I was reading through a linear algebra textbook, and I'm not sure what does the following statement refer to:
$\textbf{"For a vector space V, the eigenspaces of a transformation}$ $T: V \rightarrow V$ $\textbf{are preserved by the transformation}$ $R: V \rightarrow V$.
I was thinking that if $\lambda$ is an eigenvalue of $T$, having as a respective eigenspace $E_{\lambda}$, then the statement implies:
$E_{\lambda} = \{v \in V: T(v) = \lambda v\} = \{v \in V: (R\circ T)(v) = \lambda v\}$
Is this correct? If not, what would be the correct way to interpret the statement? Thanks for clearing up the confusion.
No, what you write is that the eigenspaces of $T$ and $R\circ T$ coincide, and in each case for the same eigenvalue$~\lambda$. That condition is very strong and unlikely to be satisfied in interesting cases: if $T$ is diagonalisable for instance it will only hold for $T$ the identity. But it cannot be right, since subspaces "being preserved by $R$" should not involve $T$ at all.
What being preserved by $R$ no doubt means is being $R$-stable: a subspace $W$ is $R$-stable is $R(W)\subseteq W$, in other words if $\forall w\in W: R(w)\in W$. With this interpretation, you can easily show that $R$ preserves each eigenspace of $T$ whenever $T$ and $R$ commute.