Solving Nilpotent Matrices: Invertibility of I+N

[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.

Homework Equations

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.

 

[Dick]

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.

 

[Dick]

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 + ...

 

[Dick]

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)

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

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

 

[Dick]

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

add them
(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?

 

[Dick]

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

The inverse of I+N is 1/(I+N). Does the MacLaurin expansion converge??

 

[angelz429]

Thanks!