# Hilbert–Schmidt operator

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

${\displaystyle \|A\|_{\operatorname {HS} }^{2}\ {\stackrel {\text{def}}{=}}\ \sum _{i\in I}\|Ae_{i}\|_{H}^{2},}$

where ${\displaystyle \{e_{i}:i\in I\}}$ is an orthonormal basis.[1][2] The index set ${\displaystyle I}$ need not be countable. However, the sum on the right must contain at most countably many non-zero terms, to have meaning.[3] This definition is independent of the choice of the orthonormal basis. In finite-dimensional Euclidean space, the Hilbert–Schmidt norm ${\displaystyle \|\cdot \|_{\text{HS}}}$ 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 ${\displaystyle \{e_{i}\}_{i\in I}}$ and ${\displaystyle \{f_{j}\}_{j\in I}}$ are such bases, then

${\displaystyle \sum _{i}\|Ae_{i}\|^{2}=\sum _{i,j}\left|\langle Ae_{i},f_{j}\rangle \right|^{2}=\sum _{i,j}\left|\langle e_{i},A^{*}f_{j}\rangle \right|^{2}=\sum _{j}\|A^{*}f_{j}\|^{2}.}$
If ${\displaystyle e_{i}=f_{i},}$ then ${\textstyle \sum _{i}\|Ae_{i}\|^{2}=\sum _{i}\|A^{*}e_{i}\|^{2}.}$ As for any bounded operator, ${\displaystyle A=A^{**}.}$ Replacing ${\displaystyle A}$ with ${\displaystyle A^{*}}$ in the first formula, obtain ${\textstyle \sum _{i}\|A^{*}e_{i}\|^{2}=\sum _{j}\|Af_{j}\|^{2}.}$ 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 ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle H}$, define ${\displaystyle x\otimes y:H\to H}$ by ${\displaystyle (x\otimes y)(z)=\langle z,y\rangle x}$, which is a continuous linear operator of rank 1 and thus a Hilbert–Schmidt operator; moreover, for any bounded linear operator ${\displaystyle A}$ on ${\displaystyle H}$ (and into ${\displaystyle H}$), ${\displaystyle \operatorname {Tr} \left(A\left(x\otimes y\right)\right)=\left\langle Ax,y\right\rangle }$.[4]

If ${\displaystyle T:H\to H}$ is a bounded compact operator with eigenvalues ${\displaystyle \ell _{1},\ell _{2},\dots }$ of ${\displaystyle |T|={\sqrt {T^{*}T}}}$, where each eigenvalue is repeated as often as its multiplicity, then ${\displaystyle T}$ is Hilbert–Schmidt if and only if ${\textstyle \sum _{i=1}^{\infty }\ell _{i}^{2}<\infty }$, in which case the Hilbert–Schmidt norm of ${\displaystyle T}$ is ${\textstyle \left\|T\right\|_{\operatorname {HS} }={\sqrt {\sum _{i=1}^{\infty }\ell _{i}^{2}}}}$.[5]

If ${\displaystyle k\in L^{2}\left(\mu \times \mu \right)}$, where ${\displaystyle \left(X,\Omega ,\mu \right)}$ is a measure space, then the integral operator ${\displaystyle K:L^{2}\left(\mu \right)\to L^{2}\left(\mu \right)}$ with kernel ${\displaystyle k}$ is a Hilbert–Schmidt operator and ${\displaystyle \left\|K\right\|_{\operatorname {HS} }=\left\|k\right\|_{2}}$.[5]

## 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

${\displaystyle \langle A,B\rangle _{\text{HS}}=\operatorname {Tr} (A^{*}B)=\sum _{i}\langle Ae_{i},Be_{i}\rangle .}$

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

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

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).[4] 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).[4]

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.[5]
• A bounded linear operator T : HH is Hilbert–Schmidt if and only if the same is true of the operator ${\textstyle \left|T\right|:={\sqrt {T^{*}T}}}$, in which case the Hilbert–Schmidt norms of T and |T| are equal.[5]
• Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.[5]
• If ${\displaystyle S:H_{1}\to H_{2}}$ and ${\displaystyle T:H_{2}\to H_{3}}$ are Hilbert–Schmidt operators between Hilbert spaces then the composition ${\displaystyle T\circ S:H_{1}\to H_{3}}$ is a nuclear operator.[3]
• If T : HH is a bounded linear operator then we have ${\displaystyle \left\|T\right\|\leq \left\|T\right\|_{\operatorname {HS} }}$.[5]
• T is a Hilbert–Schmidt operator if and only if the trace ${\displaystyle \operatorname {Tr} }$ of the nonnegative self-adjoint operator ${\displaystyle T^{*}T}$ is finite, in which case ${\displaystyle \|T\|_{\text{HS}}^{2}=\operatorname {Tr} (T^{*}T)}$.[1][2]
• If T : HH is a bounded linear operator on H and S : HH is a Hilbert–Schmidt operator on H then ${\displaystyle \left\|S^{*}\right\|_{\operatorname {HS} }=\left\|S\right\|_{\operatorname {HS} }}$, ${\displaystyle \left\|TS\right\|_{\operatorname {HS} }\leq \left\|T\right\|\left\|S\right\|_{\operatorname {HS} }}$, and ${\displaystyle \left\|ST\right\|_{\operatorname {HS} }\leq \left\|S\right\|_{\operatorname {HS} }\left\|T\right\|}$.[5] In particular, the composition of two Hilbert–Schmidt operators is again Hilbert–Schmidt (and even a trace class operator).[5]
• The space of Hilbert–Schmidt operators on H is an ideal of the space of bounded operators ${\displaystyle B\left(H\right)}$ that contains the operators of finite-rank.[5]
• If A is a Hilbert–Schmidt operator on H then
${\displaystyle \|A\|_{\text{HS}}^{2}=\sum _{i,j}|\langle e_{i},Ae_{j}\rangle |^{2}=\|A\|_{2}^{2}}$
where ${\displaystyle \{e_{i}:i\in I\}}$ is an orthonormal basis of H, and ${\displaystyle \|A\|_{2}}$ is the Schatten norm of ${\displaystyle A}$ for p = 2. In Euclidean space, ${\displaystyle \|\cdot \|_{\text{HS}}}$ is also called the Frobenius norm.