Hilbert–Schmidt Operator

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

$\, A\, ^2_ \ \stackrel\ \sum_ \, Ae_i\, ^2_H,$

where $\$ is an orthonormal basis. The index set $I$ 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 $\, \cdot\, _\text$ is identical to the Frobenius norm.

‖·‖ is well defined

The Hilbert–Schmidt norm does not depend on the choice of orthonormal basis. Indeed, if $\_$ and $\_$ are such bases, then
$\sum_i \, Ae_i\, ^2 = \sum_ \left, \langle Ae_i, f_j\rangle \^2 = \sum_ \left, \langle e_i, A^*f_j\rangle \^2 = \sum_j\, A^* f_j\, ^2.$
If $e_i = f_i,$ then $\sum_i \, Ae_i\, ^2 = \sum_i\, A^* e_i\, ^2.$ As for any bounded operator, $A = A^.$ Replacing $A$ with $A^*$ in the first formula, obtain $\sum_i \, A^* e_i\, ^2 = \sum_j\, A f_j\, ^2.$ 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 $x$ and $y$ in $H$, define $x \otimes y : H \to H$ by $(x \otimes y)(z) = \langle z, y \rangle x$, which is a continuous linear operator of rank 1 and thus a Hilbert–Schmidt operator;
moreover, for any bounded linear operator ''$A$'' on $H$ (and into $H$), $\operatorname\left( A\left( x \otimes y \right) \right) = \left\langle A x, y \right\rangle$.

If $T: H \to H$ is a bounded compact operator with eigenvalues $\ell_1, \ell_2, \dots$ of $, T, := \sqrt$, where each eigenvalue is repeated as often as its multiplicity, then $T$ is Hilbert–Schmidt if and only if $\sum_^ \ell_i^2 < \infty$, in which case the Hilbert–Schmidt norm of $T$ is $\left\, T \right\, _ = \sqrt$.

If $k \in L^2\left( \mu \times \mu \right)$, where $\left( X, \Omega, \mu \right)$ is a measure space, then the integral operator $K : L^2\left( \mu \right) \to L^2\left( \mu \right)$ with kernel $k$ is a Hilbert–Schmidt operator and $\left\, K \right\, _ = \left\, k \right\, _2$.

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

$\langle A, B \rangle_\text = \operatorname(B^* A) = \sum_i \langle Ae_i, Be_i \rangle.$

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

$H^* \otimes H,$

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 $\left, T \ := \sqrt$, 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 $S : H_1 \to H_2$ and $T : H_2 \to H_3$ are Hilbert–Schmidt operators between Hilbert spaces then the composition $T \circ S : H_1 \to H_3$ is a nuclear operator.
* If is a bounded linear operator then we have $\left\, T \right\, \leq \left\, T \right\, _$.
* is a Hilbert–Schmidt operator if and only if the trace $\operatorname$ of the nonnegative self-adjoint operator $T^ T$ is finite, in which case $\, T\, ^2_\text = \operatorname(T^* T)$.
* If is a bounded linear operator on and is a Hilbert–Schmidt operator on then $\left\, S^* \right\, _ = \left\, S \right\, _$, $\left\, T S \right\, _ \leq \left\, T \right\, \left\, S \right\, _$, and $\left\, S T \right\, _ \leq \left\, S \right\, _ \left\, T \right\,$. 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 is an ideal of the space of bounded operators $B\left( H \right)$ that contains the operators of finite-rank.
* If is a Hilbert–Schmidt operator on then $\, A\, ^2_\text = \sum_ , \langle e_i, Ae_j \rangle, ^2 = \, A\, ^2_2$ where $\$ is an orthonormal basis of ''H'', and $\, A\, _2$ is the Schatten norm of $A$ for . In Euclidean space, $\, \cdot\, _\text$ is also called the Frobenius norm.

See also

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References

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{{DEFAULTSORT:Hilbert-Schmidt Operator

Linear operators
Operator theory