Let A = ( a i j ) be an an n × n matrix such that max | a i j | &lt; 1 n . Show that I−A is invertible

Question

Let   A  =  (      a          i      j        ) be an an   n  ×  n matrix such that   max  |      a          i      j        |  &amp;lt;                                                  1                                                                  n                    . Show that I−A is invertible

Jaylyn George · Accepted Answer

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            |    &amp;lt;      1    n   we can say       R    i    &amp;lt;  1  −      1    n   and       |    1  −      a          i      i            |    ≥      |    1  −      |        a          i      i            |        |    &amp;gt;  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.

kasibug1v · Accepted Answer

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        ∣&amp;lt;  1      /    n we see       A    n   converges to 0. Giving us v=0 a contradiction.

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

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