April 25, 2024
The invariant subspace problem is “Given a linear operator on a vector space does it have a non-trivial invariant subspace?” The subspace consisting of the 0 vector and the vector space are trivially invariant.
Exercise. Show if T\colon V\to V is a linear operator then T0 = 0.
Hint: Show v + v = v implies v = 0 and T(0 + 0) = T0.
Eigenvectors determine one-dimensional invariant subspaces. If V is a vector space over the field \boldsymbol{F} and T\colon V\to V is linear then v\in V is an eignenvector with eigenvalue \lambda\in\boldsymbol{F} if v\not=0 and Tv = \lambda v. The smallest subspace containing v is {\operatorname{span}\{v\} = \{\alpha v\mid \alpha\in\boldsymbol{F}\} = \boldsymbol{F}v}.
Exercise. Show \boldsymbol{F}v is an invariant suBspace of T if v is an eigenvalue.
Not every operator on a finite-dimensional vector space has an invariant subspace.
Exercise. Show R\colon\boldsymbol{R}^2\to\boldsymbol{R}^2 defined by R(x, y) = (-y, x) has no eigenvectors.
Note the linear operator R corresponds to a \pi/2 radian counter-clockwise rotation. We use \boldsymbol{R} to denote the field of real numbers and \boldsymbol{C} for the complex numbers.
Exercise. Show R\colon\boldsymbol{C}^2\to\boldsymbol{C}^2 defined by R(x, y) = (-y, x) has two eigenvectors.
Hint: If (-y, x) = \lambda(x,y) then -y = \lambda x and x = \lambda y so -y = \lambda^2 y and x = -\lambda^2 x. The two eigenvalues are i and -i.
When looking for invariant subspaces mathematicians pick up some bad habits. Our first one is to assume the underlying field is always the complex numbers.
There are two usual suspects when looking for invariant subspaces, the kernel and the range of the operator.
The kernel of T\colon V\to V is \operatorname{ker}T = \{v\in V\mid Tv = 0\}.
Exercise. Show \operatorname{ker}T is an invariant subspace of T.
The range of T\colon V\to V is \operatorname{ran}T = \{Tv\mid v \in V\}.
Exercise. Show \operatorname{ran}T is an invariant subspace of T.
Our next bad habit to always assume \ket T = \{0\} and \operatorname{ran}T = V.
Exercise. Show v\not=0 is an eigenvector with eigenvalue \lambda if and only if v\in\operatorname{ker}(T-\lambda I).
We use I\colon V\to V to denote the identity operator Iv = v, v\in V.
Every linear operator on a vactor space over \boldsymbol{C} has an eigenvector, but this is not trivial to prove. It is useful to introduce a topology on vector spaces.
A norm on a vector space is a function \|.\|\colon V\to[0,\infty) satisfying \|v\| = 0 implies v = 0, \|\alpha v\| = |\alpha| \|v\| and \|v + w\| \le \|v\| + \|w\|. Norms define a metric d(v, w) = \|v - w\|. A vector space that is complete under this norm is a Banach space.
An inner product on a vector space is a function (.,.)\colon V\times V\to\boldsymbol{F} satisfying v \mapsto (v,w) and w\mapsto (v,w) is linear for v, w\in V. etc…
Exercise. Show \|v\| = (v, v) is a norm.
A vector space with an inner product that is complete under this norm is a Hilbert space.
There are bounded linear operators on a Hilbert space that have no eigenvectors.
The vector space of square summable sequences {\ell^2 = \{(x_j)_{j=0}^\infty\mid \sum_j |x_j|^2 < \infty, x_j\in\boldsymbol{C}\}} is a Hilbert space with inner product (x,y) = \sum_j x_j\overline{y_j}. The shift operator S\colon\ell^2\to\ell^2 defined by S(x_0, x_1, \ldots) = (0, x_0, x_1,\ldots) has no eigenvectors.
Exercise. If Sx = \lambda x then x = 0.
Hint: 0 = \lambda x_0, x_1 = \lambda x_0, rinse and repeat when \lambda\not=0. What if \lambda = 0?
The shift operator has lots of non-trivial invariant subspaces. Let \mathcal{M}_1 = \{(0, x_1,\ldots)\mid x_j\in\boldsymbol{C}\}.
Exercise. Show \mathcal{M}_1 is an invariant subspace.
It should be obvious to you now that \mathcal{M}_n = \{(0,\dots,0,x_n,\ldots)\mid x_j\in\boldsymbol{C}\} is an invariant subspace for any n \ge 0.
Arne Beurling figured out all of the invariant subspaces of the shift operator. The theory of vector spaces is quite spartan. Beurling brought in tools from functional analysis to do the heavy lifting.
He started with identifying \ell^2 with the Hardy space H^2 = \{f(z) = \sum_{j=0}^\infty c_j z^j\mid \sum_j |c_j|^2 < \infty\} of analytic functions on the unit disk D = \{z\in\boldsymbol{C}\mid |z| < 1\}. It is not trivial to prove \bar{f}(e^{i\theta}) = \lim_{r\uparrow 1} f(re^{i\theta}) exists on the boundary \partial D = \{z\mid |z| = 1\} with Lesbegure measure 1 and f(z) = ... \bar{f}.
S\mathcal{M}\subseteq\mathcal{M}. \mathcal{M}\ominus S\mathcal{M}.
T\colon\mathcal{H}\to\mathcal{H} define R\colon \mathcal{H}\to\ell^2(\mathcal{H}) by x\mapsto(x, Tx, T^2x,\ldots).
Exercise. Show \mathcal{M}_n can be identified with z^n H^2.
Beurling showed every invariant subspace of the shift operator has the form \phi H^2 where |\phi| = 1 on \partial D.