This is problem 35 of section 5.A of Axler's book Linear Algebra Done Right (3rd Edition). The problem goes as follows:
Suppose $V$ is finite-dimensional, $T$ is a operator on $V$, and $U$ is invariant under $T$. Prove that each eigenvalue of $T/U$ is an eigenvalue of $T$.
The author then adds the remark:
"The next exercise [36] asks you to verify that the hypothesis that $V$ is finite-dimensional is needed for the exercise above [35]."
Clarifying the notation for those not familiar with, $V/U$ is the quotient space defined as $$ V/U = \{v + U : v \in V \}. $$
where $$ v + U = \{ v + u : u \in U \}. $$
and the quotient operator $T/U : V/U \to V/U$ is the linear map defined as:
$$ (T/U)(v + U) = Tv + U. $$
My attempt at proving the result goes as follows:
Suppose that $\lambda$ is an eigenvalue of $T/U$, it can easily be shown that this implies that there must exist a $ v \notin U $ such that $Tv - \lambda v \in U$. Let $u_1, \ldots, u_n $ be a basis of the subspace $U$. Because $U$ is a subspace and it is invariant under $T$, it follows that $Tu_i - \lambda u_i \in U$ for $i = 1, \ldots, n$. Thus, we have a list of $n+1$ vectors in a space of dimension $n$ (because $\dim U = n$), therefore, this list has to be linearly dependent. This implies that there exists a non-trivial (not all scalars are equal to $0$) linear combination of $Tv - \lambda v, Tu_1 - \lambda u_1, \ldots, Tu_n - \lambda u_n $ that equals $0$. This implies that $$ (T-\lambda I)(\text{linear combination of } v, u_1, \ldots, u_n) = 0. $$ The linear combination of $v, u_1, \ldots, u_n $ cannot be zero because $v$ is not in $U$ and the $u_i$'s are a basis of $U$. This shows that $\lambda$ is also a eigenvalue of T. As we wanted.
My problem is that the author says that the hypothesis that $V$ is finite-dimensional is needed for the proof. But here we have only used the hypothesis that $U$ is finite-dimensional. So i suppose this proof would still be valid in a infinite-dimensional vector space $V$ as long as the invariant subspace $U$ is finite-dimensional. Is that correct?
Thanks in advance. Please feel free to ask for clarification in any passage in the proof.
