Lecture 5: Polar Decomposition, Compact Operators

tags: 224a

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Range and Kernel

Let

$\mathrm{Ker}(A)$ and

$\mathrm{Ran}(A)$ denote the kernel and range of

$A\in L(H)$. These are related as:

$$\mathrm{Ran}(A^{\ast }{)}^{\perp }=\mathrm{Ker}(A),\overline{\mathrm{Ran}(A^{\ast })}=\mathrm{Ker}(A{)}^{\perp },$$
where the second equality can be seen by taking the orthogonal complement of the first one and noting that

$(M^{\perp }{)}^{\perp }=\bar{M}$ for any subspace

$M$ (homework).

While

$\mathrm{Ker}(A)$ is always a closed subspace, this is not true of

$\mathrm{Ran}(A)$, and this distinction is important for instance because the projection theorem, orthogonal decomposition, and existence of orthonormal bases only work for closed subspaces.

Example. To see all of this in action, consider the Volterra operator

$$(Vf)(x)={\int }_0^{x}f(y)dy$$
on

$L^2(0,1)$ from the last lecture, which we showed was bounded. Its adjoint can be calculated from the definition by observing

$$\begin{array}{rl}(g,Vf)& ={\int }_0^1\overline{g(x)}({\int }_0^1\{y\le x\}f(y)dy)dx\\ & ={\int }_0^1({\int }_0^1\{y\le x\}\overline{g(x)}dx)f(y)dy\text{by Fubini}\\ & =(V^{\ast }g,f)\end{array}$$
by definition, so we must have

$$(V^{\ast }f)(x)={\int }_x^{1}f(y)dy.$$

It is now evident that

$V$ is not self-adjoint, since for any strictly positive function

$f$,

$Vf$ is strictly increasing whereas

$V^{\ast }f$ is strictly decreasing.

To compute

$\mathrm{Ker}(V)$, observe that by the fundamental theorem of calculus (or more precisely its Lebesgue version, https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem),

$(Vf)(x)$ is an almost everywhere differentiable function of

$f$, with

$$\frac{d}{dx}(Vf)(x)=f(x).$$
Thus if

$Vf=0$ a.e. we must have

$f=0$ a.e., so

$\mathrm{Ker}(V)=\{0\}$.

The above calculation also reveals that

$\mathrm{Ran}(V)\ne L^2(0,1)$ (since there are continuous functions in

$L^2$ which are not differentiable a.e.), yielding an example of a range which is not closed. However, we do know that

$$\overline{\mathrm{Ran}(V)}=\mathrm{Ker}(V^{\ast }{)}^{\perp }=L^2(0,1)$$
since

$\mathrm{Ker}(V^{\ast })=0$ as well.

Remark 1. An operator of type

$$(T_{K}f)(x):={\int }_0^{1}K(x,y)f(y)dy$$ is called an integral kernel operator, and the Fubini argument above shows that whenever such an operator is bounded we have

$$(T_K{)}^{\ast }=T_{K^{\ast }},$$
where

$K^{\ast }(x,y)=\overline{K(y,x)}$.

Projection and Unitary Operators

A projection is an operator which satisfies

$P^2=P$, and an orthogonal projection further satisfies

$P=P^{\ast }$. It is easy to check that

$\mathrm{Ran}(P)$ is closed for a projection, and there is a 1-1 correspondence between orthogonal projections and closed subspaces of

$H$ by

$H=\mathrm{Ran}(P)\oplus \mathrm{Ker}(P)$.

A partial isometry is an operator

$U:H\to H$ satisfying

$$\Vert Ux\Vert =\Vert x\Vert \mathrm{\forall }x\in \mathrm{Ker}(A{)}^{\perp },$$
which can be seen to be equivalent to

$$(Ux,Uy)=(x,y)\mathrm{\forall }x,y\in \mathrm{Ker}(A{)}^{\perp }$$
by the polarization identity. Such an operator is called unitary if

$\mathrm{Ker}(U)=\{0\}$, in which case

$U$ is a bijection.

It is easy to verify that

$U^{\ast }U=P_{\mathrm{Ker}(U{)}^{\perp }}$ and

$U{U}^{\ast }=P_{\mathrm{Ran}(U)}$. The first fact implies that the range of a unitary is always closed since a sequence

$U{x}_n$ is Cauchy if and only if

$x_n$ is Cauchy.

Intuitively, we think of unitaries as "rotations" and of partial isometries as "geometry preserving" embeddings of one subspace into another.

Example. The right shift operator on

${\ell }^2(\mathbb{N})$ is a partial isometry with

$\mathrm{Ker}=\{e_1\}$ whereas the right shift operator on

${\ell }^2(\mathbb{Z})$ is a unitary.

The Polar Decomposition

The polar decomposition allows one to express any bounded operator as a product of a partial isometry and a positive operator. Let

$|A|=\sqrt{A^{\ast }A}$ denote the absolute value of

$A$.
Theorem 1. If

$A\in L(H)$ then there is a unique partial isometry

$U$ such that

$$A=U|A|$$
with

$\mathrm{Ker}(U)=\mathrm{Ker}(A)$.

The proof is given in Reed and Simon Theorem VI.10.

Suppose we could prove that every self-adjoint operator (such as

$|A|$) can be "diagonalized":

$|A|=VD{V}^{\ast }$ for some "diagonal"

$D$ (we will prove this soon, though the definition of diagonal is different from the finite case). Then the polar decomposition gives:

$$A=U(VD{V}^{\ast })=WD{V}^{\ast }$$
for some partial isometries

$W,V$, which is a "singular value decomposition" for elements of

$L(H)$.

Compact Operators

Rather than trying to understand all self-adjoint operators, we begin by focusing on the a simpler subclass. The simplest operators in

$L(H)$ are the finite rank operators:

$$A=\sum _{n\le N}({\varphi }_n,\cdot ){\psi }_n$$
for some

${\varphi }_n,{\psi }_n\in H$. These operators always have closed range (since finite dimensional subspaces are closed), and can be understood using the methods of linear algebra.

The compact operators are a larger class which inherit many of the nice properties of finite rank operators.

Definition. An operator

$T\in L(H)$ is compact if for every bounded set

$B\subset H$, the closure of

$T(B)$ is compact.

Compactness may seem like a weird condition, but it comes in handy because it allows one to mimick finite dimensional arguments where one "optimizes" over the unit norm ball in

${\mathbb{R}}^n$ to produce e.g. an eigenvector. This is not directly possible in

$H$ because the norm ball

$B=\{x\in H:\Vert x\Vert \le 1\}$ is not compact: consider the infinitely many basis vectors

$e_1,e_2,\dots$, which have pairwise distance

$\sqrt{2}$.

A more intuitive understanding of this class is provided by the following theorem, which we will prove next time.

Theorem. An operator

$T\in L(H)$ is compact iff it is the norm limit of finite rank operators.

Note that the content of the above theorem is that it is a norm limit. It is easy to see that every bounded operator is a limit of finite rank operators in the strong topology (i.e., there are finite rank

$T_n$ such that for every

$x$,

$\Vert T_{n}x-Tx\Vert \to 0$ where the rate of convergence may depend on

$x$.)

The most obvious non-example of a compact operator is the identity, since it maps the unit ball to itself. A large class of examples is provided by integral kernel operators.

Theorem. If

$K\in L^2(0,1{)}^2$ then

$T_K$ is a compact operator.

In particular, the Volterra operator above is compact. We will prove this theorem in the next lecture.