Prove that if matrix $A$ is nilpotent, then $I+A$ is invertible.

Question

So my friend and I are working on this and here is what we have so far.

We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?

Thank you.

Duplicate of http://math.stackexchange.com/q/140348/264 and/or http://math.stackexchange.com/q/119904/264 — Zev Chonoles, Mar 09 '13 at 07:23
But $A^k=0$, what do you care ? $(I+A)B=I+A^k$ would still be $I$. You just need to make expression for $B$ correct — Aang, Mar 09 '13 at 07:25
Yes, it can be negative, but there are ways around it (see the answers), and you don't particularly care :-) BTW, I want to commend you on your way of asking homework questions. Very constructive approach to first show what you think! — Jyrki Lahtonen, Mar 09 '13 at 07:52
I have voted to reopen this question, because the OP is actually not asking for a proof, but an explanation on how his/her particular proof can be modified to work. And as Jyrki said, the OP's way of asking homework question is very constructive. — user1551, Mar 09 '13 at 11:54
@user1551: I agree with the question being a good example on how to ask a homework question, but it is still an abstract duplicate, and the accepted answer to the other question uses the exact same technique and determines the sign correctly. — tomasz, Mar 09 '13 at 13:08
@tomasz I didn't notice the second question cited by Zev. You are right, the question here has an answer there. — user1551, Mar 09 '13 at 13:11

score 8 · Answer 1 · answered Mar 09 '13 at 07:24

8

It's the usual polynomial identity $$ 1 - x^{k} = (1 - x)(1 + x + x^{2} + \dots + x^{k-1}), $$ where you are substituting $x = -A$.

answered Mar 09 '13 at 07:24

Andreas Caranti

68,873

This only works if k is odd - see Andre' Nicolas's note. – Steven Stadnicki Mar 09 '13 at 08:08
@StevenStadnicki, sorry, but I do not see your point. As noted in the answer by Julian Kuelshammer, when you substitute $x = - A$, you will get $(-1)^{k-1} A^{k-1}$ as the last term in the second term of RHS. – Andreas Caranti Mar 09 '13 at 08:11
Ahh, right; mea culpa - I was looking at the LHS, but as noted in a comment it doesn't actually matter whether you wind up with $I+A^k$ or $I-A^k$, either is just $I$ by the nilpotency. – Steven Stadnicki Mar 09 '13 at 08:14
@StevenStadnicki, ok, no problem! – Andreas Caranti Mar 09 '13 at 08:15

score 6 · Answer 2 · answered Mar 09 '13 at 07:32

6

By incrementing if necessary by $1$, we can make sure that $k$ is *odd. Then we have the familiar identity (for odd $k$) $1+x^k=(1+x)(1-x+x^2-x^3+\cdots +x^{k-1})$.

answered Mar 09 '13 at 07:32

André Nicolas

507,029

score 5 · Answer 3 · answered Mar 09 '13 at 07:23

5

As you can see already from your formula for $B=I-A\mathbf{+} A^2-\dots$ the sign can be $+$ for example if $k=2$. In fact the sign that occurs is $(-1)^{k-1}$.

answered Mar 09 '13 at 07:23

Julian Kuelshammer

9,670

Prove that if matrix $A$ is nilpotent, then $I+A$ is invertible.

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