QuantumInfo.ForMathlib.HermitianMat.Trace

Let $\alpha$ be a $\ast$-ring and $R$ be a subring of its self-adjoint elements such that $R$ is the maximal self-adjoint part of $\alpha$. For a Hermitian matrix $A \in \text{HermitianMat}_n(\alpha)$, the function $\text{trace}(A)$ returns the trace of $A$ as an element of $R$. For example, if $\alpha = \mathbb{C}$, then the trace is returned as a real number in $R = \mathbb{R}$.

Let $A$ be a Hermitian matrix with entries in $\alpha$, where $\alpha$ is an algebra over a ring $R$. Let $A.\text{mat}$ denote the underlying matrix of $A$, and $A.\text{trace}$ denote the trace of the Hermitian matrix as an element of $R$. Then the image of the Hermitian trace under the horizontal map $\text{algebraMap} : R \to \alpha$ is equal to the standard matrix trace $\text{Tr}(A.\text{mat})$.

Let $A$ be a Hermitian matrix with entries in $\alpha$, and let $r$ be a scalar from a self-adjoint ring $R$. The trace of the scalar multiplication of $r$ and $A$ is equal to the product of $r$ and the trace of $A$: $$\text{tr}(r \cdot A) = r \cdot \text{tr}(A)$$ where $\text{tr}$ denotes the trace specialized for Hermitian matrices that maps to the self-adjoint part $R$ of the base ring $\alpha$.

The trace of the zero Hermitian matrix $0 \in \text{HermitianMat}_n(\alpha)$ is equal to $0$ in the base ring $R$, where $R$ is the subring of self-adjoint elements.

Let $A$ and $B$ be Hermitian matrices of size $n$ with entries in $\alpha$. The trace of the sum of these matrices is equal to the sum of their individual traces, namely $\text{trace}(A + B) = \text{trace}(A) + \text{trace}(B)$. Here, the trace is defined such that it maps to the self-adjoint part $R$ of the scalar ring $\alpha$.

Let $A$ be a Hermitian matrix of size $n \times n$ over a star-ring $\alpha$. The trace of the negative of the matrix is equal to the negation of the trace of the matrix, i.e., $\text{tr}(-A) = -\text{tr}(A)$, where the trace is defined as a self-adjoint element of the underlying scalar structure.

For any two Hermitian matrices $A$ and $B$ of the same size, the trace of their difference is equal to the difference of their traces, i.e., $\text{tr}(A - B) = \text{tr}(A) - \text{tr}(B)$. Here, the trace of a Hermitian matrix is defined as a self-adjoint element of the underlying scalar ring (for example, a real number when the matrices are over the complex numbers).

Let $A$ be an $n \times n$ matrix over a star-ring $\alpha$. If $A$ is Hermitian, then its trace $\text{tr}(A)$ is a self-adjoint element of $\alpha$.

Let $A$ and $B$ be Hermitian matrices. Assuming a faithful scalar action of the base ring $R$ on the matrix entries $\alpha$, the trace of the Kronecker product of $A$ and $B$ is equal to the product of their individual traces, denoted as $\text{trace}(A \otimes_k B) = \text{trace}(A) \cdot \text{trace}(B)$. Here, the trace is defined such that it maps a Hermitian matrix to a self-adjoint element of its base ring.

For any Hermitian matrix $A$ with entries in $\mathbb{R}$, the value of the Hermitian trace `A.trace` is equal to the standard matrix trace $\text{tr}(A)$.

Let $A$ be a Hermitian matrix of size $n \times n$ over an `RCLike` field $\mathbb{k}$ (such as $\mathbb{R}$ or $\mathbb{C}$). The trace of $A$ as a Hermitian matrix coincides with the real part of the standard matrix trace of $A$. That is, $\text{tr}_{H}(A) = \text{Re}(\text{tr}(A))$, where $\text{tr}_{H}$ denotes the trace operation specifically defined for Hermitian matrices.

Let $n$ be a finite index set with decidable equality. For the identity Hermitian matrix $1$ over an $\mathbb{R}$-clike field $\mathbb{k}$ (such as $\mathbb{R}$ or $\mathbb{C}$), the trace of this matrix, denoted as $\text{trace}(1)$, is equal to the cardinality of the index set $n$.

Let $n$ be an index type and $\mathbb{k}$ be an `RCLike` field (such as $\mathbb{R}$ or $\mathbb{C}$). For any Hermitian matrix $A \in \text{HermitianMat}(n, \mathbb{k})$, the trace of the Hermitian matrix, $\text{tr}_{H}(A)$, is equal to the standard matrix trace of its underlying matrix, $\text{tr}(A.mat)$.

Let $T$ be a finite type and let $f: T \to \mathbb{R}$ be a function. Let $\verb|diagonal|_\mathbb{k} f$ be the Hermitian diagonal matrix over a field $\mathbb{k}$ whose diagonal entries are given by $f$. Then the trace of this Hermitian matrix is equal to the sum of the elements of $f$, i.e., $\text{trace}(\text{diag}(f)) = \sum_{i \in T} f(i)$.

