QuantumInfo.ForMathlib.Isometry

A matrix $A \in \text{Matrix}_{d \times d_2}(R)$ is an **isometry** if the product of its conjugate transpose $A^H$ and itself is the identity matrix, i.e., $A^H A = I_{d_2}$.

Let $I$ be the identity matrix of type $d \times d$ over a ring $R$. For any functions $e: d_2 \to d$ and $f: d_3 \to d$, if $e$ is bijective and $f$ is injective, then the submatrix of $I$ formed by indexing rows with $e$ and columns with $f$ (denoted as $I_{e(i), f(j)}$) is an isometry.

Let $I$ be the $d \times d$ identity matrix over a ring $R$. For any bijective function $e: d_2 \to d$, the submatrix of $I$ formed by indexing the rows with $e$ and the columns with the identity function $\text{id}$ is an isometry.

Let $I$ be the $d \times d$ identity matrix over a ring $R$. For any injective function $e : d_2 \to d$, the submatrix of $I$ formed by indexing the rows by the identity map $\text{id}$ and the columns by $e$ is an isometry.

Let $A$ be a $d \times d$ square matrix over a ring $R$. Then $A$ is an element of the unitary group $U(d, R)$ if and only if both $A$ and its conjugate transpose $A^H$ (or adjoint) are isometries.

Let $d$ be a type (indexing set) and $R$ be a ring. For any permutation $e$ of $d$, its associated permutation matrix $\text{permMatrix}_R(e)$ is an element of the unitary group $U_d(R)$.

For any isomorphisms (equivalences) of index types $e: d \simeq d_2$ and $f: d \simeq d_3$, the matrix obtained by reindexing the identity matrix $I \in \text{Matrix}_{d \times d}(R)$ using $e$ and $f$ is an isometry. Here, $(\text{reindex } e f I)_{i,j} = I_{e^{-1}(i), f^{-1}(j)}$, and a matrix $A$ is an isometry if it preserves the inner product (specifically, $A^H A = I$).

Let $e: d \simeq d_2$ be an equivalence between two index types $d$ and $d_2$. Let $1$ denote the identity matrix of size $d \times d$ over a ring $R$. The reindexing of this identity matrix by $e$ (on both rows and columns) results in a matrix that is an element of the unitary group $U_{d_2}(R)$. In terms of components, if $\delta_{i,j}$ is the Kronecker delta on $d$, the resulting matrix $M \in \text{Mat}_{d_2 \times d_2}(R)$ defined by $M_{i,j} = \delta_{e^{-1}(i), e^{-1}(j)}$ is unitary.

Let $A$ be a $d \times d$ matrix over a ring $R$, and let $e: d \simeq d_2$ be an equivalence (bijection) between index sets $d$ and $d_2$. The reindexing of $A$ by $e$, denoted as $\text{reindex}(e, e, A)$, is equal to the product $(\text{reindex}(e, \text{id}_d, 1)) \cdot A \cdot (\text{reindex}(\text{id}_d, e, 1))$, where $1$ denotes the identity matrix. This expresses the reindexed matrix as a conjugation-like product involving the reindexed identity matrices.

Let $A$ be a $d \times d$ matrix over a ring $R$, and let $\sigma$ be a permutation of the index set $d$. The reindexing of $A$ by the permutation $\sigma$ can be expressed as a conjugation by the corresponding permutation matrix: $$\text{reindex}(\sigma, \sigma, A) = P_{\sigma^{-1}} A P_{\sigma}$$ where $P_\sigma$ denotes the permutation matrix associated with $\sigma$, and both $P_{\sigma^{-1}}$ and $P_\sigma$ are elements of the unitary group $U_d(R)$.

Let $A$ and $B$ be matrices in $\text{Matrix}_d(\mathbb{k})$ where $\mathbb{k}$ is a field. Suppose $A$ is a Hermitian matrix ($A = A^H$). If every eigenvector $v \in \mathbb{k}^d$ of $A$ with eigenvalue $\lambda \in \mathbb{k}$ is also an eigenvector of $B$ with the same eigenvalue $\lambda$ (maintaining the property $Av = \lambda v \implies Bv = \lambda v$), then $A = B$.

