# Homework Help: Proving invertibility of a matrix

1. May 4, 2013

### chipotleaway

1. The problem statement, all variables and given/known data

If A is an nxn matrix such that $A^{2013}=0$, show that $A+I_n$ is invertible and find an expression for $(A+I_n)^-1$

3. The attempt at a solution

...some hints would be nice

2. May 4, 2013

### Dick

If r is a real number and |r|<1 then 1/(1+r)=1-r+r^2-r^3+r^4-... It's a geometric series. How is that proved? See if you can apply the same pattern to your matrix problem. That should be a good hint.

3. May 5, 2013

### chipotleaway

Thanks, we haven't covered that yet but it seems as though there's lots of stuff on the problem sets that we have to find for ourselves.

4. May 5, 2013

### chipotleaway

I think I need a bigger hint :p

I'm looking at $A^{2013}=0$ and wondering what kind of pattern the entries in the matrix would need to be so that the result of taking it to that power is 0 (other than all zero entries)

EDIT: Ah, found something - 'nilpotent matrices'...but the characterisations listed on WIki are mostly alien to me

Last edited: May 5, 2013
5. May 5, 2013

### Office_Shredder

Staff Emeritus
If I have a number x, and I say what is (1+x)-1 (as a 1x1 matrix), you could do the following:
$$(1+x)^{-1} = 1-x+x^2-x^3+x^4...$$

It only works if x has certain properties (otherwise the series won't converge). You can do a similar thing with matrices

6. May 5, 2013

Look at a "smaller" situation first.
Suppose you have a matrix B such that B^4 = 0

With real numbers, if x is small enough, it can be shown that

(1-x)^(-1) = 1 -x + x^2 - x^3 + x^4 -

that is,

(1-x)(1 -x + x^2 - x^3 + x^4 -...) = 1

and this expansion goes on "forever" (it is an infinite series, if you've had calculus and know that term)

At least formally, let's try the same thing with our matrix (B)

(I - B)^(-1) = I - B + B^2 - B^3 + B^4 - B^5 +...

This may look bad ("how do I work with an infinite series when the terms are matrices?") but remember, for our matrix B, B^4 = 0: that means all higher powers of B are zero also, so our candidate for (I-B)^(-1) is

(I - B)^(-1) = I - B + B^2 - B^3

just a finite sum.

Now (this is for you) work through this product:

(I-B)(I-B+B^2-B^3)

You should end up with the product equal to I - that means the inverse of (I-B) is given by (I-B)^(-1).