Finite-rank operator explained

In functional analysis, a branch of mathematics, a finite-rank operator is a bounded linear operator between Banach spaces whose range is finite-dimensional.[1]

Finite-rank operators on a Hilbert space

A canonical form

Finite-rank operators are matrices (of finite size) transplanted to the infinite dimensional setting. As such, these operators may be described via linear algebra techniques.

From linear algebra, we know that a rectangular matrix, with complex entries,

M\inCn

has rank

1

if and only if

M

is of the form

M=\alphauv*,where\|u\|=\|v\|=1and\alpha\geq0.

Exactly the same argument shows that an operator

T

on a Hilbert space

H

is of rank

1

if and only if

Th=\alpha\langleh,v\rangleuforallh\inH,

where the conditions on

\alpha,u,v

are the same as in the finite dimensional case.

Therefore, by induction, an operator

T

of finite rank

n

takes the form

Th=\sumin\alphai\langleh,vi\rangleuiforallh\inH,

where

\{ui\}

and

\{vi\}

are orthonormal bases. Notice this is essentially a restatement of singular value decomposition. This can be said to be a canonical form of finite-rank operators.

Generalizing slightly, if

n

is now countably infinite and the sequence of positive numbers

\{\alphai\}

accumulate only at

0

,

T

is then a compact operator, and one has the canonical form for compact operators.

Compact operators are trace class only if the series \sum _i \alpha _i is convergent; a property that automatically holds for all finite-rank operators.[2]

Algebraic property

The family of finite-rank operators

F(H)

on a Hilbert space

H

form a two-sided *-ideal in

L(H)

, the algebra of bounded operators on

H

. In fact it is the minimal element among such ideals, that is, any two-sided *-ideal

I

in

L(H)

must contain the finite-rank operators. This is not hard to prove. Take a non-zero operator

T\inI

, then

Tf=g

for some

f,g0

. It suffices to have that for any

h,k\inH

, the rank-1 operator

Sh,

that maps

h

to

k

lies in

I

. Define

Sh,

to be the rank-1 operator that maps

h

to

f

, and

Sg,k

analogously. Then

Sh,k=Sg,kTSh,f,

which means

Sh,

is in

I

and this verifies the claim.

Some examples of two-sided *-ideals in

L(H)

are the trace-class, Hilbert–Schmidt operators, and compact operators.

F(H)

is dense in all three of these ideals, in their respective norms.

Since any two-sided ideal in

L(H)

must contain

F(H)

, the algebra

L(H)

is simple if and only if it is finite dimensional.

Finite-rank operators on a Banach space

A finite-rank operator

T:U\toV

between Banach spaces is a bounded operator such that its range is finite dimensional. Just as in the Hilbert space case, it can be written in the form

Th=\sumin\langleui,h\rangleviforallh\inU,

where now

vi\inV

, and

ui\inU'

are bounded linear functionals on the space

U

.

A bounded linear functional is a particular case of a finite-rank operator, namely of rank one.

Notes and References

  1. Web site: Finite Rank Operator - an overview. 2004.
  2. Book: Conway, John B.. A course in functional analysis. Springer-Verlag. New York. 1990. 978-0-387-97245-9. 21195908. 267–268.