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 $T(V^{(λ)})\subseteq V^{(λ)}$?

  • $\begingroup$ Are you asking to verify your proof? $\endgroup$
    – Epsilon
    Commented Oct 10, 2015 at 13:07
  • $\begingroup$ Yes. We just learned about linear operators and T-invariant sub spaces the other day and I wanted to make sure that my logic is correct. $\endgroup$
    – yung_Pabs
    Commented Oct 10, 2015 at 16:01
  • $\begingroup$ It is indeed correct. I just proved the span$(\textbf{v}) \in V^{(\lambda)}$ part below. $\endgroup$
    – Epsilon
    Commented Oct 10, 2015 at 17:03
  • $\begingroup$ Is the converse the case, i.e., is every invariant subspace the eigenspace associated to some eigenvalue? $\endgroup$
    – MSIS
    Commented Aug 19, 2020 at 18:04
  • 1
    $\begingroup$ @MSIS No. Consider the space containing zero vector only. $\endgroup$ Commented May 26 at 7:51

1 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, $\textbf{v}_1,\textbf{v}_2 \in V^{(\lambda)} \implies \textbf{v}_1+\textbf{v}_2 \in V^{(\lambda)}$.

  • $\begingroup$ awesome! I actually was going to add something similar to my write up of the proof. thank you! $\endgroup$
    – yung_Pabs
    Commented Oct 10, 2015 at 17:08
  • $\begingroup$ Could you do the addition part? $\endgroup$
    – Epsilon
    Commented Oct 10, 2015 at 17:21
  • $\begingroup$ Yes, I understand how to do the addition part. Since $T$ is a matrix, we'd have $T(v_{1}+v_{2})=T(v_{1})+T(v_{2})=Tv_{1}+Tv_{2}=\lambda v_{1}+\lambda v_{2}=\lambda (v_{1}+v_{2})$ correct? $\endgroup$
    – yung_Pabs
    Commented Oct 10, 2015 at 17:25
  • $\begingroup$ Absolutely correct! If I cleared your doubts properly, accept the answer to make the status of the question complete. $\endgroup$
    – Epsilon
    Commented Oct 10, 2015 at 17:26

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