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

## ‖·‖$_{HS}$ is well defined

The Hilbert–Schmidt norm does not depend on the choice of orthonormal basis. Indeed, if $\{e_{i}\}_{i\in I}$ and $\{f_{j}\}_{j\in I}$ are such bases, then $\sum _{i}\|Ae_{i}\|^{2}=\sum _{i,j}\left|\langle Ae_{i},f_{j}\rangle \right|^{2}=\sum _{i,j}\left|\langle e_{i},A^{*}f_{j}\rangle \right|^{2}=\sum _{j}\|A^{*}f_{j}\|^{2}.$ If $e_{i}=f_{i},$ then ∑$_i$‖Ae$_i$‖$^2$=∑$_i$‖A$^∗$e$_i$‖$^2$.{\textstyle \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 ∑$_i$‖A$^∗$e$_i$‖$^2$=∑$_j$‖Af$_j$‖$^2$.{\textstyle \sum _{i}\|A^{*}e_{i}\|^{2}=\sum _{j}\|Af_{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 $H$), $\operatorname {Tr} \left(A\left(x\otimes y\right)\right)=\left\langle Ax,y\right\rangle$.
If $T:H\to H$ is a bounded compact operator with eigenvalues $\ell _{1},\ell _{2},\dots$ of $|T|:={\sqrt {T^{*}T}}$, where each eigenvalue is repeated as often as its multiplicity, then $T$ is Hilbert–Schmidt if and only if ∑$_{i=1}$$^∞$ℓ$_i$$^2$<∞{\textstyle \sum _{i=1}^{\infty }\ell _{i}^{2}<\infty }, in which case the Hilbert–Schmidt norm of $T$ is ‖T‖$_{HS}$=√(∑$_{i=1}$$^∞$ℓ$_i$$^2$){\textstyle \left\|T\right\|_{\operatorname {HS} }={\sqrt {\sum _{i=1}^{\infty }\ell _{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
$\langle A,B\rangle _{\text{HS}}=\operatorname {Tr} (B^{*}A)=\sum _{i}\langle Ae_{i},Be_{i}\rangle .$
The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on H. They also form a Hilbert space, denoted by B$_{HS}$(H) or B$_2$(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 : 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 |T|:=√(T$^∗$T){\textstyle \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 : H → H is a bounded linear operator then we have $\left\|T\right\|\leq \left\|T\right\|_{\operatorname {HS} }$.
* T is a Hilbert–Schmidt operator if and only if the trace $\operatorname {Tr}$ of the nonnegative self-adjoint operator $T^{*}T$ is finite, in which case $\|T\|_{\text{HS}}^{2}=\operatorname {Tr} (T^{*}T)$.
* If T : H → H is a bounded linear operator on H and S : H → H 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.
* If A is a Hilbert–Schmidt operator on H then $\|A\|_{\text{HS}}^{2}=\sum _{i,j}|\langle e_{i},Ae_{j}\rangle |^{2}=\|A\|_{2}^{2}$ where $\{e_{i}:i\in I\}$ is an orthonormal basis of H, and $\|A\|_{2}$ is the Schatten norm of $A$ for p = 2. In Euclidean space, $\|\cdot \|_{\text{HS}}$ is also called the Frobenius norm.