Proving an eigenspace is a T-Invariant subspace.

Question

I want to know if I'm going about this proof the correct way.

Problem Statement: Let $T$ be a linear operator on a vector space $V$, and let $λ$ be a scalar. The eigenspace $V^{(λ)}$ is the set of eigenvectors of $T$ with eigenvalue $λ$, together with $\textbf{0}$. Prove that $V^{(λ)}$ is a $T$-invariant subspace.

So I need to show that $T(V^{(λ)})\subseteq V^{(λ)}$.

Since $V^{(λ)}$ is the set of eigenvectors of the matrix $T$ corresponding to $λ$, that meants that for any $\textbf{v}\in V^{(λ)}$, we have $T\textbf{v}=λ\textbf{v}$.

Clearly, $T\textbf{v}\in T(V^{(λ)})$. Then we know that for any $\textbf{v}\in V^{(λ)}$, $\text{span}(\textbf{v})\in V^{(λ)}$. Thus, $λ\textbf{v}\in V^{(λ)}$. Since $T\textbf{v}=λ\textbf{v}$, then $T\textbf{v}\in V^{(λ)}$. So then $T(V^{(λ)})\subseteq V^{(λ)}$?

Answer 1

Your proof is correct. Just a small thing to add.

If $\textbf{v}\in V^{(λ)}$ then $T \textbf{v}= \lambda \textbf{v}$.

Claim: If $\textbf{v} \in V^{(\lambda)}$ then, $k\textbf{v} \in V^{(\lambda)}$, for any scalar $k$.

Proof: $T(k\textbf{v})= k\ (T(\textbf{v}))=k\ (\lambda\textbf{v})= \lambda (k\textbf{v}) \implies k\textbf{v}\in V^{(\lambda)}$

Similarly you can show that, $\textbf{v}_1,\textbf{v}_2 \in V^{(\lambda)} \implies \textbf{v}_1+\textbf{v}_2 \in V^{(\lambda)}$.