In mathematics, a Hilbert–Schmidt operator, named after David Hilbert and Erhard Schmidt, is a bounded operator <math> A \colon H \to H </math> that acts on a Hilbert space <math> H </math> and has finite Hilbert–Schmidt norm

<math display="block">\|A\|^2_{\operatorname{HS \ \stackrel{\text{def{=}\ \sum_{i \in I} \|Ae_i\|^2_H,</math>

where <math>\{e_i: i \in I\}</math> is an orthonormal basis. The index set <math> I </math> need not be countable. However, the sum on the right must contain at most countably many non-zero terms, to have meaning. This definition is independent of the choice of the orthonormal basis.

In finite-dimensional Euclidean space, the Hilbert–Schmidt norm <math>\|\cdot\|_\text{HS}</math> is identical to the Frobenius norm.

‖·‖ is well defined

The Hilbert–Schmidt norm does not depend on the choice of orthonormal basis. Indeed, if <math>\{e_i\}_{i\in I}</math> and <math>\{f_j\}_{j\in I}</math> are such bases, then

<math display="block">

\sum_i \|Ae_i\|^2 = \sum_{i,j} \left| \langle Ae_i, f_j\rangle \right|^2 = \sum_{i,j} \left| \langle e_i, A^*f_j\rangle \right|^2 = \sum_j\|A^* f_j\|^2.

</math>

If <math>e_i = f_i, </math> then <math display="inline"> \sum_i \|Ae_i\|^2 = \sum_i\|A^* e_i\|^2. </math> As for any bounded operator, <math> A = A^{**}. </math> Replacing <math> A </math> with <math> A^* </math> in the first formula, obtain <math display="inline"> \sum_i \|A^* e_i\|^2 = \sum_j\|A f_j\|^2. </math> The independence follows.

Examples

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 <math>x</math> and <math>y</math> in <math>H</math>, define <math>x \otimes y : H \to H</math> by <math>(x \otimes y)(z) = \langle z, y \rangle x</math>, which is a continuous linear operator of rank 1 and thus a Hilbert–Schmidt operator;

moreover, for any bounded linear operator <math>A</math> on <math>H</math> (and into <math>H</math>), <math>\operatorname{tr}\left( A\left( x \otimes y \right) \right) = \left\langle A x, y \right\rangle</math>.

If <math>T: H \to H</math> is a bounded compact operator with eigenvalues <math>\ell_1, \ell_2, \dots</math> of <math>|T| := \sqrt{T^*T}</math>, where each eigenvalue is repeated as often as its multiplicity, then <math>T</math> is Hilbert–Schmidt if and only if <math display="inline">\sum_{i=1}^{\infty} \ell_i^2 < \infty</math>, in which case the Hilbert–Schmidt norm of <math>T</math> is <math display="inline">\left\| T \right\|_{\operatorname{HS = \sqrt{\sum_{i=1}^{\infty} \ell_i^2}</math>.

If <math>k \in L^2\left( X \times X \right)</math>, where <math>\left( X, \Omega, \mu \right)</math> is a measure space, then the integral operator <math>K : L^2\left( X \right) \to L^2\left( X \right)</math> with kernel <math>k</math> is a Hilbert–Schmidt operator and <math>\left\| K \right\|_{\operatorname{HS = \left\| k \right\|_2</math>. Conversely, if a bounded operator on <math>L^2\left( X \right)</math> is Hilbert-Schmidt then it may be written as an integral operator of this form.

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

<math display="block">\langle A, B \rangle_\text{HS} = \operatorname{tr}(B^* A) = \sum_i \langle Ae_i, Be_i \rangle.</math>

The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on .

They also form a Hilbert space, denoted by or , which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces

<math display="block">H^* \otimes H,</math>

where is the dual space of .

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, is finite-dimensional.

Properties

  • Every Hilbert–Schmidt operator is a compact operator.
  • A bounded linear operator is Hilbert–Schmidt if and only if the same is true of the operator <math display="inline">\left| T \right| := \sqrt{T^* T}</math>, in which case the Hilbert–Schmidt norms of <math>T</math> and <math display="inline">\left| T \right|</math> are equal.
  • Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.
  • If <math>S : H_1 \to H_2</math> and <math>T : H_2 \to H_3</math> are Hilbert–Schmidt operators between Hilbert spaces then the composition <math>T \circ S : H_1 \to H_3</math> is a nuclear operator.
  • If is a bounded linear operator then we have <math>\left\| T \right\| \leq \left\| T \right\|_{\operatorname{HS</math>.
  • is a Hilbert–Schmidt operator if and only if the trace <math>\operatorname{tr}</math> of the nonnegative self-adjoint operator <math>T^{*} T</math> is finite, in which case <math>\|T\|^2_\text{HS} = \operatorname{tr}(T^* T)</math>.