Prove $T(Aut(\lambda))\subset Aut(\lambda)$ where $A\in Mat_n(\mathbb{K})$ and $\lambda$ eigenvalue of $A$

Question

Let $T_A:\mathbb{K}^n\rightarrow\mathbb{K}^n$ the linear operator associated to $A$. Prove $T(Aut(\lambda))\subset Aut(\lambda)$ where $A\in Mat_n(\mathbb{K})$ and $\lambda$ eigenvalue of $A$

Note: $Aut(\lambda)=\{X\in\mathbb{K}^n:AX=\lambda X\}$

I'm too stuck trying to solve this exercise. Can someone help me?

Answer 1

If $X\in\operatorname{Aut}(\lambda)$, then$$A(AX)=A(\lambda X)=\lambda AX$$and therefore $AX\in\operatorname{Aut}(\lambda)$.

Answer 2

Let $X\in \text{aut}(\lambda)$. Then $T(X)=AX=\lambda X$ by definition. We need to show that $T(X)\in \text{aut}(\lambda)$. Now $A(T(X))=A(\lambda X)=\lambda AX=\lambda^2X=\lambda (\lambda X)=\lambda(T(X))$. Hence $T(X)\in \text{aut}(\lambda)$. In fact that's the nice thing about eigenspaces, they are invariant subspaces. The same is true for generalized eigenspaces.