# Hilbert–Schmidt operator

In mathematics, a Hilbert–Schmidt operator, named for David Hilbert and Erhard Schmidt, is a bounded operator A on a Hilbert space H with finite Hilbert–Schmidt norm

$\|A\|_{\operatorname {HS} }^{2}=\sum _{i\in I}\|Ae_{i}\|^{2},$ where $\|\cdot \|$ is the norm of H, $\{e_{i}:i\in I\}$ 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 $\|\cdot \|_{\text{HS}}$ is identical to the Frobenius norm.

## Definition

Suppose that $\left(H,\|\cdot \|\right)$ is a Hilbert space. If $\{e_{i}:i\in I\}$ is an orthonormal basis of H then for any linear operator A on H define:

$\left\|A\right\|_{\operatorname {HS} }^{2}\equiv \sum _{i\in I}\left\|Ae_{i}\right\|^{2}$ where this sum may be finite or infinite. Note that this value is actually independent of the orthonormal basis $\{e_{i}:i\in I\}$ 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 $\left\|Ae_{i}\right\|^{2}$ are non-zero (even if I is uncountable). If A is a bounded linear operator then we have $\left\|A\right\|\leq \left\|A\right\|_{\operatorname {HS} }$ .

A bounded operator A on a Hilbert space $\left(H,\|\cdot \|\right)$ is a Hilbert–Schmidt operator if $\left\|A\right\|_{\operatorname {HS} }^{2}$ is finite. Equivalently, A is a Hilbert–Schmidt operator if the trace $\operatorname {Tr}$ of the nonnegative self-adjoint operator $A^{*}A$ is finite, in which case $\|A\|_{\text{HS}}^{2}=\operatorname {Tr} (A^{*}A)$ .

If A is a Hilbert–Schmidt operator on H then

$\|A\|_{\text{HS}}^{2}=\sum _{i,j}|A_{i,j}|^{2}=\|A\|_{2}^{2}$ where $\{e_{i}:i\in I\}$ is an orthonormal basis of H, $A_{i,j}:=\langle e_{i},Ae_{j}\rangle$ , and $\|A\|_{2}$ is the Schatten norm of $A$ for p = 2. In Euclidean space, $\|\cdot \|_{\text{HS}}$ is also called the Frobenius norm.

## 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 $H$ ), $\operatorname {Tr} \left(A\left(x\otimes y\right)\right)=\left\langle Ax,y\right\rangle$ .

If $T:H\to H$ is a bounded compact operator with eigenvalues $l_{1},l_{2},\dots$ , where each eigenvalue is repeated as often as its multiplicity, then $T$ is Hilbert–Schmidt if and only if $\sum _{i=1}^{\infty }l_{i}^{2}<\infty$ , in which case the Hilbert–Schmidt norm of $T$ is $\left\|T\right\|_{\operatorname {HS} }={\sqrt {\sum _{i=1}^{\infty }l_{i}^{2}}}$ .

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 $\left\|K\right\|_{\operatorname {HS} }=\left\|k\right\|_{2}$ .

## 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

$\langle A,B\rangle _{\text{HS}}=\operatorname {Tr} (A^{*}B)=\sum _{i}\langle Ae_{i},Be_{i}\rangle .$ The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on 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

$H^{*}\otimes H,$ 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.

## Properties

• Every Hilbert–Schmidt operator T : HH is a compact operator.
• A bounded linear operator T : HH is Hilbert–Schmidt if and only if the same is true of the operator $\left|T\right|:={\sqrt {T^{*}T}}$ , 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 T : HH is a bounded linear operator then we have $\left\|T\right\|\leq \left\|T\right\|_{\operatorname {HS} }$ .
• If T : HH is a bounded linear operator on H and S : HH is a Hilbert–Schmidt operator on H then $\left\|S^{*}\right\|_{\operatorname {HS} }=\left\|S\right\|_{\operatorname {HS} }$ , $\left\|TS\right\|_{\operatorname {HS} }\leq \left\|T\right\|\left\|S\right\|_{\operatorname {HS} }$ , and $\left\|ST\right\|_{\operatorname {HS} }\leq \left\|S\right\|_{\operatorname {HS} }\left\|T\right\|$ . 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 H is an ideal of the space of bounded operators $B\left(H\right)$ that contains the operators of finite-rank.