Let $\mathbb{k}$ be a field with a structure of $\mathbb{R}$-complete like field (such as $\mathbb{R}$ or $\mathbb{C}$). Let $A$ be an $n \times n$ Hermitian matrix over $\mathbb{k}$. Suppose there exists an $n \times m$ matrix $U$ (where $n$ and $m$ are finite index sets) that is a two-sided isometry, satisfying $U U^H = I_n$ and $U^H U = I_m$. If $A$ can be diagonalized by $U$ as $A = U D U^H$, where $D$ is a diagonal matrix $\text{diag}(D_1, \dots, D_m)$ with real entries $D_i \in \mathbb{R}$, then for any function $f: \mathbb{R} \to \mathbb{R}$, the continuous functional calculus (CFC) of $A$ with respect to $f$ is given by: $$f(A) = U \text{diag}(f(D_1), \dots, f(D_m)) U^H$$

Let $\mathbf{A}$ be an $n \times n$ Hermitian matrix over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). Suppose $\mathbf{A}$ can be diagonalized by a unitary matrix $\mathbf{U}$ such that $\mathbf{A} = \mathbf{U} \mathbf{D} \mathbf{U}^\dagger$, where $\mathbf{D} = \text{diag}(d_1, d_2, \dots, d_n)$ is a diagonal matrix of real eigenvalues $d_i \in \mathbb{R}$. For any continuous function $f: \mathbb{R} \to \mathbb{R}$, the continuous functional calculus (CFC) of $\mathbf{A}$ applied to $f$ is given by $$f(\mathbf{A}) = \mathbf{U} \text{diag}(f(d_1), f(d_2), \dots, f(d_n)) \mathbf{U}^\dagger$$ where $\mathbf{U}^\dagger$ denotes the conjugate transpose of $\mathbf{U}$.

Let $A$ be a matrix and $f: \mathbb{R} \to \mathbb{R}$ be a function. For any matrix $u$ such that both $u$ and its conjugate transpose $u^H$ are isometries (i.e., they preserve the norm), the continuous functional calculus (CFC) of the conjugated matrix satisfies: $$\text{cfc } f (u A u^H) = u (\text{cfc } f A) u^H$$ where $u^H$ denotes the conjugate transpose of $u$.

Let $A$ be a square matrix and $u$ be a unitary matrix from the unitary group $U(d, \mathbb{k})$. For any continuous function $f: \mathbb{R} \to \mathbb{R}$, applying the continuous functional calculus (CFC) to the conjugation of $A$ by $u$ satisfies $$f(u A u^{-1}) = u f(A) u^{-1}$$ where $u^{-1}$ is the inverse of the unitary matrix $u$.

Let $A$ be a matrix and $f: \mathbb{R} \to \mathbb{R}$ be a function. For any unitary matrix $u$ in the unitary group $U(d, \mathbb{k})$, the continuous functional calculus (CFC) satisfies: $$f(u^\dagger A u) = u^\dagger f(A) u$$ where $u^\dagger$ denotes the conjugate transpose (Hermitian adjoint) of $u$, and $u$ is the matrix representation of the unitary element.

Let $A$ be a matrix and $f: \mathbb{R} \to \mathbb{R}$ be a function. For any equivalence $e : d \simeq d_2$ between the index types $d$ and $d_2$, the continuous functional calculus of the reindexed matrix satisfies: $$\text{cfc } f (\text{reindex } e \ e \ A) = \text{reindex } e \ e \ (\text{cfc } f \ A)$$ where $\text{reindex } e \ e \ A$ denotes the matrix $A$ with its rows and columns permuted or renamed according to $e$.

Let $A$ and $B$ be matrices. If $A$ and $B$ commute (i.e., $AB = BA$), then their corresponding linear maps on the Euclidean space, denoted by $A.\text{toEuclideanLin}$ and $B.\text{toEuclideanLin}$, also commute.

