Artin's Algebra Exercise 1.1.16

Question

The question goes $A^k = 0$, for some $k>0$ ($A$ is square). Prove that $A+I$ is invertible.

I did $A^k = 0 \implies A^k+I=I$. So, $(A+I)(A^{k-1} +.... + I) = I$. So it's invertible with the inverse given above.

I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $\det{A^k}=(\det{A})^k=0.$ And doing $A^{k-1}$ during factorisation doesn't feel right since you are multiplying with $A^{-1}$.

Can you please direct me to the correct direction and tell me why the thing I did is incorrect?

Thanks in advance.

Edit: I see that my factorisation is incorrect (plus/minus signs) but the question remains the same.

Answer 1

What you did is almost correct. In fact:$$(A+\operatorname{Id})\bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+\cdots-A+\operatorname{Id}\bigr)=\operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+\cdots-A+\operatorname{Id}$ is the inverse of $A+\operatorname{Id}$. What does it matter that $\det(A^k)=(\det A)^k=0$? This has nothing to do with determinants.

Answer 2

The identity: $\;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\dots-xy^{n-2}+y^{n-1})$ $\;(n\text{ odd})$ is valid in any ring provided $x$ and $y$ commute.

Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.

Answer 3

There is another way to solve the problem. Suppose that $\det(A+I) = 0$. Then $A$ must have an eigenvalue of $-1$, since $\lambda= -1$ is a root of the characteristic equation $\det(A+\lambda I) = 0$. Therefore, $\exists \> v \in V, v\neq 0$ such that $Av = -v$. But now, by linearity, $A^kv = (-1)^k v \neq 0$ , contradicting the assumption that $A^k = 0$.

Your solution method is still the proper algebraic method (assuming the correct factorization technique).