This is homework, I'll post what I've done so far.
Let A be a square matrix.
Given A^2 = 0, show that (IA) 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(IA) 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!
Inverse matrices and the determinant
Moderators: gmalivuk, Moderators General, Prelates
Re: Inverse matrices and the determinant
Here's my hint: eigenvalues of A.

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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 (IA). So you're basically thinking what could I put in place of X so that X*(IA)=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?
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?
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}{1p}[/imath].
Re: Inverse matrices and the determinant
Thank you guys.
I have figured that (I+A)(IA) = I^2  A^2 = I
so the inverse of (IA) in the A^2 = 0 case is (I+A)
similarly in the A^3 case (I+A+A^2)(IA) = I
so the inverse is the geometric series 1 + A + A^2 + ... + A^(n1)
I have figured that (I+A)(IA) = I^2  A^2 = I
so the inverse of (IA) in the A^2 = 0 case is (I+A)
similarly in the A^3 case (I+A+A^2)(IA) = I
so the inverse is the geometric series 1 + A + A^2 + ... + A^(n1)
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