers

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 ${R}_{i}=\sum _{j\ne i}|{a}_{ij}|$. Using the fact that $\underset{i,j}{max}|{a}_{ij}|<\frac{1}{n}$ we can say ${R}_{i}<1-\frac{1}{n}$ and $|1-{a}_{ii}|\ge |1-|{a}_{ii}||>1-\frac{1}{n}$. 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.

kasibug1v

Beginner2022-10-10Added 4 answers

The series $\sum _{i=0}^{\mathrm{\infty}}{A}^{i}$ 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 ${A}^{n}v=v$ 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.

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 ${A}^{n}v=v$ 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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