# Hilbert–Schmidt operator

Short description: Nuclear operator of order 2; a bounded operator A on a Hilbert space H such that tr(A*A) is finite

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

$\displaystyle{ \|A\|^2_{\operatorname{HS}} \ \stackrel{\text{def}}{=}\ \sum_{i \in I} \|Ae_i\|^2_H, }$

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 $\displaystyle{ \sum_i \|Ae_i\|^2 = \sum_i\|A^* e_i\|^2. }$ As for any bounded operator, $\displaystyle{ A = A^{**}. }$ Replacing $\displaystyle{ A }$ with $\displaystyle{ A^* }$ in the first formula, obtain $\displaystyle{ \sum_i \|A^* e_i\|^2 = \sum_j\|A f_j\|^2. }$ 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 $\displaystyle{ x }$ 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 $\displaystyle{ H }$), $\displaystyle{ \operatorname{Tr}\left( A\left( x \otimes y \right) \right) = \left\langle A x, y \right\rangle }$.[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 $\displaystyle{ \sum_{i=1}^{\infty} \ell_i^2 \lt \infty }$, in which case the Hilbert–Schmidt norm of $\displaystyle{ T }$ is $\displaystyle{ \left\| T \right\|_{\operatorname{HS}} = \sqrt{\sum_{i=1}^{\infty} \ell_i^2} }$.[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 $\displaystyle{ \left| T \right| := \sqrt{T^* T} }$, 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 $\displaystyle{ \|T\|^2_\text{HS} = \operatorname{Tr}(T^* T) }$.[1][2]
• If T : HH is a bounded linear operator on H and S : HH is a Hilbert–Schmidt operator on H then $\displaystyle{ \left\| S^* \right\|_{\operatorname{HS}} = \left\| S \right\|_{\operatorname{HS}} }$, $\displaystyle{ \left\| T S \right\|_{\operatorname{HS}} \leq \left\| T \right\| \left\| S \right\|_{\operatorname{HS}} }$, and $\displaystyle{ \left\| S T \right\|_{\operatorname{HS}} \leq \left\| S \right\|_{\operatorname{HS}} \left\| T \right\| }$.[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 $\displaystyle{ \|A\|^2_\text{HS} = \sum_{i,j} |\langle e_i, Ae_j \rangle|^2 = \|A\|^2_2 }$ 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.