1
$\begingroup$

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 ?

$\endgroup$
1
  • $\begingroup$ The easiest examples I can think of have the new eigenvalue $0$. That means you need a $v \in V\setminus U$ with $Tv \in U$. Taking $V$ to be a sequence space, like an $\ell^p(\mathbb{N})$, makes finding examples relatively easy. $\endgroup$ – Daniel Fischer Feb 14 '15 at 14:03
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.