Finite rank operator

In functional analysis, a finite rank operator is a bounded linear operator between Banach spaces whose range is finite dimensional.

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, they can be described via linear algebra techniques

From linear algebra, we know that a rectangular matrix, with complex entries, "M" ∈ C^{"n" × "m"} has rank 1 if and only if "M" is of the form

:$M\; =\; alpha\; cdot\; u\; v^*,\; quad\; mbox\{where\}\; quad\; |u\; |\; =\; |v|\; =\; 1\; quad\; mbox\{and\}\; quad\; alpha\; geq\; 0\; .$

Exactly the same argument shows that an operator "T" on a Hilbert space "H" is rank 1 if and only if

:$T\; h\; =\; alpha\; langle\; h,\; v\; angle\; u\; quad\; mbox\{for\; all\}\; quad\; h\; in\; H\; ,$

where the conditions on "α", "u", and "v" are the same as in the finite dimensional case.

Therefore, by induction, an operator "T" of finite rank "n" takes the form

:$T\; h\; =\; sum\; \_\{i\; =\; 1\}\; ^n\; alpha\_i\; langle\; h,\; v\_i\; angle\; u\_i\; quad\; mbox\{for\; all\}\; quad\; h\; in\; H\; ,$

where {"u_i"} and {"v_i"} 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 {"α_i"} accumulate only at 0, "T" is then a compact operator, and one has the canonical form for compact operators.

If the series ∑_"i" "α_i" is convergent, "T" is a trace class operator.

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" ∈ "I", then "Tf" = "g" for some "f, g" ≠ 0. It surffices to have that for any "h, k" ∈ "H", the rank-1 operator "S"_{"h, k"} that maps "h" to "k" lies in "I". Define "S"_{"h, f"} to be the rank-1 operator that maps "h" to "f", and "S"_{"g, k"} analogously. Then

:$S\_\{h,k\}\; =\; S\_\{g,k\}\; T\; S\_\{h,f\},\; ,$

which means "S"_{"h, k"} 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

Finite rank operator $T:U\; o\; V$ 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

:$T\; h\; =\; sum\; \_\{i\; =\; 1\}\; ^n\; alpha\_i\; langle\; h,\; v\_i\; angle\; u\_i\; quad\; mbox\{for\; all\}\; quad\; h\; in\; U\; ,$

where now $u\_iin\; V$, and $v\_iin\; U\text{'}$ 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.