Let $A$ be a Hermitian matrix with entries in an `RCLike` field $\mathbb{k}$ (such as $\mathbb{R}$ or $\mathbb{C}$). The sum of the eigenvalues of $A$ is equal to the trace of $A$, where the trace is considered as an element of the real numbers $\mathbb{R}$. Formally, $\sum_{i} \lambda_i = \text{tr}(A)$, where $\text{tr}(A)$ is the specialized trace for Hermitian matrices that maps to the self-adjoint part of the base field.

Let $A$ be a Hermitian matrix with entries in an `RCLike` field $\mathbb{k}$ (such as $\mathbb{R}$ or $\mathbb{C}$). The real-valued trace of $A$, denoted as `A.trace`, is equal to $0$ if and only if the standard matrix trace of $A$, denoted as `A.mat.trace`, is equal to $0$.

Let $n$ be an index type and $\mathbb{k}$ be an `RCLike` field (such as $\mathbb{R}$ or $\mathbb{C}$). For any $n \times n$ Hermitian matrix $A$ over $\mathbb{k}$, the self-adjoint trace $\text{tr}_{\text{sa}}(A)$ is equal to $1$ if and only if the standard matrix trace $\text{tr}(A)$ is equal to $1$. Here, the self-adjoint trace $\text{tr}_{\text{sa}}(A)$ is defined to be the real-valued trace (specifically, the real part of the matrix trace in the case of complex matrices).

Let $A$ be a Hermitian matrix with complex entries indexed by $n$. Given a bijection (equivalence) $e: n \simeq m$ between index sets $n$ and $m$, let $A'$ be the Hermitian matrix obtained by reindexing $A$ by $e$. Then the trace of $A'$ is equal to the trace of $A$. Specifically, since the trace of a Hermitian matrix is defined as a real number, this equality holds in $\mathbb{R}$.

Given a Hermitian matrix $A$ over a star-ring $\alpha$ with indices in the product space $m \times n$, the left partial trace $\text{traceLeft}(A)$ is a Hermitian matrix with indices in $n$. It is obtained by taking the left partial trace of the underlying matrix $A_{\text{mat}}$, where the result remains Hermitian. For an entry indexed by $(j_1, j_2) \in n \times n$, the operation is defined by summing over the first subsystem: $$(\text{traceLeft}(A))_{j_1, j_2} = \sum_{i \in m} (A_{\text{mat}})_{(i, j_1), (i, j_2)}$$

Given a Hermitian matrix $A \in \mathcal{H}_{m \times n}(\alpha)$ over a star-ring $\alpha$, where the index set is a product $m \times n$, the right partial trace $\text{traceRight}(A)$ is a Hermitian matrix in $\mathcal{H}_m(\alpha)$. It is defined by applying the matrix right partial trace to the underlying matrix of $A$: $$(\text{traceRight}(A))_{i, j} = \sum_{k \in n} A_{(i, k), (j, k)}$$ The result is guaranteed to be a Hermitian matrix of size $m \times m$ if the original matrix $A$ is Hermitian.

Let $A$ be a Hermitian matrix with indices in $m \times n$. Let $A.\text{mat}$ denote the underlying matrix in $M_{m \times n \times m \times n}(\alpha)$. Then the underlying matrix of the left partial trace of $A$ is equal to the left partial trace of the underlying matrix $A$. That is, $$(\text{traceLeft}(A)).\text{mat} = \text{traceLeft}(A.\text{mat})$$ where the left partial trace for a matrix $M$ is defined as $(\text{traceLeft}(M))_{i_1, j_1} = \sum_{i_2 \in m} M_{(i_2, i_1), (i_2, j_1)}$.

Let $A$ and $B$ be Hermitian matrices with row and column indices in $d \times n$ over a star-additive monoid. The left partial trace of their sum is equal to the sum of their individual left partial traces: $$\text{Tr}_L(A + B) = \text{Tr}_L(A) + \text{Tr}_L(B)$$ where the left partial trace $\text{Tr}_L$ of a Hermitian matrix $M$ is defined such that its underlying matrix $\text{mat}(\text{Tr}_L(M))$ coincides with the standard matrix partial trace over the first subsystem.

For any Hermitian matrix $A$ whose row and column indices are structured as products $d \times d_1$ and $d \times d_2$ respectively, the left partial trace operation commutes with negation: $(-A)^{\text{traceLeft}} = -(A^{\text{traceLeft}})$. Here, $\text{traceLeft}$ denotes the partial trace taken over the first component $d$ of the index products, resulting in a Hermitian matrix of size $d_1 \times d_2$.

