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I have this following problem:

let $0\neq A\in M_n(F)$ such that $A^k=0$ for integer $k>1$

a. what are the eigenvalues of A?

b. show that $B=\alpha\cdot I_n-A$ invertible for every scalar $\alpha\in F, \alpha \neq0.$

I know that the answer for a is that zero is the only eigenvalue but I dont know how to explain it well.

and I dont have a clue for b.

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If $Av = \lambda v$ for some $v \ne 0$, then $0 = A^kv = \lambda^k v$, so $\lambda = 0$.

Therefore $\sigma(A) = \{0\}$. Hence, every scalar $\alpha \ne 0$ is not an eigenvalue of $A$ so $\alpha I - A$ is invertible.

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For part B, let $\lambda$ be an eigenvalue of $B$. Then,$ \left|B -\lambda I_n\right|=\left|\alpha\cdot I_n-A-\lambda I_n\right|=\left|(\alpha-\lambda)\cdot I_n -A\right|=0$,

Which implies $\lambda=\alpha$, because $|kI_n -A|=0$ iff $|A-kI_n|=0$ iff $k=0$.

Now, $\alpha \neq 0$ is the only eigenvalue of $B$. So, $|B|=$product of eigenvalues of $B \neq 0$. Therefore, $B$ is invertible.

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