title: Functional Calculus
author: Keith A. Lewis
institution: KALX, LLC
email: [email protected]
classoption: fleqn
abstract: Apply a function to a linear transformation
...
\newcommand\RR{\bm{R}}
\newcommand\CC{\bm{C}}
\newcommand\ker{\operatorname{ker}}
\newcommand\ran{\operatorname{ran}}
If $T\colon V\to V$ is a linear transformation on the vector space
$V$ then $p(T)$ is well-defined for any polynomial $p$. This is the
polynomial functional calculus. Recall the spectrum $σ(T)$ is the
set of complex numbers $λ\in\CC$ such that $T - λ I$ is not invertable,
where $I$ is the identity operator. If $v$ is an eigenvector with
eigenvalue $λ$ then $λ\in σ(T)$ since $T - λ I$ has $v \not= 0$
in its kernel.
Exercise. If $V$ is finite dimensional the spectrum of $T$ is the set of eigenvalues of $T$.
Scalar multiples of an eigenvector form a one-dimensional invariant subspace
but it is not trivial to find eigenvectors.
The polynomial functional calculus can be used to find invariant subspaces.
For any polynomial $p$ the kernel of $p(T)$ is an invariant subspace for $T$.
A vector $v$ is in the kernel of $p(T)$ if and only if $p(T)v = 0$.
Since $p(T)T = Tp(T)$ we have $p(T)Tv = Tp(T)v = 0$ so $Tv$ is in the kernel of $p(T)$.
If $p(T)$ is invertible then $\ker p(T) = {0}$ is trivially invariant.
The spectral mapping theorem can tell us when $p(T)$ is not invertable.
Theorem. For any polynomial $p$, $σ(p(T)) = p(σ(T))$.
Solution
For any $λ\in\CC$ and any polynomial $p$ we have $p(z) - p(λ) = (z - λ)q(z)$
for some polynomial $q$ so $p(T) - p(λ)I = (T - λI)q(T)$.
If $λ\in σ(T)$ then $T - λI$ is not invertable so $p(T) - p(λ)I$ is
not invertable. This shows $p(λ)\in σ(p(T))$.
We also have $p(z) - p(λ) = (z - λ)^k q(z)$ where $q(λ)\not=0$ for some $k$
since $p$ is a polynomial.
If $0\in σ(p(T))$ then $p(T)$ is not invertable and its kernel is
an invariant subspace.
For any $λ\in σ(T)$ the polynomial
$p(z) = (z - λ)q(z)$, where $q$ is a polynomial, satisfies $p(λ) = 0$
so $\ker p(T)$ is an invariant subspace.
we can find a
polynomial $e_λ$ with $e_λ(λ) = 1$ and $e_λ(μ) = 0$ if $μ\in σ(T)$
and $μ \not= λ$.
If $V$ has a norm we can use that to define a norm on operators $|T| = \sup_{|v| = 1}|Tv|$.
If $T$ is bounded then so is $σ(T)$ since $(T - λ I)^{-1} = \sum_{n\ge0} T^n/λ^{n+1}$
converges if $λ > |T|$. Since $λ\mapsto (T - λ I)^{-1}$ is continuous the complement
of the spectrum is an open set.
Let little ell two be the vector space of square-summable sequences
$\ell^2 = {(z_n):\sum_n |z_n|^2 < \infty}$ where $z_n\in\CC$ are complex numbers.
This is a Hilbert space with inner product $z·w = \sum_n z_n\overline{w_n}$.
Define the right shift operator $S\colon\ell^2\to\ell^2$ by
$S(z_0, z_1, \ldots) = (0, z_0, z_1,\ldots)$.
Exercise. Show the right shift operator does not have any eigenvectors.
Hint: If $Sz = λz$ then $0 = λz_0$, $z_1 = λz_0$, $\ldots$.
The adjoint of the right shift operator is the left shift operator
$S^*(z_0, z_1, \ldots) = (z_1, z_2,\ldots)$. It has lots of eigenvectors.
If $z = (λ^n)$ then $S^*z = λz$ and $z\in\ell^2$ if $|λ| < 1$.
The polynomial functional calculus gives us a subspace for each point in the spectrum.
Lemma. (Spectral mapping theorem) If $p$ is a polynomial then $p(σ(T)) = σ(p(T))$.