# Nilpotent matrices

angelz429
[SOLVED] Nilpotent matrices

## Homework Statement

Use the McLaurin series for 1/(1+x) to show that I + N is invertible where N is a nilpotent matrix.

n/a

## The Attempt at a Solution

It has something to do with the inverse of 1+x is 1/(1+x), but I'm lost. I'm not sure where to start nor what exactly I need to do.

Homework Helper
You might want to start with the hint and find the McLaurin series for 1/(1+x).

angelz429
Ok, I understand that I need to start with the maclaurin series, and I've done that, but I'm not sure what that shows me.

Homework Helper
You might want to show us what you found for the MacLaurin series so we can make sure you are finding the correct thing. Then substitute N for x. What else? What does the nilpotency of N tell you about the infinite MacLaurin series?

angelz429
ok well we know:
P(n) = 1 + x + ... + x^n
xP(n) = x + x^2 + ... + x^n + x^(n+1)

(x-1)P(n)=x^(n+1)-1

P(n) = x^(n+1)-1/(x-1) goes to -1/(x-1) = 1/(1-x) <== close to 1/(1+x)

if you use -x, P(n)= 1/(1+x)

because the MacLaurin series for 1/(1+x)
f(x) = f(0) + f'(0)x + f''(0)x^2 + ...
f(x) = 1 + x + x^2 + ...

Homework Helper
The MacLaurin series for 1/(1-x)=1+x+x^2+x^3+x^4+..., yes. Doesn't that make the series for 1/(1+x)=1-x+x^2-x^3+x^4-...?? Taking x->-x, just as you said?

angelz429
yes...
Let P(n) = 1-x+x^2-x^3+x^4-...+x^n
xP(n) = x-x^2+.................-x^n+x^(n+1)

(x+1)P(n)=1+x^(n+1)

therefore P(n)=1+x^(n+1)/(x+1) which goes to 1/(1+x)

Homework Helper
yes...
Let P(n) = 1-x+x^2-x^3+x^4-...+x^n
xP(n) = x-x^2+.................-x^n+x^(n+1)

(x+1)P(n)=1+x^(n+1)

therefore P(n)=1+x^(n+1)/(x+1) which goes to 1/(1+x)

Ok, so what happens when x is the nilpotent matrix N??

angelz429
P(n) approaches 1/(1+0) = 1

angelz429
so how does this show that I + N is invertible?