PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.Cayley.Inverse

Let $H$ be a complex Hilbert space. For a unitary operator $u: H \cong H$, this definition represents $u$ as a partially defined linear map (a `LinearPMap`) from $H$ to $H$ whose domain is the entire space $H$.

Let $H$ be a complex Hilbert space and let $u: H \to H$ be a unitary operator. When $u$ is viewed as a partially defined linear map, its domain is the entire space $H$ (denoted by $\top$).

Let $H$ be a complex Hilbert space and let $u: H \to H$ be a unitary operator. When $1 - u$ is viewed as a partially defined linear map, its domain is the entire space $H$ (denoted by $\top$).

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. The partial inverse Cayley transform of $u$ is the partial linear map from $H$ to $H$ defined by $i(I + u)(I - u)^{-1}$, where $I$ is the identity operator and $i$ is the imaginary unit. The domain of this operator is the range of $I - u$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. The domain of the partial inverse Cayley transform of $u$ is equal to the range of the operator $I - u$, where $I$ is the identity operator on $H$.

Let $H$ be a complex Hilbert space and let $u: H \to H$ be a unitary operator. If the kernel of the operator $I - u$ is trivial (i.e., $\ker(I - u) = \{0\}$, meaning $1$ is not an eigenvalue of $u$), then the range of $I - u$ is dense in $H$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partially defined linear operator on $H$. The unitary operator obtained via the Cayley transform of $T$, denoted $\text{cayleyUnitary}(T)$, is equal to the Cayley transform of $T$ defined as a partial linear map, denoted $\text{cayleyPMap}(T)$, when the former is represented as a partial linear map.

Let $H$ be a complex Hilbert space. If $T$ is a self-adjoint partially defined linear operator on $H$ and $U$ is the unitary operator defined as the Cayley transform of $T$, then the kernel of the operator $1 - U$ is trivial: $$\ker(1 - U) = \{0\}.$$

Let $Q$ be a partially defined linear operator on a complex Hilbert space $H$. For any non-zero complex scalar $a \in \mathbb{C}$, the range of the operator $a Q$ is equal to the range of $Q$, i.e., $\text{range}(a Q) = \text{range}(Q)$.

Let $H$ be a Hilbert space over $\mathbb{C}$ and let $Q$ be a linear partial map on $H$. If the kernel of $Q$ is trivial, i.e., $\ker Q = \{0\}$, then for any vector $y$ in the domain of the inverse partial map $Q^{-1}$ (which is equivalent to the range of $Q$), applying $Q$ to the image of $y$ under $Q^{-1}$ yields $y$: $$Q(Q^{-1}y) = y$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Let $u = (T - iI)(T + iI)^{-1}$ be the Cayley transform of $T$. Then the domain of the inverse Cayley transform of $u$ (defined as $i(I + u)(I - u)^{-1}$) is equal to the domain of $T$.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. Suppose that the operator $I - u$ has a trivial kernel, where $I$ is the identity operator (meaning $1$ is not an eigenvalue of $u$). For any vector $a$ in the domain of $I - u$, let $x = (I - u)a$ be a vector in the range of $I - u$. Then the inverse Cayley transform of $u$, defined as $i(I + u)(I - u)^{-1}$, maps $x$ to $i(I + u)a$: $$(i(I + u)(I - u)^{-1})((I - u)a) = i(I + u)a$$ where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Let $u$ be the Cayley transform of $T$ (denoted `cayleyUnitary T`). Then the partial inverse Cayley transform of $u$, defined as $i(I + u)(I - u)^{-1}$, is equal to $T$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary linear isometry. For any vectors $a, b \in H$, the following identity holds: $$\langle i(a + u(a)), b - u(b) \rangle = \langle a - u(a), i(b + u(b)) \rangle$$ where $i$ is the imaginary unit and $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary linear isometry. If the operator $I - u$ is injective (i.e., its kernel is trivial, $\ker(I - u) = \{0\}$), then the inverse Cayley transform of $u$, defined as the partial linear operator $i(I + u)(I - u)^{-1}$ with domain $\operatorname{ran}(I - u)$, is symmetric. That is, for all $x, y \in \operatorname{ran}(I - u)$, the inner product satisfies $$\langle i(I + u)(I - u)^{-1}x, y \rangle = \langle x, i(I + u)(I - u)^{-1}y \rangle.$$

