question about eigenvalue on quotient space (from Linear Algebra done Right) I'am study the algebra,but have some question for the exercise5.A


  
*Suppose $V$ is finite dimensional and $\def\L{\mathcal L}T\in\L(V)$, and $U$ is invariant under$~T$. Prove that each eigenvalue of $T/U$ is an eigenvalue of$~T$.
  
  
  [The exercise below asks you to verify that the hypothesis that $V$ is finite-dimensional is needed for the exercise above.]

  
*Given an example of a vector space $V$, and operator $T\in\L(V)$ and a subspace $U$ of$~V$ that is invariant under $T$ such that $T/U$ has an eigenvalue that is not an eigenvalue of$~T$.
  

How to solve the question? I'm puzzling how to define eigenvalue of $T/U$, especially $T/U(v+U) = λv+U$ when $v$ also in $U$?
THANKS!
 A: The (not entirely standard) notation $T/U$ means the action induced by $T$ on the quotient space $V/U$, in other words $v+U\mapsto T(v)+U$, which is well defined because $T(v+u)\in T(v)+U$ for any $u\in U$ since $U$ is $T$-stable.
An eigenvector of $T/U$ for an eigenvalue $\lambda$ is a coset $v+U$, distinct from $0+U=U$ (since eigenvectors cannot be zero; so you are not allowed to take $v\in U$ as you say in the question), such that $T/U(v+U)=\lambda(v+U)=\lambda(v)+U$. This says that $T(v)+U=\lambda v+U$; note that this does not imply that $v$ itself is an eigenvector of$~T$, so the question is not entirely trivial.
So suppose $v\notin U$ and $\lambda$ are given with $T(v)+U=\lambda v+U$.
Note that this means that not just $U$, but also the sum $V'=\langle v\rangle\oplus U$ is $T$-stable: $T$ sends both any vector in $U$ and the vector $v$ into this sum$~V'$. I will show that $V'$ already contains an eigenvector for$~\lambda$ of$~T$ (so $V$ certainly does too).
Consider the minimal polynomial $\mu$ of the restriction$~T|_U$ of $T$ to $U$. If it has $\lambda$ as root, then the restriction already has an eigenvector for$~\lambda$ of$~T$, and we are done without even using the fact that $v+U$ is an eigenvector of$~T/U$. Suppose henceforth that $\lambda$ is not an eigenvalue of$~T|_U$, so that $\mu[\lambda]$ is a nonzero scalar. Since $T/U$ acts on the coset $v+U$ as multiplication by$~\lambda$, we get that $\mu[T/U]$ acts on that coset as multiplication by that nonzero scalar $\mu[\lambda]$, sending it to the coset $\mu[\lambda]v+U\neq U$; so in particular $\mu[T](v)\neq0$, and $\mu$ cannot (also) be the minimal polynomial $\mu'$ of the restriction$~T|_{V'}$ of $T$ to$~V'$. On the other hand $\mu'$ must be a polynomial multiple of $\mu$ (since $\mu'[T]$ must in particular kill vectors of $U$), and $\mu(X-\lambda)$, is an annihilating polynomial for$~T|_{V'}$, because $\mu[T]\circ(T-\lambda I)$ kills both $U$ and $v$ (the latter because $T(v)+U=\lambda v+U$ means $T(v)-\lambda v\in U$ which is killed by $\mu[T]$). But then $\mu(X-\lambda)$ is minimal polynomial $\mu'$ of$~T|_V'$, and since it has $\lambda$ as root, it follows that $\lambda$ is an eigenvalue of the restriction $T|_{V'}$ (since any root of the minimal polynomial is an eigenvalue), and therefore also of$~T$. (Indeed once easily shows that the nonzero vector $\mu[T](v)\in V'$ is an eigenvector for$~\lambda$ of$~T$.)

The above proof may seem difficult, so I will propose some alternatives.
If you choose a basis for $V$ that starts with a basis for $U$ and completes that, then expressed on this bases $T$ has a block upper triangular form
$$ \pmatrix{A&*\\0&B},$$
where $A$ is the matrix of the restriction $T|_U$, and $B$ is the matrix of $T/U$, with respect to the basis of $V/U$ obtained from the second part of our basis for$~V$. Now by a property of determinants for block triangular matrices, the characteristic polynomial $\chi_T$ of $T$ is the product of the characteristic polynomials $\chi_A$ of $A$ and $\chi_B$ of $B$. By hypothesis the latter has a root$~\lambda$, hence so does $\chi_T$.
I did not propose this proof, because Axler is well known not to like determinants, and correspondingly should detest having to use characteristic polynomials.

If you know about left- (rather than the usual right-) eigenvectors, this is almost immediate. Whenever $\lambda$ is an eigenvalue for$~T$ acting on a finite dimensional vector space, there is not only some eigenvector $v\in V$ for$~\lambda$, but also a left eigenvector: a linear form $\phi\in V^*$ such that $\phi\circ T=\lambda\phi$. But a left eigenvector of $T/U$ for $\lambda$, which is a linear form $\phi:V/U\to K$ immediately gives a left eigenvector $\tilde\phi=\phi\circ\pi$ where $\pi:V\to V/U$ is the canonical projection, as the following manipulation shows:
$$\tilde\phi\circ T=\phi\circ\pi\circ T
 =\phi\circ T/U\circ\pi=\lambda\phi\circ\pi=\lambda\tilde\phi.
$$
But the fact that whenever there is a right eigenvector for$~\lambda$ there is also a left eigenvector, and vice versa, is a result that is about equivalent to the question at hand, so this answer is not really satisfying.

For an example of what goes wrong in infinite dimension, take one of the easiest cases where injectivity and surjectivity of a linear operator are not the same: the shift-down and shift-up operators on infinite sequences of scalars. It suffices to consider finitely supported sequences (finitely many nonzero terms), which can be modelled by polynomials, so take $V=K[X]$; the shift-down operator $S$ consist of removing the constant term and then dividing by$~X$, while the shift-up operator $T$ is multiplication by$~X$. Clearly $S$ is not injective (constant polynomials are killed) but it is surjective, while $T$ is injective (no polynomial is killed by multiplying it by$~X$) but not surjective (any image has zero constant term). Being injective means that $T$ has no eigenvalue $\lambda=0$ (nor does it has any other eigenvalue for that matter). However $T$ has plenty of stable subspaces, for instance $U$ could be the set of polynomials divisible by $X^7$. Since every polynomial enters into $U$ after at most $7$ applications of$~T$, the induced operator $T/U$ is nilpotent, and does have an eigenvector for $\lambda=0$ (indeed $\overline v=X^6+U$ has $T(\overline v)=X^7+U=U$, the zero vector in $V/U$), answering the second question.
I did not do anything with the shift-down operator $S$, but it provides an example of the existence of a right eigenvector for $\lambda=0$ (any nonzero constant polynomial) without any corresponding left eigenvector (a linear form $\phi$ on the polynomials killed by right-composition with $S$ would have to vanish on the image$~S(V)=V$, implying $\phi=0$). So that is why the last argument above requires finite dimension; the others use minimal and characteristic polynomials, which only exist (in general) for linear operators on finite dimensional spaces.
A: Hint: Use (and prove) that $\chi_T(X) = \chi_{T/U}(X)\cdot \chi_{T|_U}(X)$, where $\chi_f(X)$ denotes the characteristic polynomial of the linear map $f$.
