QuantumInfo.ForMathlib.HermitianMat.Rpow

Given a Hermitian matrix $A \in \text{HermitianMat}_d(\mathbb{K})$ and a real number $r \in \mathbb{R}$, the matrix power $A^r$ is defined via the continuous functional calculus (CFC). Specifically, if $A$ is diagonalized as $A = U \Lambda U^\dagger$, then $A^r = U \Lambda^r U^\dagger$, where the power function is applied elementwise to the diagonal entries of $\Lambda$ using the real power function $x \mapsto x^r$. Note that if an eigenvalue is $0$ and $r < 0$, the result follows the convention $0^r = 0$.

For a finite-dimensional Hilbert space $\mathbb{k}^d$, let $\text{Herm}_d(\mathbb{k})$ denote the set of $d \times d$ Hermitian matrices over the field $\mathbb{k} \in \{\mathbb{R}, \mathbb{C}\}$. This definition provides an instance of the real power operator $\text{^} : \text{Herm}_d(\mathbb{k}) \times \mathbb{R} \to \text{Herm}_d(\mathbb{k})$. For a Hermitian matrix $A$ and a real exponent $r$, $A^r$ is defined via the functional calculus (specifically `HermitianMat.rpow`), such that if $A$ has spectral decomposition $\sum_i \lambda_i P_i$, then $A^r = \sum_i \lambda_i^r P_i$.

Let $A$ be a Hermitian matrix of dimension $d$ over the field $\mathbb{K}$, and let $U$ be a unitary matrix (an element of the unitary group $d$). For any real exponent $r \in \mathbb{R}$, the $r$-th power of the conjugated matrix $U A U^*$ is equal to the conjugation of the $r$-th power of $A$ by $U$. That is, $(U A U^*)^r = U (A^r) U^*$.

For a Hermitian matrix $A$ and a real number $r$, the power operation $A^r$ is equivalent to the matrix power function `rpow` applied to $A$ and $r$.

Let $A$ be a Hermitian matrix and $r$ be a real number. Then the matrix power $A^r$ is equal to the result of applying the continuous functional calculus to $A$ with the real power function $x \mapsto x^r$.

Let $f: d \to \mathbb{R}$ be a function and $M = \text{diag}(f)$ be the diagonal matrix whose entries are given by $f(i)$ for $i \in d$. For any real exponent $r \in \mathbb{R}$, the $r$-th power of the diagonal matrix is the diagonal matrix of the $r$-th powers of its entries, i.e., $(\text{diag}(f))^r = \text{diag}(i \mapsto f(i)^r)$.

Let $\mathbb{C}^{d \times d}$ be the space of $d \times d$ complex matrices, and let $\text{HermitianMat}(d, \mathbb{C})$ denote the subspace of Hermitian matrices. For any real number $r \geq 0$, the function that maps a Hermitian matrix $A$ to its matrix power $A^r$, defined via the continuous functional calculus, is continuous.

Let $A$ be a non-singular Hermitian matrix in $\mathbb{C}^{d \times d}$. Then the map $r \mapsto A^r$ is continuous for all $r \in \mathbb{R}$, where $A^r$ denotes the matrix power with real exponent.

Let $A$ be a Hermitian matrix in $\mathbb{C}^{d \times d}$. Then the map $x \mapsto A^x$ is continuous on the interval $(0, \infty)$, where $A^x$ denotes the matrix power of $A$ with real exponent $x$.

Let $A$ be a Hermitian matrix in $\mathbb{C}^{d \times d}$. The function $x \mapsto A^x$, which maps a real number $x$ to the $x$-th power of the matrix $A$, is continuous on the interval $(-\infty, 0)$.

For any Hermitian matrix $A$, raising $A$ to the power of the real number $1$ results in the matrix $A$ itself, i.e., $A^1 = A$.

For any Hermitian matrix $A \in \text{HermitianMat}_d(\mathbb{k})$, the square root of $A$ (defined via the continuous functional calculus of $\sqrt{x}$) is equal to the continuous functional calculus of the function $f(x) = x^{1/2}$ applied to $A$.

Let $I$ denote the identity matrix in the space of Hermitian matrices of dimension $d$ over the field $\mathbf{k}$. For any real number $r \in \mathbb{R}$, the $r$-th power of the identity matrix is equal to the identity matrix. That is, $$I^r = I$$ where the power is defined via the continuous functional calculus for Hermitian matrices.

For any Hermitian matrix $A$, the real power of $A$ with exponent $0$ is the identity matrix $I$. That is, $$A^0 = I$$ where $I$ denotes the identity matrix of the same dimension as $A$.

