Linear Algebra: Proof involving basic properties

[SNOOTCHIEBOOCHEE]

Homework Statement

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.

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.

 

[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?

 

[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?

 

[HallsofIvy]

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 A^k= 0, you are done.

 

[SNOOTCHIEBOOCHEE]

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 A^k= 0, you are done.

How does this statement prove that it is invertible?

 

[gabbagabbahey]

How does this statement prove that it is invertible?

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