Let A=(a_(ij)) be an an n xx n matrix such that max|a_(ij)|<1/n. Show that I−A is invertible

Lisantiom 2022-10-08 Answered
Let A = ( a i j ) be an an n × n matrix such that max | a i j | < 1 n . Show that I−A is invertible
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Answers (2)

Jaylyn George
Answered 2022-10-09 Author has 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 i i , R i ) where R i = j i | a i j | . Using the fact that max i , j | a i j | < 1 n we can say R i < 1 1 n and | 1 a i i | | 1 | a i i | | > 1 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.
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kasibug1v
Answered 2022-10-10 Author has 4 answers
The series i = 0 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 A v = v as a result A n v = v for all n. But as m a x a i j ∣< 1 / n we see A n converges to 0. Giving us v=0 a contradiction.
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