$(I-A)^k=0 \text{ implies that } \exists A^{-1} \text{ s.t. }AA^{-1}=I $ I think this proposition is right. If this is not right, could you provide a counter example? However, this is definitely right for the $\mathbb{R}^3 $ case. Here is how I proved it. Is it right and if so rigorous? Even if yes, are there more ways to prove this, I'm really curious. I put two ways, I'm not sure if either is right, or maybe one of them is rigorous and the other one is not. Could you please point out flaws in the proofs if the idea is right but it is not rigorously explained? I'm new to proofs and it's summer so I can't annoy my professors. Thanks.
$\\(I-A)^k=0 \\
det(I-A)^k=0 \Rightarrow det(I-A)=0 \\
\text{Therefore there exists a non-zero vector x, }\\
(2) \quad(I-A)x=Ix-Ax=0 \\
\text{There exists non trivial x s.t. } Ax=\lambda_{2}x \\
\text{So now } (I-A)x=x-\lambda_2x=(1-\lambda_2)x \\
\text{Multiplying both sides by (A-I) k times, we get } 0=(1-\lambda)^kx \\
\text{Since x is non-trivial, }1-\lambda=0 \Rightarrow \lambda=1 \\
\therefore \text{A has eigenvalue 1 of multiplicity k, and so A is invertible}
$

Let's try this with minimal polynomial argument:
Consider $p(x)=(1-x)^k$ . Then $A$ is a matrix which satisfies the polynomial $p(x)$. Now the minimal polynomial of $A$ has to be a divisor of $p(x)$( This fact can be easily proven using the divison algorithm). So the minimal polynomial of $A$ has to be of the form $m(x)=(1-x)^r$ where $r\leq k$. So the minimal polynomial of $A$ has roots 1 with multiplicity $r$. Now the roots of the minimal polynomial of $A$ are exactly the eigen values od $A$. So eigen values of $A$ is 1 and this is the only eigen value. So $0$ is not an eigen value of $A$ hence $A$ is invertible.
