PhyslibAlpha.AlgebraicFramework.HilbertSpace.TraceClass.Basic

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. $T$ is said to be **trace-class** if there exists a Hilbert basis $\{e_i\}_{i \in I}$ of $H$ such that the sum $$\sum_{i \in I} \langle e_i, |T| e_i \rangle$$ converges, where $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus.

Let $H$ be a complex Hilbert space. For any Hilbert basis $\{e_i\}_{i \in I}$ of $H$ and any vector $y \in H$, the series of the squared magnitudes of the inner products between the basis vectors and $y$ converges to the squared norm of $y$: $$\sum_{i \in I} |\langle e_i, y \rangle|^2 = \|y\|^2.$$ This is known as **Parseval's identity**.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is trace-class, then for any Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the collection of real values $\text{Re}(\langle e_i, |T| e_i \rangle)$ is summable. Here, $|T| = \sqrt{T^* T}$ denotes the absolute value of $T$ defined via the continuous functional calculus.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. $T$ is a trace-class operator if and only if for every Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the sum $$\sum_{i \in I} \langle e_i, |T| e_i \rangle$$ converges, where $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus.

Let $H$ be a finite-dimensional complex Hilbert space. Then every continuous linear operator $T : H \to H$ is a trace-class operator.

Let $H$ be a finite-dimensional complex Hilbert space and $T: H \to H$ be a continuous linear operator. For any Hilbert basis $\{e_i\}_{i \in w}$ of $H$, the linear-algebraic trace of $T$ is equal to the sum of its diagonal inner products: $$\text{Tr}(T) = \sum_{i \in w} \langle e_i, T e_i \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a trace-class operator. The **trace norm** (or 1-norm) of $T$, denoted as $\|T\|_1$, is the real number defined by the sum: $$\|T\|_1 = \sum_{i \in I} \langle e_i, |T| e_i \rangle$$ where $\{e_i\}_{i \in I}$ is a Hilbert basis of $H$ and $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus. Although the formal definition utilizes a witness basis from the proof that $T$ is trace-class, the resulting value is independent of the choice of basis.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $h_1$ and $h_2$ are two proofs (or witnesses) that $T$ is trace-class, then the value of the trace norm $\|T\|_1$ defined using $h_1$ is equal to the value defined using $h_2$.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator that is trace-class. The trace of $T$, denoted as $\text{Tr } T$, is defined as the sum of the diagonal inner products $$\text{Tr } T = \sum_{i \in I} \langle e_i, T e_i \rangle$$ where $\{e_i\}_{i \in I}$ is the Hilbert basis of $H$ used to satisfy the definition of $T$ being trace-class. While the definition uses a specific witness basis, for a trace-class operator, this sum converges unconditionally and is independent of the choice of basis.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a trace-class bounded linear operator. For any Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the trace norm $\|T\|_1$ is equal to the sum of the diagonal inner products of the absolute value operator $|T| = \sqrt{T^* T}$: $$\|T\|_1 = \sum_{i \in I} \langle e_i, |T| e_i \rangle$$ Note that since $|T|$ is a positive operator, the inner products $\langle e_i, |T| e_i \rangle$ are non-negative real numbers.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a trace-class bounded linear operator. The trace norm of $T$, denoted as $\|T\|_1$, is non-negative: $$\|T\|_1 \geq 0$$

Let $H$ be a complex Hilbert space. If $T : H \to H$ is a non-negative ($T \geq 0$) trace-class operator, then for any Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the trace of $T$ is equal to the sum of its diagonal entries with respect to that basis: $$\text{Tr } T = \sum_{i \in I} \langle e_i, T e_i \rangle$$

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is self-adjoint and trace-class, then for any Hilbert basis $\{b_i\}_{i \in I}$ of $H$, the trace of $T$ is equal to the sum of the diagonal inner products: $$\text{Tr } T = \sum_{i \in I} \langle b_i, T b_i \rangle.$$

Let $H$ be a complex Hilbert space and $p: H \to H$ be a bounded linear operator. The operator $p$ has **finite multiplicity** if it is a star-projection (meaning $p^2 = p$ and $p^* = p$) and it is a trace-class operator. This condition is the Murray–von Neumann finiteness condition, which for a projection is equivalent to its range being finite-dimensional.

Let $H$ be a complex Hilbert space and $p: H \to H$ be a bounded linear operator. If $p$ has finite multiplicity (meaning $p$ is a trace-class star-projection), then the range of $p$ is finite-dimensional over $\mathbb{C}$.

Let $H$ be a complex Hilbert space and $p: H \to H$ be a bounded linear operator. If $p$ has finite multiplicity (meaning it is a star-projection and a trace-class operator), then its trace $\text{Tr } p$ is equal to the dimension of its range, $\text{ran } p$, over the complex numbers, written as $\text{Tr } p = \dim_{\mathbb{C}}(\text{ran } p)$.

Let $H$ be a finite-dimensional complex Hilbert space. For any bounded linear operator $p: H \to H$ with finite multiplicity, the trace of $p$ is equal to the dimension of its range as a complex vector space: $$\text{Tr } p = \dim_{\mathbb{C}}(\text{ran } p)$$