Finite-rank operator

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
- 1.1 A canonical form
- 1.2 Algebraic property
2 Finite-rank operators on a Banach space

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 ∈ 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

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

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

$T h = \sum _{i = 1} ^n \alpha_i \langle h, v_i\rangle u_i \quad \mbox{for all} \quad h \in H ,$

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 suffices 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.