Give an example of a vector space $V$, an operator $T \in \mathcal L(V)$ and a $T$-$\space$invariant subspace $U$ of $V$ such that $T/U$ has an eigenvalue that is not an eigenvalue of $T$.
Attempt: I can show that if $V$ is finite dimensional, and $\lambda$ is an eigenvalue of $T/U$ then it is also an eigenvalue of $T$:
$$\exists v \notin U\space: T/U(v+U)=T(v)+U=\lambda v+U$$
$$\Rightarrow T(v)-\lambda v\in U\quad(1)\\$$
Consider $T-I\lambda\in \mathcal L(V)$, if this linear map is invertible then its nullspace is zero and $(T-I\lambda)|_U \in \mathcal L(U)$ is also invertible $\\(2)$ by fundamental theorem of finite dimensional linear map.
Since $(2)$ contradicts $(1)$ in the injectivity of $T-I\lambda$ on $V$, we can conclude that $T-I\lambda$ is not invertible and $\lambda$ is an eigenvalue of $T$.
Now I know that $V$ and $U$ has to be infinite dimensional so that $(2)$ doesn't hold, and the restriction $(T-I\lambda)|_U$ is injective but not surjective. So far I have tried several maps on polynomials or $F^{\infty}$ but no luck, $T$ always ended up having all the eigenvalues of $T/U$ !
Is there any particular way of thinking to deal with this type of question ?
Best Answer
Since an example must by infinite dimensional, consider in the $K$ vector space $K[X]$ the operator $\phi$ defined by multiplication by some non-constant polynomial$~P$. It cannot have any eigenvalues, since an eigenvector $A$ for eigenvalue$~\lambda$ would be a nonzero polynomial$~Q$ such that $(P-\lambda)Q=0$, which is impossible.
However by appropriately choosing a quotient space, one can make the image of any nonzero polynomial $Q$ be an eigenvector for any chosen eigenvalue$~\lambda$. Just take the ring theoretic quotient $K[X]/I$ where $I$ is the ideal generated by $(P-\lambda)Q$; note that all ideals are $\phi$-stable, and that the image of $Q$ in the quotient is nonzero because of the degree.