eigen value of $A$ if $(A-\lambda I)^{k}v=0$

Question

For an $n×n$ matrix $A$ and $\lambda \ \in \ \mathbb{R}$ suppose $(A- \lambda I)^{k} v=0$ for some non zero $v \ \in \mathbb{R}^{n}$ where $k>1$ then is $\lambda$ necessarily an eigen value of $A$?

Answer 1

Hint $(A- \lambda I)^{k}$ not invertible implies $$\left( \det(A- \lambda I)\right)^{k}=\det \left((A- \lambda I)^{k} \right)=0 $$

Answer 2

Consider the smallest $k$ such that $(A - \lambda I)^k v = 0$. Since $v \neq 0$, it must be at least $1$. Let $w = (A - \lambda I)^{k - 1}v$. Note that $w \neq 0$, otherwise this would contradict the definition of $k$.

Then, $$(A - \lambda I)w = (A - \lambda I)(A - \lambda I)^{k-1}v = (A - \lambda I)^k v = 0.$$ Therefore, $$Aw - \lambda Iw = 0 \implies Aw = \lambda w.$$ As $w \neq 0$, we know that $w$ is an eigenvector for $A$, with eigenvalue $\lambda$.