Eigenvalues of an operator induced in a quotient space 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 ? 
 A: Let $H=\ell^2({\mathbb Z})$. Denote by $e_k$ $(k\in {\mathbb Z}$ the standard orthonormal basis in $H$ and let $T$ be the backward shift: $Te_k=e_{k-1}$. Then $T$ is a unitary operator and its spectrum is the unit circle. The closed subspace $K$ of $H$ which is spanned by vectors $e_k$ $k\leq 0$ is obviously invariant for $T$. If $\lambda \in {\mathbb C}$ is
such that $|\lambda|<1$, then $x=\sum_{k=0}^{\infty}\lambda^k e_k$ is in $H$ and
$$ T/K: x+K \mapsto \lambda (x+K) $$
which means that $\lambda$ is an eigenvalue of $T/K$. Hence, each number in the open unit circle is an eigenvalue for $T/K$. But, of course, it is not in the spectrum of $T$.
A: 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.
A: I know this has answers, but I figured I'd give one that may be a bit simpler.
Let $T:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$ (if you're unsure what $\ell^2(\mathbb{N})$ is, just consider the vector space $\mathbb{C}^\infty $) be the forward shift operator, i.e. $$T((z_1,z_2,z_3,...))=(0,z_1,z_2,....).$$ If $ Tz=\lambda z$, then $\lambda z_1=0 $ and for every $k\geq 2 $, $\lambda z_k=z_{k-1}$. In each case $\lambda\neq0 $ or $\lambda=0$, we see that these equalities necessitate $z=0$, whence $ T $ has no eigenvalues. Let $ U=\text{sp}\{e_2,e_3,e_4,...\} $ and notice $ T[\ell^2(\mathbb{N})]\subset U $, so that $ T/U=0 $, i.e. $ 0 $ is an eigenvalue of the map $ T/U $.