Let $\mathbb{k}$ be a field with a structure of `RCLike` (either $\mathbb{R}$ or $\mathbb{C}$), and let $E$ be an inner product space over $\mathbb{k}$. Let $A$ and $B$ be symmetric linear operators on $E$. Then the family of subspaces formed by the intersection of the eigenspaces of $A$ and $B$, indexed by pairs of eigenvalues $(\mu_1, \mu_2) \in \text{Eigenvalues}(A) \times \text{Eigenvalues}(B)$, is an orthogonal family. Specifically, for any pair of eigenvalues $\mu_{12} = (\mu_1, \mu_2)$, the subspace $V_{\mu_{12}} = \text{ker}(A - \mu_1 I) \cap \text{ker}(B - \mu_2 I)$ is orthogonal to $V_{\mu'_{12}}$ whenever $\mu_{12} \neq \mu'_{12}$.

Let $\alpha$ be a complete lattice, and let $f: \iota \to \alpha$ and $g: \iota' \to \alpha$ be two families of elements in $\alpha$. If for every $i \in \iota$, either $f(i)$ is the bottom element $\bot$ or there exists some $i' \in \iota'$ such that $f(i) \le g(i')$, then it holds that: $$\sup_{i \in \iota} f(i) \le \sup_{i' \in \iota'} g(i')$$

Let $\mathbf{k}$ be a field which is either $\mathbb{R}$ or $\mathbb{C}$ (an `RCLike` field), and let $E$ be a finite-dimensional inner product space over $\mathbf{k}$. Given two symmetric linear operators $A, B: E \to E$ that commute (i.e., $AB = BA$), the collection of intersections of their eigenspaces, $\text{ker}(A - \mu_1 I) \cap \text{ker}(B - \mu_2 I)$ for all pairs of eigenvalues $(\mu_1, \mu_2) \in \text{Spec}(A) \times \text{Spec}(B)$, forms a direct sum decomposition of the vector space $E$.

Let $\mathbb{k}$ be a field with a structure of $\mathbb{R}$ or $\mathbb{C}$ (an `RCLike` field), and let $E$ be a finite-dimensional inner product space over $\mathbb{k}$. Suppose $A$ and $B$ are symmetric linear operators on $E$ ($A, B: E \to E$) that commute with each other ($AB = BA$). Then $E$ decomposes into an internal direct sum of the intersections of their eigenspaces. Specifically, $E = \bigoplus_{(\mu_1, \mu_2) \in \sigma(A) \times \sigma(B)} (E_{\mu_1}(A) \cap E_{\mu_2}(B))$, where $\sigma(A)$ and $\sigma(B)$ denote the sets of eigenvalues of $A$ and $B$ respectively, and $E_{\mu}(M)$ denotes the eigenspace of operator $M$ corresponding to eigenvalue $\mu$.

Let $\mathbf{k}$ be $\mathbb{R}$ or $\mathbb{C}$. For two symmetric linear operators $A, B$ on the Euclidean space $\mathbf{k}^d$ that commute (i.e., $AB = BA$), there exists an orthonormal basis of $\mathbf{k}^d$ consisting of vectors that are simultaneous eigenvectors for both $A$ and $B$. This basis is constructed by taking the direct sum of orthonormal bases for the intersections of the eigenspaces of $A$ and $B$, specifically $\bigoplus_{(\mu, \nu)} (E_A(\mu) \cap E_B(\nu))$, where $E_A(\mu)$ and $E_B(\nu)$ denote the eigenspaces of $A$ and $B$ corresponding to eigenvalues $\mu$ and $\nu$ respectively.

Given two symmetric linear operators $A$ and $B$ on a $d$-dimensional Euclidean space $\mathbb{K}^d$ that commute ($AB = BA$), let $\{v_i\}_{i=1}^d$ be their shared orthonormal eigenbasis (provided by `LinearMap.sharedEigenbasis`). For each index $i \in \{1, \dots, d\}$, this function returns the real part of the inner product $\langle v_i, A v_i \rangle_{\mathbb{K}}$, which corresponds to the eigenvalue of $A$ associated with the $i$-th vector of the shared basis.