Let $d$ be a finite index set and $a: d \to \mathbb{R}$ be a function representing the diagonal entries of a square matrix. For any real exponent $r \in \mathbb{R}$, the $r$-th power of the diagonal matrix with entries $a_i$ is equal to the diagonal matrix whose entries are the $r$-th powers of the original entries. That is, $$(\text{diag}(a))^r = \text{diag}(a^r)$$ where the $i$-th diagonal entry of the resulting matrix is $a_i^r$.

Let $A$ be a Hermitian matrix indexed by $d$, and let $e: d \simeq d_2$ be a bijection between index sets. For any real exponent $r \in \mathbb{R}$, the $r$-th power of the matrix $A$ reindexed by $e$ is equal to the reindexing of the $r$-th power of $A$. That is, $$(A_{\text{reindexed by } e})^r = (A^r)_{\text{reindexed by } e}$$ where the reindexing operation permutes the rows and columns of the matrix according to the equivalence $e$.

Let $A$ be a semidefinite Hermitian matrix ($A \ge 0$). For any real numbers $p$ and $q$ satisfying $p + q \neq 0$, the $(p+q)$-th power of $A$ is equal to the product of the $p$-th power of $A$ and the $q$-th power of $A$. That is, $$A^{p+q} = A^p A^q$$ provided that the sum of the exponents is non-zero.

Let $A$ be a Hermitian matrix such that $0 \le A$ (i.e., $A$ is positive semidefinite). For any real numbers $p$ and $q$, the matrix powers satisfy the identity $A^{p \cdot q} = (A^p)^q$. Here, $A^r$ denotes the power of a Hermitian matrix with a real exponent $r$.

Let $A$ be a positive semidefinite Hermitian matrix ($0 \le A$). For any real numbers $q$ and $r$ such that $q \neq 0$ and $r + 2q \neq 0$, the conjugation of $A^q$ by $A^r$, defined as $(A^r) (A^q) (A^r)^*$, is equal to $A^{r + 2q}$.

Let $A$ be a positive semi-definite Hermitian matrix (denoted $0 \le A$). Then the product of the matrix power $A^{1/2}$ with itself is equal to $A$, i.e., $A^{1/2} \cdot A^{1/2} = A$.

Let $A$ be a Hermitian matrix in $\mathbb{C}^{d \times d}$ (or $\mathbb{R}^{d \times d}$). If $A$ is positive definite (denoted $A > 0$), then for any real power $p \in \mathbb{R}$, the matrix power $A^p$ is also positive definite, namely $A^p > 0$.

Let $A$ be a Hermitian matrix such that $A \ge 0$ (i.e., $A$ is positive semi-definite). Then for any real power $p \in \mathbb{R}$, the matrix power $A^p$ is also positive semi-definite, denoted as $A^p \ge 0$.

Let $A$ be a Hermitian matrix. If $A$ is positive definite (denoted as $A > 0$), then its inverse matrix $A^{-1}$ is equal to its real power with exponent $-1$, namely $A^{-1} = A^{-1}_{\mathbb{R}}$.

Let $B$ be a positive definite Hermitian matrix. Then the conjugation of $B$ by the matrix $B^{-1/2}$ results in the identity matrix, i.e., $B^{-1/2} B (B^{-1/2})^* = I$.

Let $A$ be a Hermitian matrix in $\mathbb{K}^{d \times d}$ and let $p$ be a real number. If $A$ is positive definite ($A > 0$), then the inverse of the $p$-th power of $A$ is equal to $A$ raised to the power of $-p$, i.e., $(A^p)^{-1} = A^{-p}$.

Let $A$ and $B$ be Hermitian matrices of dimension $d$ over a field $\mathbb{k}$. If $B$ is a positive definite matrix and $A \le B$ in the Loewner order (meaning $B - A$ is positive semidefinite), then the "sandwich" operation of $A$ by $B^{-1/2}$ satisfies $$B^{-1/2} A B^{-1/2} \le I$$ where $I$ is the identity matrix and $B^{-1/2}$ denotes the real power of the Hermitian matrix with exponent $-1/2$.

Let $A$ be a Hermitian matrix and assume that $A$ is positive definite. For any real number $p$, the product of the matrix raised to the power $-p$ and the matrix raised to the power $p$ is the identity matrix, i.e., $A^{-p} A^p = I$.

Let $A$ be a Hermitian matrix and $p$ be a real number. If $A$ is positive definite, then the matrix power $A^p$ is an invertible matrix.

