Linear Algebra: Proof involving basic properties

1. Sep 25, 2008

SNOOTCHIEBOOCHEE

1. The problem statement, all variables and given/known data

A square matrix A is called nilpotent if A^k =0 for some k>0. Prove that if A is nilpotent then I+A is invertible.

3. The attempt at a solution

My guess would be to do a proof by induction (on the size of the matrix)

So for the trivial cases:

Let A be a 2x2 Nilpotent matrix... thus it is of the form

[0 x] [0 0]
[0 0] Or [x 0]

Clearly when we add I to A in this case, we get get a matrix, whose det =/= 0

Im having trouble doing the more general cases, seeing as that i cannot mentally see what nilpotent matrices of a larges size look like.

Is this even the best way to approach this problem?

Any help is appreciated.

2. Sep 25, 2008

SNOOTCHIEBOOCHEE

Would using the fact that the trace of a nilpotent matrix is equal to 0 help in any way? perhaps coupled with the trace of I= n?

3. Sep 26, 2008

Defennder

I can think of one way to do this. If I+A is invertible, then det(I-A) is non-zero, and that implies that 1 is not an eigenvalue of A. So we need to prove that last statement. Do it by contradiction:

Suppose 1 is an eigenvalue of A, that implies there is a nonzero vector v such that Av=v. Given that A is nilpotent, can such a non-zero vector v exist?

Unfortunately this method isn't constructive; while it does show that I+A is invertible, it cannot come up with the inverse of I+A itself. So does anyone else have a better idea?

4. Sep 26, 2008

HallsofIvy

Staff Emeritus
Oh, I think it can be done directly.

Remember that $$1- x^k= (1+ x)(1- x+ x^2+ \cdot\cdot\cdot+ (-1)^k x^{k-1})$$? If that still works for matrices, then taking x= A and k such that Ak= 0, you are done.

5. Sep 29, 2008

SNOOTCHIEBOOCHEE

How does this statement prove that it is invertible?

6. Sep 29, 2008

gabbagabbahey

If $I+A$ is invertible then $I=(I+A)(I+A)^{-1}$ ...does that look anything like what Hallsofly posted?