Proving that for nilpotent A, A+I is invertible

[Identity]

Suppose A is nilpotent, i.e. $$A^k=0$$, $$k>0$$. Show that $$A+I$$ is invertible.

Here's my go at it... although I'm unsure about step marked with asterisks... is it ok? I'm not sure how to show that it's true for sure.

Contrapositive: Suppose $$A+I$$ isn't invertible. Then $$(A+I)X =0$$ has a non-zero solution, say $$X = X_0$$.
Then

$$(A+I)X_0=0$$

$$AX_0+X_0=0$$

$$A^kX_0 +A^{k-1}X_0=0$$

If $$A^k=0$$ then $$A^{k-1}X_0=0$$... but $$X_0$$ is not a solution of $$A^{k-1}X=0$$.************

Hence, $$A$$ must not nilpotent, QED

Thanks

 

[Office_Shredder]

but $$X_0$$ is not a solution of $$A^{k-1}X=0$$.************

Why?

 

[Identity]

Ok... I was bluffing. But I don't know how else to proceed!

I just assumed that since $$X_0$$ was the solution of one equation it couldn't solve another different equation.

 

[Office_Shredder]

It helps to gather some intuition from the one dimensional case here.

What is the inverse of (1+x) when x is small (x is a real number)?

 

[Identity]

If you mean multiplicative inverse isn't it

$$\frac{1}{1+x} = 1-x+x^2-...$$

Can this be used for matrices too...?

EDIT: Heh, I guess it can, thanks office_shredder

 

[Office_Shredder]

Of course you have to be worried about convergence, but in the nilpotent case that's not a big deal

 

[dmatador]

I am a little slow, so could someone please humor me: just using 1/(1 + x) = 1 - x + x^2 - ...
and seeing x as A and 1 as I, isn't it apparent that this series converges since it is geometric and can be easily calculated? At what point does the nilpotency of A matter, and how can you use it in proof using the mentioned series?

 

[Mark44]

For some integer power k, A^k = 0 (the zero matrix).

1/(1 + x) in the reals corresponds to (I + A)^-1.

 

[Office_Shredder]

and seeing x as A and 1 as I, isn't it apparent that this series converges since it is geometric and can be easily calculated? At what point does the nilpotency of A matter, and how can you use it in proof using the mentioned series?

If A is just a real number and A=2, the series doesn't converge. The mere fact it is geometric isn't enough; you need your matrix to somehow be "small enough". What exactly small enough is is a lot less obvious for matrices than for real numbers, but if a matrix is nilpotent it qualifies

 

[Petr Mugver]

$$(A+I)v=0$$

$$Av=-v$$

$$A^2v=v$$

$$\dots$$

$$A^kv=(-)^{k+1}v=0$$

$$v=0$$

$$v=0$$ is the only vector in $$\textrm{Ker}(A+I)$$

$$A+I$$ is invertible.

 

[Petr Mugver]

$$Exp(X) = X+Id$$

Its not true. If X^k=0

$$e^X=Id+A+A^2/2+\dots A^{k-1}/(k-1)!$$

 

[vigvig]

Its not true. If X^k=0

$$e^X=Id+A+A^2/2+\dots A^{k-1}/(k-1)!$$

First, X^k = 0 is equivalent to being nilpotent. Second, you may want to replace all your As by Xs, but you are right. Sometimes you do get Exp(X)=X+id but not always. I forgot you could have something like a Heisenberg matrix where exponentiation puts ones on the diagonal but does alter one entry.

 

[trambolin]

Suppose (I+A) is not invertible. Then there exists a vector v such that (I+A)v = 0.

Then v=-Av. Then -1 is an eigenvalue which is a contradiction since A is nilpotent i.e. all eigenvalues are zero.