Let $A$ and $B$ be Hermitian matrices. Suppose $B$ is positive definite. Then the inverse of the congruence transformation of $A$ by $B^{-1/2}$ is equal to the congruence transformation of $A^{-1}$ by $B^{1/2}$. That is, $$((B^{-1/2})^* A (B^{-1/2}))^{-1} = (B^{1/2})^* A^{-1} (B^{1/2})$$ where $M^*$ denotes the conjugate transpose and $X \mapsto S^* X S$ is the sandwich (congruence) operator.

Let $A$ be a $d \times d$ Hermitian matrix over the complex numbers $\mathbb{C}$. If $A$ is positive semi-definite (i.e., $A \geq 0$) and $r$ is a non-zero real number, then the kernel of the matrix power $A^r$ is equal to the kernel of $A$. In symbols, $\ker(A^r) = \ker(A)$.

Let $A$ be a Hermitian matrix of dimension $d$ over the complex numbers $\mathbb{C}$. If $A$ is positive semi-definite (i.e., $A \ge 0$), then the kernel of the matrix power $A^r$ is a subspace of the kernel of $A$, denoted as $\ker(A^r) \subseteq \ker(A)$.

Let $A$ be a $d \times d$ matrix and $B$ be a $d_2 \times d_2$ matrix over a field $\mathbb{k}$. Let $C$ and $D$ be Hermitian matrices of dimensions $d$ and $d_2$ respectively. Then the congruence transformation of the Kronecker product $C \otimes D$ by the Kronecker product $A \otimes B$ is equal to the Kronecker product of the individual congruence transformations. That is, $$(A \otimes B) (C \otimes D) (A \otimes B)^* = (A C A^*) \otimes (B D B^*)$$ where $\text{conj } M P = M P M^*$ denotes the congruence transformation of $P$ by $M$, and $\otimes_k$ denotes the Kronecker product.

Let $A$ and $B$ be complex Hermitian matrices of dimensions $d \times d$ and $d_2 \times d_2$ respectively. If $A$ and $B$ are positive semi-definite (i.e., $A \ge 0$ and $B \ge 0$ in the Loewner order), then for any real number $r$, the $r$-th power of their Kronecker product is equal to the Kronecker product of their individual $r$-th powers. That is, $$(A \otimes B)^r = A^r \otimes B^r$$ where $\otimes$ denotes the Kronecker product and $M^r$ denotes the matrix power with a real exponent.

Let $K \subset \mathbb{R}$ be a compact set. Let $I = (0, \infty)$ be the set of positive real numbers. The function that maps an exponent $r \in I$ to the power function $f_r(t) = t^r$ is continuous on $I$ with respect to the topology of uniform convergence on $K$. Specifically, the map $r \mapsto (t \mapsto t^r)$ is continuous from the subset $(0, \infty)$ of the real numbers to the space of functions on $K$ endowed with the uniform norm.

Let $X$ be a topological space and $S \subseteq X$ a subset. Suppose $A: X \to \mathcal{H}_d(\mathbb{C})$ is a continuous mapping from $S$ into the space of $d \times d$ complex Hermitian matrices, and $p: X \to \mathbb{R}$ is a continuous real-valued function on $S$. If $p(x) > 0$ for all $x \in S$, then the mapping $x \mapsto A(x)^{p(x)}$ is continuous on $S$, where $A^p$ denotes the matrix power defined via the continuous functional calculus.

Let $A$ be a Hermitian matrix of dimension $d$ over the complex numbers $\mathbb{C}$, and assume that $A$ is positive semidefinite ($0 \le A$). For any real number $p > 0$, the continuous functional calculus of $A^2$ applied to the function $f(x) = x^{p/2}$ is equal to the continuous functional calculus of $A$ applied to the function $g(x) = x^p$. That is, $(A^2)^{p/2} = A^p$.

For any complex Hermitian matrix $A$ of dimension $d$ and any real number $p$, the trace of the $p$-th power of $A$ is equal to the sum of the $p$-th powers of its eigenvalues. Specifically, if $\lambda_i$ denotes the $i$-th eigenvalue of $A$ for $i \in \{1, \dots, d\}$, then $$\text{Tr}(A^p) = \sum_{i=1}^d \lambda_i^p.$$

For a complex Hermitian matrix $A \in \text{Herm}_d(\mathbb{C})$ and real numbers $q, T \in \mathbb{R}$, we define the finite integral approximation: $$\text{rpowApprox}(A, q, T) = \int_0^T t^q \left( \frac{1}{1+t} I - (A + t I)^{-1} \right) \, dt$$ where $I$ denotes the identity matrix. This expression serves as a truncated integral representation used to prove the operator monotonicity of the power function $A \mapsto A^q$.

