Invariant subspace of a diagonal matrix

Question

Suppose $T:V\to V$ is a diagonalizable linear transformation with no repeated eigenvalues, and denote $\{e_i:i\in I\}$ an eigenbasis of $V$. Suppose $W$ is an invariant subspace of $T$. Is it true that $W=\mathrm{span}\{e_i:J\subset I\}$ for some $J$? Why?

Answer 1

Suppose $\{\lambda_i \mid i \in I\}$ are the eigenvalues that go along with the eigenbasis $\{e_i \mid i \in I\}$. Take any $w \in W$, which may be written relative to the eigenbasis as $w = a_1 e_1 + \cdots + a_n e_n$. Here we also assume that all the $a_i$ are nonzero. Now $$\begin{aligned} (T - \lambda_n) w &= a_1 (\lambda_1 - \lambda_n) e_1 + \cdots + a_{n-1} (\lambda_{n-1} - \lambda_n) e_{n-1} + a_n (\lambda_n - \lambda_n) e_n \\ &= b_1 e_1 + \cdots + b_{n-1} e_{n-1} \end{aligned} $$ where each $b_i$ is nonzero, by the assumption that all eigenvalues are distinct. It is clear that by iterating this process, we stay inside $W$, and so each eigenvector $e_1, \ldots, e_n$ belongs to $W$. So your claim is true.

Answer 2

$\textbf{Answer to originally stated question.}$

This is not true. Let $T$ be the identity transformation on $\mathbb{R}^2$. Then the standard basis: $e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is an eigenbasis.

Note that $W = \text{span}\left(\begin{bmatrix}1 \\1 \end{bmatrix}\right)$ is an invariant subspace that does not satisfy the condition you listed.