If $A$ is an $n \times n$ nilpotent matrix show that $I-A$ is invertible then find the spectrum of I-A ?
for part one i've shown that $I-A$ is invertible by finding its inverse using that $A$ is a nilpotent. My problem is in the next part, will anybody help me please?
Without using the Cayely-Hamilton theorem:
For all $\lambda\ne0$ we see that $\frac1\lambda A$ is also nilpotent hence the matrix $I-\frac1\lambda A$ is invertible hence $$\chi_A(\lambda)=\det(\lambda I-A)\ne0$$ hence for all $\lambda\ne0$, $\lambda$ is not an eigenvalue of $A$ but since $\operatorname{sp}(A)\ne\emptyset$ then $0$ is the only eigenvalue of $A$.
Edit: To find the eigenvalues of $I-A$: $$\chi_{A-I}(\lambda)=\det(\lambda I-(I-A))=\det((\lambda-1)I+A)=(-1)^n\chi_A(1-\lambda)$$ and since $0$ is the only root of $\chi_A$ then $1$ is the only root of $\chi_{A-I}$.
Hint: Let $(0,v)$ be an eigenpair of $A$. Prove that $(1,v)$ is an eigenpair of $I-A$.
