Linear Algebra Matrix Inverse Proof

RoKr93
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Homework Statement


Let A be a square matrix.

a. Show that (I-A)^-1 = I + A + A^2 + A^3 if A^4 = 0.

b. Show that (I-A)^-1 = I + A + A^2 + ... + A^n if A^(n+1) = 0.


Homework Equations


n/a


The Attempt at a Solution


I thought I'd want to use the fact that the multiplication of a matrix and its inverse is equal to I. So I started with (I-A)*(I + A + A^2 + A^3) = I. But that doesn't seem like the right direction...I'm not sure where to go from there.
 
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RoKr93 said:

Homework Statement


Let A be a square matrix.

a. Show that (I-A)^-1 = I + A + A^2 + A^3 if A^4 = 0.

b. Show that (I-A)^-1 = I + A + A^2 + ... + A^n if A^(n+1) = 0.


Homework Equations


n/a


The Attempt at a Solution


I thought I'd want to use the fact that the multiplication of a matrix and its inverse is equal to I. So I started with (I-A)*(I + A + A^2 + A^3) = I. But that doesn't seem like the right direction...I'm not sure where to go from there.
What exactly do you mean you started with (I-A)*(I + A + A^2 + A^3) = I? Did you show the lefthand side equals the righthand side, or did you simply assert it?
 
I asserted it and wanted to start the proof from there, ie eventually get the same value on each side.
 
Just multiply (I-A)*(I + A + A^2 + A^3) out. What do you get?
 
Oh...wow. Guess it's been staring me in the face this whole time.

Thank you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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