Show exp(A) is invertible for a matrix A

[chipotleaway]

Homework Statement

Define [itex]e^A=I_n+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+...+\frac{1}{2012!}A2012[/itex]
where A is an nxn matrix such that [itex]A^{2013}=0[/itex]. Show that [itex]e^A[/itex] is invertible and find an expression for [itex](e^A)^{-1}[/itex] in terms of A.

The Attempt at a Solution

To first show that it's invertible, can I just show the function [itex]e^x[/itex] has an inverse?

 

[Dick]

Homework Statement

Define [itex]e^A=I_n+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+...+\frac{1}{2012!}A2012[/itex]
where A is an nxn matrix such that [itex]A^{2013}=0[/itex]. Show that [itex]e^A[/itex] is invertible and find an expression for [itex](e^A)^{-1}[/itex] in terms of A.

The Attempt at a Solution

To first show that it's invertible, can I just show the function [itex]e^x[/itex] has an inverse?

Where x is a real number? That would work fine if A were diagonal. I don't think assuming ##A^{2013}=0## makes it any easier. I would just try to show ##e^{sA}e^{tA}=e^{(s+t)A}## first. It's really just the binomial theorem. Then put s=1 and t=(-1).

 

[HallsofIvy]

Do you know that [itex]x^n- 1= (x- 1)(1+ x+ x^2+ \cdot\cdot\cdot+ x^{n-1})[/itex]?

If not try proving it by induction on n. Notice that proof works for matrices as well as numbers.

 

[statdad]

Try it first for a matrix B with B^4 = 0 (to make the problem smaller): if you can do it with one where you can see all the terms in the sum and see what happens, writing down the work for your case will just require a little careful writing.