Let M be invertible and N be nilpotent. Show that M-N is invertible

Question

I have the following task:

Let $M, N \in M_n(\mathbb{R})$ be two matrices s.t. $MN=NM$.

(1) Show that if $M$ is invertible and $N$ be nilpotent of order k, that $M-N$ is invertible.

(2) Let $M$ and $N$ be nilpotent of order k. Show that $M+N$ is nilpotent. Of which order?

(3) Is the commuting hypothesis for the first two points necessary?

My Solution:

(1) I can write $(M^n - N^n) = (M-N) \sum \limits_{i=0}^{n-1}M^iN^{n-1-i}$. Since the matrices are commutative. Now if I consider the case $n=k$:

$M^k = (M-N) \sum \limits_{i=0}^{k-1}M^iN^{k-1-i}$. Now by multiplying $M^{-k}$ from the right, I get:

$Id = (M-N) (\sum \limits_{i=0}^{k-1}M^iN^{k-1-i})M^{-k}$.

Is that enought to show that $M-N$ is invertible?

(2) Is there a general way for rewriting the sum of commutative matrices to the power of some exponent? Ex. :$(M+N)^f = ...$ For a random exponent $f$

(3) They have to be commutative. I've found a counterexample if they are not commutative.

Answer 1

Good attempt that you have constructed the inverse explicitly. Just a minor typo.

$$(M^n-N^n)=(M-N)\sum_{i=0}^{n-1}M^{\color{red}i}N^{n-1-i}$$ $$(M^k-N^k)=M^k-0=(M-N)\sum_{i=0}^{k-1}M^{\color{red}i}N^{k-1-i}$$

$$I=(M-N)\left( \sum_{i=0}^{k-1}M^{\color{red}i}N^{k-1-i}\right)M^{-k}$$

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By Binomial Theorem,

$$(M+N)^{l}=\sum_{i=0}^{l}\binom{l}{i}M^{l-i}N^{i}$$

$$(M+N)^{2k}=\sum_{i=0}^{2k}\binom{2k}{i}M^{2k-i}N^{i}$$

Since $\max(2k-i, i\} \ge k$, we have $(M+N)^{2k}=0$.

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Answer 2

The correct equalities are:

$M^n - N^n = (M-N) \sum \limits_{i=0}^{n-1}M^iN^{n-1-i}$. $M^k= (M-N) \sum \limits_{i=0}^{k-1}M^iN^{k-1-i}$.

You proved that there exists the matrix

$A =\left(\sum \limits_{i=0}^{k-1}M^iN^{k-1-i}\right)\cdot M^{-k}$

such that

$Id = (M-N)\cdot A$

Since the matrices $M$ and $N$ are commutative, it follows that also

$Id =A\cdot (M-N)$

Actually the hypothesis of commutativity is not necessary to obtain the last equality because if there exists a matrix $A$ such that $ B\cdot A = Id $, it is possible to prove that $ A\cdot B = Id$ .

So what you wrote is sufficient in order to prove that $M-N$ is invertible.