Inverse of $I + A$

Question

I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$.

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.

Here's my attempt.

I claim that the inverse of $I + A$ is $$(I + A)^{-1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n.$$ We prove this is a right inverse. \begin{align*} (I + A)\sum\limits_{n=0}^{k-1} (-1)^n A^n & = \sum\limits_{n=0}^{k-1} (-1)^n (I + A)A^n \\ & = \sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1}) \end{align*} If $n$ is even, then $n + 1$ is odd and vice-versa. Hence, the series telescopes: \begin{align*} & = (A^0 + A^{1}) - (A^1 + A^2) + (A^2 + A^3) + \ldots \pm (A^{k-1} + A^k) \\ & = I \pm A^k \\ & = I \end{align*} It seems that the sign of the final term is dependent on whether $k$ is even or odd. However, it shouldn't matter, because if $A^k = 0$, then $-A^k = 0$.

Is there a better way to formalize this fact?

Answer 1

What you did is basically the factorization $x^n - y^n$ for commuting $x,y$ in the same ring $R$, which you can prove using induction. And obviously the two matrices commute since $I$ commutes with every matrix.

To formalize this, you can just claim that $I = I^k = I^k - O = I^k - (-A)^k$ can be factored in such way, so that the second factor $B$ satisfies $(I+A)B = I$, as desired.

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For instance, $x^n - y^n = (x-y)\left(\sum\limits_{i=0}^{n-1} x^i y^{n-i-1}\right)$. Holds for $n=2$ (direct proof); suppose it holds for an arbitrary $n\geq 2$. Then $$ x^{n+1} - y^{n+1} = (x+y)(x^n-y^n)- xy(x^{n-1} - y^{n-1}).$$ So you know how to proceed.

Answer 2

I think a neater way to demonstrate that the series telescopes is to split the sum into its two parts and replace the dummy variable in the second part with its increment by one: \begin{align} \sum_{n=0}^{k-1}(-1)^n\left(A^n+A^{n+1}\right)&=I+\sum_{n=1}^{k-1} (-1)^nA^n-\sum_{n=0}^{k-1} (-1)^{n+1}A^{n+1}\\ &= I+\sum_{n=1}^{k-1} (-1)^nA^n-\sum_{m=1}^k (-1)^mA^m\\ &=I-(-1)^kA^k\\ &=I\ . \end{align}

Answer 3

You could use $$ \sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1}) =\sum\limits_{n=0}^{k-1} (-1)^n A^n +\sum\limits_{n=0}^{k-1} (-1)^n A^{n+1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=0}^{k-1} (-1)^{n+1} A^{n+1} =\sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=1}^{k} (-1)^n A^n = I - (-1)^kA^k $$

Answer 4

We know that, if $|x|<1$, then

$$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 -x^5 + \cdots$$

For just any square matrix, $\mathbf A$, it is not true that

$$\dfrac{1}{1+\mathbf A} = 1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 -\mathbf A^5 + \cdots$$

But, since $\mathbf A$ is nilpotent, this becomes the finite sum

$$(1+\mathbf A)^{-1} = 1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 + \cdots + (-1)^{k-1}\mathbf A^{k-1}$$

This can be easily verified by multiplying the right side by $1+\mathbf A$.