Let $A$ and $B$ be complex $d \times d$ Hermitian matrices. Suppose $A$ and $B$ are positive definite (denoted $A \succ 0$ and $B \succ 0$). If $A \le B$ in the Loewner order (meaning $B - A$ is positive semidefinite), then for any real number $q \ge 0$ and any real parameter $T > 0$, the approximating operators satisfy $rpowApprox(A, q, T) \le rpowApprox(B, q, T)$.

Given real numbers $q$, $T$, and $x$, this function is defined as the integral: $$f(q, T, x) = \int_{0}^{T} t^q \left( \frac{1}{1+t} - \frac{1}{x+t} \right) dt$$ This function serves as the scalar basis for the approximation of the real power $A^q$ of a Hermitian matrix $A$ via the Continuous Functional Calculus (CFC).

Let $A$ be a positive definite Hermitian matrix of dimension $d$ over $\mathbb{C}$. For any real numbers $q$ and $T$ such that $q \ge 0$ and $T > 0$, the approximated matrix power $A^q_T$ (denoted by `rpowApprox A q T`) is equal to the result of applying the scalar function $x \mapsto f_{q,T}(x)$ (denoted by `scalarRpowApprox q T`) to the matrix $A$ via the continuous functional calculus.

For a real number $q$, the function returns the value of the integral $$\int_0^\infty \frac{u^{q-1}}{1 + u} \, du$$ as a real number. This constant appears in the integral representation of the power function $x^q$ and is equal to $\frac{\pi}{\sin(\pi q)}$ for $0 < q < 1$.

Let $q$ be a real number such that $0 < q < 1$. Then the function $f(u) = \frac{u^{q-1}}{1+u}$ is integrable on the interval $(0, \infty)$.

Let $q$ be a real number such that $0 < q < 1$. Then the constant $C_q$ (defined as `rpowConst q`) is strictly positive.

For any real numbers $x$, $q$ such that $x > 0$, $0 < q < 1$, the scalar function $f(T) = \text{scalarRpowApprox}(q, T, x)$ converges to $C_q \cdot (x^q - 1)$ as $T \to \infty$, where $C_q$ denotes the constant $\text{rpowConst}(q)$.

Let $A$ be a positive definite Hermitian matrix of dimension $d$ over $\mathbb{C}$. For any real number $q$ such that $0 < q < 1$, the matrix-valued function $T \mapsto \text{rpowApprox}(A, q, T)$ converges to $C_q \cdot (A^q - I)$ as $T \to \infty$ in the topology of the normed space of Hermitian matrices, where $C_q$ denotes the constant $\text{rpowConst}(q)$ and $I$ is the identity matrix.

Let $A$ and $B$ be complex Hermitian matrices of dimension $d$. If $A$ is positive definite ($A > 0$), $A$ is less than or equal to $B$ in the Loewner order ($A \le B$), and $q$ is a real number such that $0 < q \le 1$, then $A^q \le B^q$.

Let $A$ and $B$ be complex Hermitian matrices of dimension $d$. If $0 \le A \le B$ and $0 < q \le 1$, then $A^q \le B^q$, where the inequalities denote the Loewner partial order.

Let $A$ and $B$ be positive semi-definite $d \times d$ complex Hermitian matrices (i.e., $A \ge 0$ and $B \ge 0$). For any real number $r$ such that $0 < r \le 1$, the following inequality of Lieb-Thirring type holds for the trace: $$\text{Tr}\left( B^r A^r B^r \right) \le \text{Tr}\left( (B A B)^r \right)$$ Here, $X^r$ denotes the $r$-th power of a Hermitian matrix $X$, and for a Hermitian matrix $H$ and matrix $M$, $H.\text{conj}(M)$ represents the congruent transformation $M H M^*$. In the case where $M = B^r$ or $M = B$, the trace terms correspond to $\text{Tr}(B^r A^r B^r)$ and $\text{Tr}((B A B)^r)$ respectively.

Let $A$ and $B$ be complex Hermitian matrices of dimension $d$. If $A$ and $B$ are positive semi-definite (i.e., $A \ge 0$ and $B \ge 0$) and $p$ is a real number such that $0 < p \le 1$, then the trace of the sum raised to the power $p$ is less than or equal to the sum of the traces of the individual matrices raised to the same power: $$\text{tr}((A + B)^p) \le \text{tr}(A^p) + \text{tr}(B^p)$$