Finding the Inverse of (A-I) Given A^2=0

[transgalactic]

i am given a square matrix called A(they don't specify how it looks)
i am given that
A^2=0

I-unit matrix

find (A-I)^-1
??-1 is the inverse command

i can pick many matrices which A^2=0
and do this easy subtraction and inverse function

but i need here a general solution
??

 

[CompuChip]

Maybe this question can be of help to you

Or, if you want another approach, notice the real number identity
\frac{1}{x - 1} = \frac{1}{x - 1} \times \frac{x + 1}{x + 1}

 

[Mark44]

I believe that where they're heading with this problem is expressing (A - I)^(-1) as an infinite series. Another poster had a similar problem yesterday.

A problem that is somewhat analogous is the series representation for 1/(1 -a) = 1 + a + 1/2! * a^2 + 1/3! * a^3 + ... + 1/n! * a^n + ...

You're trying to find (A - I)^(-1), which is (-(I - A))^(-1). This latter expression can be shown (I think) to be equal to -(I - A)^(-1), which should be equal to
-[I + A + 1/2! * A^2 + 1/3! * A^3 + ... + 1/n! * A^n + ...]

As a sanity check, I picked a 2 X 2 matrix A = [0 1; 0 0] (shown row by row), and things worked out, which gave me some evidence that I was on the right track.


BTW, I is usually called an identity matrix.

 

[CompuChip]

I believe that where they're heading with this problem is expressing (A - I)^(-1) as an infinite series. Another poster had a similar problem yesterday.

Yes, only it won't become infinite. And I posted a link to that question in my earlier post.

 

[Mark44]

Yes, only it won't become infinite.

Didn't want to give everything away

 

[transgalactic]

thanks