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Let $\lambda$ be an eigenvalue of a square matrix $A$.

Show the null space of $(A-\lambda I)^j$ is an A-Invariant subspace of $\mathbb{C}^n$ for all positive integers $j$.

Proof without requiring transformations.

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Let $E_{\lambda,j}=\text{ker} (A-\lambda I)^j$ and $y\in E_{\lambda,j}$. You have to show that $Ay\in E_{\lambda,j}$, i.e. that $(A-\lambda I)^jAy=0$.

Just use the fact that $y\in E_{\lambda,j}$ and that $(A-\lambda I)^jA=A(A-\lambda I)^j$ (this last point should be justified).

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Actually, we have that $\operatorname{ker}p(A)$ is $A$-invariant for any polnomial $p \in \mathbb C[x]$.

The proof goes like this: Notice that $A$ commutes with any polnomial expression of $A$, i.e. $Ap(A)=p(A)A$.

Now, let $v \in \operatorname{ker}p(A)$.

We have $p(A)Av=Ap(A)v=0$, hence $Av \in \operatorname{ker}p(A)$, which is the desired result.

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