Let $A$ and $B$ be symmetric linear operators on a finite-dimensional Euclidean space $\mathbb{K}^d$ (where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$), and let $v_i$ denote the $i$-th vector of the shared orthonormal eigenbasis for $A$ and $B$, which exists because $A$ and $B$ commute. For each index $i \in \{1, \dots, d\}$, the $i$-th eigenvalue of $B$ associated with this basis is defined as the real part of the inner product $\langle v_i, B v_i \rangle$. Because $B$ is symmetric, this value is indeed real and satisfies $B v_i = \lambda_i v_i$.

Let $A$ and $B$ be symmetric linear operators on a finite-dimensional Euclidean space $\mathbb{K}^d$ (where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$). If $A$ and $B$ commute, then for every index $i \in \{1, \dots, d\}$, the $i$-th vector of their shared orthonormal eigenbasis $v_i$ is a simultaneous eigenvector of both $A$ and $B$. That is, there exist eigenvalues $\mu$ of $A$ and $\nu$ of $B$ such that $v_i$ lies in the intersection of the eigenspace of $A$ associated with $\mu$ and the eigenspace of $B$ associated with $\nu$.

Let $A$ and $B$ be symmetric linear operators on a finite-dimensional Euclidean space $\mathbb{R}^d$ (or $\mathbb{C}^d$) denoted by $E_d$. If $A$ and $B$ commute, then for any index $i \in \{1, \dots, d\}$, the action of the operator $A$ on the $i$-th vector of their shared orthonormal eigenbasis $v_i$ is given by $A(v_i) = \lambda_i v_i$, where $\lambda_i$ is the $i$-th eigenvalue of $A$ associated with this basis.

Let $A$ and $B$ be symmetric linear operators on a finite-dimensional Euclidean space $\mathbb{k}^d$. If $A$ and $B$ commute, then for any index $i \in \{1, \dots, d\}$, the action of the operator $B$ on the $i$-th vector of their shared orthonormal eigenbasis $v_i$ is given by $B(v_i) = \nu_i v_i$, where $\nu_i$ is the $i$-th eigenvalue of $B$ associated with this basis.

Let $A$ and $B$ be two $d \times d$ Hermitian matrices over the field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$) that commute, such that $AB = BA$. The function `Matrix.sharedEigenbasis` constructs an orthonormal basis for the Euclidean space $\mathbb{k}^d$ consisting of vectors that are simultaneous eigenvectors for both $A$ and $B$ (viewed as linear operators on $\mathbb{k}^d$).

Given two Hermitian matrices $A, B \in \text{Mat}_d(\mathbb{k})$ that commute ($AB = BA$), let $U$ be the matrix whose columns are the vectors of the shared orthonormal eigenbasis of $A$ and $B$. Then $U$ is an element of the unitary group $U(d, \mathbb{k})$. This matrix represents the change of basis from the shared eigenbasis to the standard basis of the Euclidean space $\mathbb{k}^d$.

Let $A, B \in \text{Mat}_d(\mathbb{k})$ be two Hermitian matrices that commute ($AB = BA$). Let $U$ be the unitary matrix `sharedEigenvectorUnitary hA hB hAB` whose columns are the shared orthonormal eigenvectors of $A$ and $B$, and let $\{v_j\}_{j \in d}$ be the shared orthonormal eigenbasis `sharedEigenbasis hA hB hAB`. For any index $j \in d$, the product of the matrix $U$ and the $j$-th standard basis vector $e_j$ (represented as `Pi.single j 1`) is equal to the $j$-th vector $v_j$ in the shared eigenbasis.

