April 25, 2024
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\bm{C} 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 \operatorname{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)).
For any λ\in\bm{C} 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 \operatorname{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\bm{C} 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)).