Hilbert-Schmidt operator ($T\in {\mathcal{S}}_2$)

A bounded linear operator on a Hilbert space is Hilbert-Schmidt when ${\sum }_n\Vert T{e}_n{\Vert }^2<\infty$ for one, equivalently every, orthonormal basis. Every Hilbert-Schmidt operator is a compact operator.

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.

An everywhere-defined linear operator $T$ on an infinite-dimensional separable Hilbert space need not be bounded merely because ${\sum }_k\Vert T{e}_k{\Vert }^2<\infty$ for a Hilbertian basis $(e_k)$. Let $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$, extend $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis, and let the linear functional $\varphi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\varphi (x)e_1$ has $T{e}_k=0$ for every $k$ but $Tv=e_1$. For $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k$, $v-v_N\to 0$ while $T(v-v_N)=e_1$, proving discontinuity. Thus the definition of a Hilbert-Schmidt operator must require boundedness, or first define the operator on finite linear combinations and take its unique continuous extension.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

For $(Tg)(v)=\int k(v,w)g(w)dw$, the Cauchy-Schwarz inequality gives $\Vert Tg{\Vert }_2\le \Vert k{\Vert }_{L^2(v,w)}\Vert g{\Vert }_2$. The kernel norm is the Hilbert-Schmidt norm, which bounds the operator norm. Uniform kernel bounds apply also to spatially parametrized operators after integrating over position.

The Hilbert-Schmidt norm of a Hilbert-Schmidt operator is the square root of the displayed sum. The value is independent of the orthonormal basis, by the Parseval identity. For a finite matrix, it is the Frobenius norm.

The Frobenius norm is the finite-matrix Hilbert-Schmidt norm. It equals $\sqrt{\mathrm{tr}(A^{\ast }A)}$, is unchanged by unitary multiplication on either side, and is the square root of the sum of squared singular values. For unit vectors $u,v$, $\Vert u{u}^{\ast }-v{v}^{\ast }{\Vert }_F^2=2(1-|u^{\ast }v{|}^2)$, a useful sign-invariant measure of distance between rank-one orthogonal projection matrices.

For Hilbert-Schmidt operators $A$ and $B$, define the Hilbert-Schmidt inner product by $⟨A,B{⟩}_{\mathrm{HS}}={\sum }_k⟨A{e}_k,B{e}_k⟩$ for a Hilbertian basis $(e_k)$. With an inner product linear in its first argument this equals $\mathrm{tr}(B^{\ast }A)$; with an inner product conjugate-linear in its first argument it equals $\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm, independently of that convention.

If $\rho$ is a density operator with decreasing eigenvalues ${\lambda }_j$ and $T$ has matrix rank at most $k$, let $\Pi$ project onto the image of $T$. The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product givesThe Hermitian effect variational principle bounds $\mathrm{Tr}(\rho \Pi )$ by ${\sum }_{j<k}{\lambda }_j$, proving the claimed estimate without assuming $\rho$ is invertible.

The Hilbert-Schmidt distance is $d_{\mathrm{HS}}(A,B)=\Vert A-B{\Vert }_{\mathrm{HS}}$.

A bounded operator $T$ on a separable Hilbert space is trace class when $|T{|}^{1/2}$ is Hilbert--Schmidt. Its trace norm is

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

For bounded operators with trace-class commutator, the multiplicative commutator differs from identity by a trace-class operator. Along $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$, the trace of its logarithmic derivative is the constant $\mathrm{Tr}[A,B]$. This follows by writing $e^{B}A{e}^{-B}-A$ as the integral of conjugates of $[B,A]$ and using trace invariance under bounded similarities. Integrating the Fredholm determinant derivative from $F(0)=I$ proves the formula. Individual traces of $A$ and $B$ need not exist.

An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\Vert T{\Vert }_1=\Vert A{\Vert }_{\mathrm{HS}}\Vert B{\Vert }_{\mathrm{HS}}$ by using the polar decomposition of a bounded operator.

The Schatten norm Hölder inequality gives $\Vert AB{\Vert }_r\le \Vert A{\Vert }_p\Vert B{\Vert }_q$ when $1/r=1/p+1/q$. In particular, the product of two Hilbert-Schmidt operators is trace class and

For a trace-class operator, the series $\mathrm{tr}T={\sum }_n⟨T{e}_n,e_n⟩$ converges absolutely and is independent of the orthonormal basis. It satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

The trace-class operator condition makes this operator trace well-defined; the class of eligible operators and the trace functional are distinct concepts.

Every bounded operator $S$ defines a functional $T\mapsto \mathrm{tr}(ST)$ on the trace-class operators, andFinite-rank density in ${\mathcal{S}}_1$ makes every continuous functional arise uniquely this way.

For vectors $x,y$ in a Hilbert space, the rank-one operator $(x\otimes y)z=⟨z,y⟩x$ has $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $\mathrm{tr}(x\otimes y)=⟨x,y⟩$.

For the rank-one operator $Tf=⟨f,v⟩u$ on a Hilbert space with a countable Hilbertian basis $(e_k)$, the operator norm and Hilbert-Schmidt norm both equal $\Vert u\Vert \Vert v\Vert$. The Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, with equality at $f=v/\Vert v\Vert$ when $v\ne 0$. The Parseval identity for a Hilbertian basis gives ${\sum }_k\Vert T{e}_k{\Vert }^2=\Vert u{\Vert }^2{\sum }_k|⟨e_k,v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$. The zero-vector cases are immediate.

Let $Kx=a\phi (x)$ on a vector space, where $\phi$ is a linear functional. Its image of a linear map lies in the span of $a$. Applying $K$ twice gives $K^{2}x=a\phi (a)\phi (x)=\phi (a)Kx$. Mathematical induction then proves the displayed power formula. If $\phi (a)=0$, $K^2=0$, even when $K\ne 0$; this is a nilpotent operator. If $\phi (a)\ne 0$, then $K/\phi (a)$ is an idempotent linear map. In a real inner product space, the special case $\phi (x)=b\cdot x$ gives $\phi (a)=a\cdot b$.

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.