Let $A$ and $B$ be Hermitian matrices whose row and column indices are structured as the product of two index sets $d \times n$. The left partial trace operation on Hermitian matrices, denoted $\text{traceLeft}$, is distributive over subtraction: $$(A - B).\text{traceLeft} = A.\text{traceLeft} - B.\text{traceLeft}$$ where the left partial trace of a Hermitian matrix is defined such that its entries are the sums over the first index component $i \in d$, maintaining the Hermitian property in the resulting $n \times n$ matrix.

Let $A$ be a Hermitian matrix with row indices in $m \times n$ and column indices in $m \times n$. The matrix representation of the right partial trace of $A$, denoted as $(\text{traceRight}(A)).\text{mat}$, is equal to the right partial trace of the matrix representation of $A$, denoted as $A.\text{mat}.\text{traceRight}$. Here, the right partial trace $(\text{traceRight}(M))_{i, j}$ of a matrix $M \in M_{(m \times n) \times (m \times n)}$ is defined as the sum over the second index: $\sum_{k \in n} M_{(i, k), (j, k)}$.

For any Hermitian matrices $A$ and $B$ whose row and column indices are structured as $m \times n$, the right partial trace satisfies the additivity property $(A + B)^{\text{traceRight}} = A^{\text{traceRight}} + B^{\text{traceRight}}$. Here, the right partial trace $\text{traceRight}$ denotes the operation of taking the trace over the second component $n$ of the index products, resulting in a Hermitian matrix with indices in $m$.

Let $A$ be a Hermitian matrix with row and column indices in $m \times n$. The right partial trace of the negation of $A$ is equal to the negation of the right partial trace of $A$, i.e., $\text{traceRight}(-A) = -\text{traceRight}(A)$.

Let $A$ and $B$ be Hermitian matrices whose row and column indices are structured as $m \times n$. The right partial trace $\text{traceRight}$ satisfies the distributive property over subtraction: $$(A - B)^{\text{traceRight}} = A^{\text{traceRight}} - B^{\text{traceRight}}$$ where $A^{\text{traceRight}}$ denotes the Hermitian matrix obtained by taking the partial trace over the second subsystem (the $n$ component of the index product).

Let $A$ be a Hermitian matrix with indices in $m \times n$, and let $r$ be a real number. The left partial trace of the scalar product $r \cdot A$ is equal to the scalar product of $r$ and the left partial trace of $A$, denoted as $\text{Tr}_{\text{left}}(r \cdot A) = r \cdot \text{Tr}_{\text{left}}(A)$. Here, the operation $\text{Tr}_{\text{left}}$ maps a Hermitian matrix to a smaller Hermitian matrix by summing over the first subsystem's indices.

Let $A$ be a Hermitian matrix in $M_{(n \times m) \times (n \times m)}(\alpha)$ and $r$ be a real number. The right partial trace of the scalar multiplication of $A$ by $r$ is equal to $r$ times the right partial trace of $A$, that is, $\text{traceRight}(r \cdot A) = r \cdot \text{traceRight}(A)$.

Let $A$ be a Hermitian matrix with row and column indices in $d_1 \times d_2$. The trace of the left partial trace of $A$ is equal to the trace of $A$: $$\text{tr}(\text{traceLeft}(A)) = \text{tr}(A)$$ where $\text{traceLeft}(A)$ denotes the partial trace of the Hermitian matrix over the first subsystem, and $\text{tr}$ denotes the trace of a Hermitian matrix (which, for scalars in an `RCLike` field like $\mathbb{C}$, corresponds to the real part of the standard matrix trace).

Let $A$ be a Hermitian matrix with row and column indices in $d_1 \times d_2$. The trace of its right partial trace $\text{traceRight}(A)$ is equal to the trace of $A$, i.e., $$\text{tr}(\text{traceRight}(A)) = \text{tr}(A)$$ where the trace of a Hermitian matrix is take in the sense of `HermitianMat.trace` (mapping to the maximal self-adjoint part of the scalars).

Let $A$ be a Hermitian matrix of size $m \times m$ and $B$ be a Hermitian matrix of size $n \times n$ over an $\mathbb{R}$-clike field $\mathbb{k}$. Assume the index set for $m$ is finite. Then the left partial trace of the Kronecker product $A \otimes B$ is equal to the product of the trace of $A$ and the matrix $B$, i.e., $$\text{traceLeft}(A \otimes B) = \text{Tr}(A) \cdot B$$ where $\text{Tr}(A)$ is the real-valued trace of the Hermitian matrix $A$.

Let $A$ be an $m \times m$ Hermitian matrix and $B$ be an $n \times n$ Hermitian matrix. Let $A \otimes_k B$ denote the Kronecker product of $A$ and $B$, which is a Hermitian matrix over the index set $(m \times n)$. Then the right partial trace of the Kronecker product is given by: $$\text{traceRight}(A \otimes_k B) = \text{trace}(B) \cdot A$$ where $\text{trace}(B)$ is the trace of $B$ (viewed as an element of the self-adjoint part of the base ring) and $\cdot$ denotes scalar multiplication.