How to show I_n + A is invertible

[HappyN]

Let A be an n x n matrix such that A^k=0_n,n (the n x n zero matrix) for some natural integer k. How would you show that I_n + A is invertible?

 

[lurflurf]

Suppose that (I+A)^-1 exists and compute it.

 

[AlephZero]

Think about the expansion of (1+x)^-1 by the Binomial theorem.

@lurflurf, this works fine when A is singular. For example if n = 2 and A =
0 1
0 0

 

[lurflurf]

^Oops clearly A is singular, I meant suppose A+I is nonsingular, that is needed, and in fact always true.

 

[monea83]

Hi all,

I tried to prove this for myself, but did not get anywhere :-(

Think about the expansion of (1+x)^-1 by the Binomial theorem.

@lurflurf, this works fine when A is singular. For example if n = 2 and A =
0 1
0 0

I don't quite get this... how does the Binomial theorem help here? According to http://en.wikipedia.org/wiki/Binomial_theorem the expansion is only defined for non-negative integers?

 

[AlephZero]

I meant what the Wiki page calls "Newton's generalized binomial theorem".

This gives an infinite series expansion in general, and the series may not converge.

But you know that A^k = 0, and therefore A^m = 0 for any integer m > k, so in this case the series has a finite number of non-zero terms.

 

[mathwonk]

If you know how to multiply (x-y)(x+y)? or even (1-x)(1+x)? you can do this.

 

[Deveno]

we know that A^k is the 0-matrix, right?

well, let's look at the product:

(I + A)(I - A + A^2 - A^3 +...+ (-1)^(k-1)A^(k-1)) =

I - A + A^2 - A^3 +...+ (-1)^(k-1)A^(k-1) + A - A^2 + A^3 -...+(-1)^(k-1)A^k

= I + (-1)^(k-1)A^k = I