where is the norm of H, an orthonormal basis of H. Note that the index set need not be countable; however, at most countably many terms will be non-zero. These definitions are independent of the choice of the basis. In finite-dimensional Euclidean space, the Hilbert–Schmidt norm is identical to the Frobenius norm.
where this sum may be finite or infinite. Note that this value is actually independent of the orthonormal basis of H that is chosen. Moreover, if the Hilbert–Schmidt norm is finite then the convergence of the sum necessitates that at most countably many of the terms are non-zero (even if I is uncountable). If A is a bounded linear operator then we have .
A bounded operator A on a Hilbert space is a Hilbert–Schmidt operator if is finite. Equivalently, A is a Hilbert–Schmidt operator if the trace of the nonnegative self-adjoint operator is finite, in which case .
If A is a Hilbert–Schmidt operator on H then
An important class of examples is provided by Hilbert–Schmidt integral operators. Every bounded operator with a finite-dimensional range (these are called operators of finite rank) is a Hilbert–Schmidt operator. The identity operator on a Hilbert space is a Hilbert–Schmidt operator if and only if the Hilbert space is finite-dimensional. Given any and in , define by , which is a continuous linear operator of rank 1 and thus a Hilbert–Schmidt operator; moreover, for any bounded linear operator on (and into ), .
If is a bounded compact operator with eigenvalues , where each eigenvalue is repeated as often as its multiplicity, then is Hilbert–Schmidt if and only if , in which case the Hilbert–Schmidt norm of is .
If , where is a measure space, then the integral operator with kernel is a Hilbert–Schmidt operator and .
Space of Hilbert–Schmidt operators
The product of two Hilbert–Schmidt operators has finite trace-class norm; therefore, if A and B are two Hilbert–Schmidt operators, the Hilbert–Schmidt inner product can be defined as
The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on H. They also form a Hilbert space, denoted by BHS(H) or B2(H), which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces
where H∗ is the dual space of H. The norm induced by this inner product is the Hilbert–Schmidt norm under which the space of Hilbert–Schmidt operators is complete (thus making it into a Hilbert space). The space of all bounded linear operators of finite rank (i.e. that have a finite-dimensional range) is a dense subset of the space of Hilbert–Schmidt operators (with the Hilbert–Schmidt norm).
The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, H is finite-dimensional.
- Every Hilbert–Schmidt operator T : H → H is a compact operator.
- A bounded linear operator T : H → H is Hilbert–Schmidt if and only if the same is true of the operator , in which case the Hilbert–Schmidt norms of T and |T| are equal.
- Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.
- If and are Hilbert–Schmidt operators between Hilbert spaces then the composition is a nuclear operator.
- If T : H → H is a bounded linear operator then we have .
- If T : H → H is a bounded linear operator on H and S : H → H is a Hilbert–Schmidt operator on H then , , and . In particular, the composition of two Hilbert–Schmidt operators is again Hilbert–Schmidt (and even a trace class operator).
- The space of Hilbert–Schmidt operators on H is an ideal of the space of bounded operators that contains the operators of finite-rank.