QuantumInfo.ForMathlib.MatrixNorm.TraceNorm

For a matrix $A \in \text{Mat}_{m,n}(R)$, the trace norm (also known as the nuclear norm or Schatten 1-norm) is defined as the real part of the trace of the square root of the product of its conjugate transpose $A^\dagger$ and itself. That is, $\|A\|_{\text{tr}} = \text{Re}(\text{Tr}[\sqrt{A^\dagger A}])$.

The trace norm of the zero matrix $0 \in \mathbb{M}_{m \times n}(R)$ is equal to $0$. Here, $\|0\|_{\text{tr}} = 0$.

For any $m \times n$ matrix $A$ over a ring $R$, the trace norm of its negative is equal to the trace norm of the matrix itself, denoted as $\| -A \|_{\text{tr}} = \| A \|_{\text{tr}}$.

Let $A$ be a positive semi-definite $n \times n$ matrix over $R$ ($0 \le A$), and let $u$ be an $m \times n$ isometry matrix such that $u^H u = I_n$. Then the square root of the conjugation of $A$ by $u$ is equal to the conjugation of the square root of $A$ by $u$. That is, $$\sqrt{u A u^H} = u \sqrt{A} u^H,$$ where $\sqrt{\cdot}$ denotes the functional calculus square root for matrices.

Let $A$ be an $n \times m$ matrix over $R$ and $u$ be a $k \times n$ matrix over $R$, where the index set of $k$ is finite. If $u$ is an isometry (i.e., $u^H u = I_n$), then the trace norm of the product $u A$ is equal to the trace norm of $A$, denoted as $\|u A\|_{\text{tr}} = \|A\|_{\text{tr}}$.

Let $A \in \text{Matrix}_{n \times m}(R)$ and $u \in \text{Matrix}_{k \times m}(R)$. If $u$ is an isometry (i.e., $u^H u = I_m$), then the trace norm of the product of $A$ and the conjugate transpose of $u$ is equal to the trace norm of $A$: $\|A u^H\|_{\text{tr}} = \|A\|_{\text{tr}}$.

Let $A$ be a square matrix of size $n \times n$ over $R$. For any $m \times n$ matrices $u$ and $v$ that are isometries (i.e., $u^H u = I_n$ and $v^H v = I_n$), the trace norm of the product $u A v^H$ is equal to the trace norm of $A$: $$\|u A v^H\|_{\text{tr}} = \|A\|_{\text{tr}}$$ where $v^H$ denotes the conjugate transpose of $v$.

Let $R$ be a ring and $A \in \text{Matrix}_{n \times n}(R)$ be a square matrix. For any unitary matrix $U$ in the unitary group $U(n, R)$, the trace norm of the conjugated matrix $U A U^H$ is equal to the trace norm of $A$, where $U^H$ denotes the conjugate transpose of $U$.

Let $A$ be an $n \times n$ Hermitian matrix with entries in $\mathbb{R}$ or $\mathbb{C}$. The trace norm of $A$, denoted $\|A\|_{\text{tr}}$, is equal to the sum of the absolute values of its eigenvalues $\lambda_i$: $$\|A\|_{\text{tr}} = \sum_{i} |\lambda_i|$$

For any $m \times n$ matrix $A$ over a ring $R$ (where $R$ is typically $\mathbb{R}$ or $\mathbb{C}$), the trace norm $\|A\|_{\text{tr}}$ is non-negative, i.e., $0 \leq \|A\|_{\text{tr}}$.

Let $A$ be an $m \times n$ matrix over a field $R$. The trace norm of $A$, denoted $\|A\|_{\text{tr}}$, is equal to zero if and only if $A$ is the zero matrix ($A = 0$).

For any matrix $A \in \text{Matrix}_{m \times n}(R)$ and any scalar $c \in R$, the trace norm of the scalar product $c \cdot A$ is equal to the absolute value (or norm) of $c$ multiplied by the trace norm of $A$: $\|c \cdot A\|_{\text{tr}} = \|c\| \cdot \|A\|_{\text{tr}}$.

For any square matrix $A \in M_n(R)$, the trace norm $\|A\|_{\text{tr}}$ is the maximum value of the set of real numbers obtained by the traces of the product of $A$ and any unitary matrix $U$. Specifically, $\|A\|_{\text{tr}}$ is the greatest element of the set $\{\text{Tr}(UA) \mid U \in \mathcal{U}(n)\}$, where $\mathcal{U}(n)$ denotes the unitary group of degree $n$.

Let $A$ and $B$ be $n \times n$ square matrices over a ring $R$ (typically $\mathbb{R}$ or $\mathbb{C}$). The trace norm $\|\cdot\|_{\text{tr}}$ satisfies the triangle inequality: $$\|A + B\|_{\text{tr}} \leq \|A\|_{\text{tr}} + \|B\|_{\text{tr}}$$ where $\|M\|_{\text{tr}}$ denotes the trace norm (also known as the nuclear norm or Ky Fan $n$-norm) of a matrix $M$.

For any $n \times n$ matrices $A$ and $B$ with entries in $R$, the trace norm of their difference is less than or equal to the sum of their trace norms: $$\|A - B\|_{\text{tr}} \le \|A\|_{\text{tr}} + \|B\|_{\text{tr}}$$ where $\|\cdot\|_{\text{tr}}$ denotes the trace norm (also known as the nuclear norm).

Let $A$ be a positive semidefinite matrix of size $m \times m$ over a ring $R$. Then the trace norm of $A$, denoted $\|A\|_{\text{tr}}$, is equal to its trace, $\operatorname{Tr}(A)$.

For any two matrices $M, N \in \text{Mat}_{n \times n}(R)$ and any real number $0 \le \lambda \le 1$, the trace norm $\|\cdot\|_{\text{tr}}$ satisfies the convexity inequality: $$\|\lambda M + (1 - \lambda) N\|_{\text{tr}} \le \lambda \|M\|_{\text{tr}} + (1 - \lambda) \|N\|_{\text{tr}}$$ where the trace norm of a matrix $A$ is denoted by $A.\text{traceNorm}$.