## Inverse matrices and the determinant

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edgey
Posts: 14
Joined: Sat Oct 10, 2009 6:34 pm UTC

### Inverse matrices and the determinant

This is homework, I'll post what I've done so far.

Let A be a square matrix.
Given A^2 = 0, show that (I-A) is invertible.
(then extend this to A^n)

I can see that if A^n = 0 then det(A) must be zero (as det0 = 0) and so A isn't invertible.
How can I go from here to showing that det(I-A) is nonzero?

Ideas I have are that I is orthogonal and so adding it to my matrix A I get a matrix which is no longer composed of linearly dependent parts. This seems messy though.
Any hints would be lovely!

jml104
Posts: 2
Joined: Sat Nov 13, 2010 10:02 pm UTC

### Re: Inverse matrices and the determinant

Here's my hint: eigenvalues of A.

greengiant
Posts: 272
Joined: Thu Jul 09, 2009 9:26 am UTC

### Re: Inverse matrices and the determinant

There are probably a few different ways to attack this question, but I wouldn't personally use the determinant. Instead, why not try to actually find the inverse matrix of (I-A). So you're basically thinking what could I put in place of X so that X*(I-A)=I (bearing in mind that A*A=0)

If this isn't helpful enough I'm sure someone else will be happy to give you further hints. And by the way, I'm assuming you've covered the fact that matrix multiplication is distributive over addition?

Tyl.
Posts: 32
Joined: Thu Sep 11, 2008 9:23 pm UTC

### Re: Inverse matrices and the determinant

Hint: we know for real number [imath]p[/imath] with less than unit modulus (i.e. [imath]|p| < 1[/imath]), that [imath]\sum\limits_{k=0}^\infty p^k = \frac{1}{1-p}[/imath].

edgey
Posts: 14
Joined: Sat Oct 10, 2009 6:34 pm UTC

### Re: Inverse matrices and the determinant

Thank you guys.

I have figured that (I+A)(I-A) = I^2 - A^2 = I
so the inverse of (I-A) in the A^2 = 0 case is (I+A)

similarly in the A^3 case (I+A+A^2)(I-A) = I

so the inverse is the geometric series 1 + A + A^2 + ... + A^(n-1)
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