Given two Hermitian matrices $A, B \in M_d(\mathbb{k})$ that commute (i.e., $AB = BA$), let $\{v_j\}_{j=1}^d$ be their shared orthonormal eigenbasis. For a fixed index $j \in \{1, \dots, d\}$, this definition returns the eigenvalue $\lambda_j^{(A)} \in \mathbb{R}$ of the matrix $A$ corresponding to the shared eigenvector $v_j$. This is computed by treating the matrix $A$ as a symmetric linear map on the Euclidean space $\mathbb{k}^d$.

Given two symmetric (Hermitian) matrices $A, B \in M_d(\mathbb{K})$ that commute ($AB = BA$), let there be a common orthonormal eigenbasis $\{v_j\}_{j \in d}$. The function `sharedEigenvalueB` returns the eigenvalue $\lambda_j \in \mathbb{R}$ corresponding to the matrix $B$ for the $j$-th basis vector, such that $B v_j = \lambda_j v_j$.

Let $A$ and $B$ be two $d \times d$ Hermitian matrices over $\mathbb{k}$ that commute ($AB = BA$). Let $\{v_1, \dots, v_d\}$ be their shared orthonormal eigenbasis and $\lambda_j^{(A)}$ be the $j$-th eigenvalue of $A$ corresponding to the vector $v_j$. Then, for any index $j \in \{1, \dots, d\}$, the matrix-vector product of $A$ and the $j$-th basis vector $v_j$ satisfies $A v_j = \lambda_j^{(A)} v_j$.

Given two Hermitian matrices $A, B \in M_d(\mathbb{K})$ that commute ($AB = BA$), let $\{v_j\}_{j \in d}$ be their shared orthonormal eigenbasis and $\lambda_j^{(B)}$ be the $j$-th eigenvalue of $B$ corresponding to this basis. For any index $j$, the matrix-vector product of $B$ and the $j$-th basis vector $v_j$ satisfies $B v_j = \lambda_j^{(B)} v_j$.

Let $A, B \in \text{Mat}_d(\mathbb{k})$ be two Hermitian matrices that commute ($AB = BA$). Let $U$ be the unitary matrix whose columns are the shared orthonormal eigenbasis of $A$ and $B$. Then the matrix $U^\dagger A U$ is a diagonal matrix.

Let $A$ and $B$ be Hermitian matrices in $M_d(\mathbb{k})$ that commute ($AB=BA$). Let $U$ be the unitary matrix whose columns are the vectors of the shared orthonormal eigenbasis of $A$ and $B$. Then the matrix product $U^\dagger B U$ is a diagonal matrix.

Let $A$ and $B$ be Hermitian matrices in $M_d(\mathbb{K})$ (where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$). If $A$ and $B$ commute, then there exists a unitary matrix $U \in \text{U}(d, \mathbb{K})$ such that both $U A U^\dagger$ and $U B U^\dagger$ are diagonal matrices.

For any matrix $U$ in the unitary group $\text{U}(d, \mathbb{K})$, $U$ is an invertible element within that group.

Let $U$ be an element of the unitary group $U(d, \mathbb{K})$, where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$. Then the underlying matrix $U_{\text{val}} \in M_d(\mathbb{K})$ is invertible, with its inverse being its adjoint $U^\dagger$ (or conjugate transpose).

Let $M$ be a Hermitian matrix in $\text{Mat}_d(\mathbb{k})$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$) and $U$ be a unitary matrix. If $U M U^\dagger$ is a diagonal matrix, then for any bijection $e : d \simeq \{0, \dots, n-1\}$ (where $n$ is the cardinality of $d$), there exists a real-valued function $f: \mathbb{R} \to \mathbb{R}$ such that $M$ can be expressed via the continuous functional calculus (CFC) as: $$M = \text{cfc}\left(f, U^\dagger D U\right)$$ where $D$ is the diagonal matrix defined by $D_{ii} = e(i)$ for $i \in d$.

Let $A$ and $B$ be Hermitian matrices that commute with each other. Then there exists a common matrix $C$ such that both $A$ and $B$ can be expressed as a continuous functional calculus (CFC) of $C$. That is, there exist functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ such that $A = f(C)$ and $B = g(C)$.