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator such that the kernel of $I - u$ is trivial. For any $a \in H$, let $x = (I - u)a$, which is an element of the domain of the inverse Cayley transform $\Phi(u) = i(I + u)(I - u)^{-1}$. Then the following identity holds: $$\Phi(u)x + i x = 2ia$$ where $I$ is the identity operator and $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator such that $1$ is not an eigenvalue of $u$ (i.e., the kernel of $I - u$ is trivial). Let $C^{-1}(u) = i(I + u)(I - u)^{-1}$ be the inverse Cayley transform of $u$, which is a partially defined linear map with domain $\text{ran}(I - u)$. For any $a \in H$, let $x = (I - u)a$. Then the following identity holds: $$C^{-1}(u)x - ix = 2i u(a)$$ where $i$ is the imaginary unit and $I$ is the identity operator.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. If the range of the operator $I - u$ (where $I$ is the identity operator) is dense in $H$, then the domain of the partial inverse Cayley transform of $u$, defined as $i(I + u)(I - u)^{-1}$, is also dense in $H$.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. If the kernel of $I - u$ is trivial (i.e., $\ker(I - u) = \{0\}$), where $I$ is the identity operator, then the range of the operator $K + iI$ is the entire space $H$, where $K = i(I + u)(I - u)^{-1}$ is the partial inverse Cayley transform of $u$ and $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. Suppose that $1$ is not an eigenvalue of $u$, meaning the kernel of $I - u$ is trivial, where $I$ is the identity operator. Let $K_u$ be the inverse Cayley transform of $u$, defined as the partial linear map $i(I + u)(I - u)^{-1}$ with domain equal to the range of $I - u$. Then the range of the operator $K_u - iI$ is the entire space $H$, where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. Suppose that $1$ is not an eigenvalue of $u$ (i.e., the kernel $\ker(I - u) = \{0\}$) and that the range of $I - u$ is dense in $H$. Then the inverse Cayley transform of $u$, defined as the partial linear map $i(I + u)(I - u)^{-1}$ with domain $\text{ran}(I - u)$, is a self-adjoint operator.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. Suppose that $1$ is not an eigenvalue of $u$ (i.e., $\ker(I - u) = \{0\}$) and that the range of $I - u$ is dense in $H$. Let $K_u = i(I + u)(I - u)^{-1}$ be the inverse Cayley transform of $u$, which is a self-adjoint partially defined linear operator. Then the Cayley transform of $K_u$, defined as the bounded linear operator $\mathcal{C}(K_u) = (K_u - iI)(K_u + iI)^{-1}$, satisfies $\mathcal{C}(K_u)x = u(x)$ for all $x \in H$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. Suppose that $1$ is not an eigenvalue of $u$ (i.e., $\ker(I - u) = \{0\}$) and that the range of $I - u$ is dense in $H$. Let $T$ be the inverse Cayley transform of $u$, defined as the partial linear map $T = i(I + u)(I - u)^{-1}$. Then the Cayley transform of $T$, defined as the bounded linear operator $C(T) = (T - iI)(T + iI)^{-1}$, is equal to $u$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. If the kernel of $I - u$ is trivial (i.e., $\ker(I - u) = \{0\}$), then the inverse Cayley transform of $u$, defined as the partial linear map $i(I + u)(I - u)^{-1}$, is self-adjoint.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. If $1$ is not an eigenvalue of $u$, then the inverse Cayley transform of $u$, defined as the partial linear map $i(I + u)(I - u)^{-1}$ on the range of $I - u$, is a self-adjoint operator.