mjhjmtu

the question goes $A^k = 0$, for some $k>0$. 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 with inverse given above.

I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $detA^k=(detA)^k=0.$ And doing $A^k-1$ during factorisation doesn't feel right.

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.

What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.

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

Here $I$ and $A$ commute, and we may suppose, 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.