QuantumInfo.ForMathlib.HermitianMat.Schatten

For a matrix $A \in \text{Mat}_{d \times d}(\mathbb{C})$ and an exponent $p \in \mathbb{R}$, the Schatten $p$-norm $\Vert A \Vert_{S^p}$ is defined as $$\Vert A \Vert_{S^p} = \left( \text{Tr}\left[ (A A^*)^{p/2} \right] \right)^{1/p}$$ where $A^*$ denotes the conjugate transpose of $A$, and the power of the positive semi-definite matrix $A A^*$ is defined via the continuous functional calculus. The result is returned as the real part of the trace.

Let $A$ be a Hermitian matrix of dimension $d$ over the complex numbers $\mathbb{C}$. If $A$ is positive semidefinite ($0 \le A$) and $p$ is a positive real number ($p > 0$), then the Schatten $p$-norm of the underlying matrix of $A$ (denoted $A.mat$) is given by $$\|A\|_p = (\text{Tr}(A^p))^{1/p},$$ where $A^p$ is defined via the functional calculus for Hermitian matrices and $\text{Tr}$ denotes the matrix trace.

For any $n \times n$ complex matrix $A$ and any real number $p$, the Schatten $p$-norm of $A$ is non-negative, i.e., $\|A\|_{S^p} \geq 0$.

Let $A$ be a positive semidefinite Hermitian matrix of size $d \times d$ over $\mathbb{C}$. For any positive real numbers $p > 0$ and $k > 0$, the Schatten $p$-norm of the matrix power $A^k$ is equal to the $k$-th power of the Schatten $(kp)$-norm of $A$. That is, $$\|A^k\|_{S^p} = (\|A\|_{S^{kp}})^k.$$

Let $A$ be a $d \times d$ complex Hermitian matrix such that $A$ is positive semi-definite ($0 \le A$). For any real number $r > 0$, the trace of the $r$-th power of $A$ is equal to the $r$-th power of its Schatten $r$-norm, denoted as $\text{tr}(A^r) = \|A\|_{S^r}^r$.

Let $A$ be a $d \times d$ complex matrix and let $p \in \mathbb{R}$. Let $M = A^* A$ be the Hermitian matrix formed by the product of the conjugate transpose of $A$ and $A$ itself. The real part of the trace of the functional calculus of $M$ applied to the power function $x \mapsto x^{p/2}$ is equal to the sum of the $p/2$-th powers of the eigenvalues of $M$: $$\text{Re}\left(\text{Tr}\left((A^* A)^{p/2}\right)\right) = \sum_{i=1}^d \lambda_i(A^* A)^{p/2}$$ where $\lambda_i(A^* A)$ denotes the $i$-th eigenvalue of the Hermitian matrix $A^* A$.

For any $d \times d$ complex matrix $A$ and any positive real number $p > 0$, the $p$-th power of the Schatten $p$-norm of $A$ is equal to the sum of the $p$-th powers of its singular values, i.e., $$\|A\|_{S^p}^p = \sum_{i \in d} \sigma_i(A)^p$$ where $\sigma_i(A)$ denotes the $i$-th singular value of $A$.

Let $A$ be a square complex matrix of size $d \times d$ and $p$ be a positive real number ($p > 0$). The Schatten $p$-norm of $A$, denoted $\|A\|_{S^p}$, is equal to the $p$-th root of the sum of the $p$-th powers of its singular values $\sigma_i(A)$: $$\|A\|_{S^p} = \left( \sum_{i \in d} \sigma_i(A)^p \right)^{1/p}$$

Let $A$ be a $d \times d$ complex matrix and $p$ be a positive real number. Let $\|A\|_{S^p}$ denote the Schatten $p$-norm of $A$, and let $\sigma^\downarrow_i(A)$ denote the $i$-th singular value of $A$ sorted in non-increasing order. Then the $p$-th power of the Schatten $p$-norm equals the sum of the $p$-th powers of the sorted singular values: $$\|A\|_{S^p}^p = \sum_{i=0}^{n-1} (\sigma^\downarrow_i(A))^p$$ where $n$ is the dimension of the matrix.

Let $A$ and $B$ be complex $d \times d$ Hermitian matrices that are positive semidefinite ($A \ge 0, B \ge 0$). For any real numbers $p, q > 1$ satisfying the conjugacy condition $\frac{1}{p} + \frac{1}{q} = 1$, the real inner product of $A$ and $B$ satisfies the Young inequality: $$\langle A, B \rangle_{\mathbb{R}} \leq \frac{\text{Tr}(A^p)}{p} + \frac{\text{Tr}(B^q)}{q}$$ where $\langle A, B \rangle_{\mathbb{R}} = \text{Tr}(AB)$ and $A^p, B^q$ are defined via functional calculus.

Let $A$ and $B$ be Hermitian matrices of size $d \times d$ over $\mathbb{C}$. If $A$ is positive semidefinite ($0 \le A$), then the product of the conjugate transpose of $(A^{1/2} B)$ and the matrix $(A^{1/2} B)$ is equal to the matrix $B A B$. Specifically, let $C = A^{1/2} B$, then $C^* C = B A B$.

Let $A$ and $B$ be $d \times d$ Hermitian matrices over $\mathbb{C}$ such that $A$ is positive semidefinite ($A \ge 0$). For any real number $\alpha > 0$, the Schatten $2\alpha$-norm of the product $A^{1/2}B$ raised to the power $2\alpha$ is equal to the trace of the $\alpha$-th power of the matrix $A B A$, denoted here as $(A.conj \ B.mat)^\alpha$. Specifically: $$\|A^{1/2} B\|_{S^{2\alpha}}^{2\alpha} = \text{tr}((A^{1/2} B^* B A^{1/2})^\alpha)$$ where $A^{1/2}$ is defined via functional calculus for Hermitian matrices.

Let $A$ and $B$ be complex $d \times d$ matrices. For any positive real numbers $r, p, q > 0$ satisfying the relation $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$, the Schatten $r$-norm of the product $AB$ is bounded by the product of the Schatten $p$-norm of $A$ and the Schatten $q$-norm of $B$: $$\|AB\|_{S^r} \le \|A\|_{S^p} \|B\|_{S^q}$$ where the Schatten $p$-norm is defined as $\|M\|_{S^p} = \left( \sum_{i} \sigma_i(M)^p \right)^{1/p}$ for singular values $\sigma_i(M)$. This inequality holds for both the norm $(p, q, r \ge 1)$ and quasi-norm $(0 < p, q, r < 1)$ cases.

Let $A$ and $B$ be positive semidefinite Hermitian matrices of size $d \times d$ over $\mathbb{C}$. Let $\alpha, p, q$ be positive real numbers such that $\frac{1}{2\alpha} = \frac{1}{p} + \frac{1}{q}$. Then the trace of the power of the matrix $A^{1/2}BA^{1/2}$ satisfies the inequality $$\text{Tr}((A^{1/2} B A^{1/2})^\alpha) \le \left( \text{Tr}(A^{p/2})^{1/p} \cdot \text{Tr}(B^q)^{1/q} \right)^{2\alpha},$$ where $A^r$ denotes the $r$-th power of a Hermitian matrix defined via functional calculus and $A. \text{conj } B. \text{mat}$ denotes the congruence action $A^{1/2} B A^{1/2}$.