PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.Cayley.Inverse

Given a unitary operator $u : H \cong H$ on a complex Hilbert space $H$, this definition represents $u$ as a linear partial map $u : \mathcal{D}(u) \to H$ where the domain $\mathcal{D}(u)$ is the entire space $H$.

For any unitary operator $u : H \cong H$ on a complex Hilbert space $H$, the domain of its representation as a linear partial map is the entire space $H$.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. The domain of the partial linear map $I - u$, where $I$ is the identity operator, is the entire space $H$.

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 $i(I + u)(I - u)^{-1}$, where $I$ is the identity operator on $H$ and $(I - u)^{-1}$ denotes the inverse of the operator $I - u$ defined on its range.

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 $u$, defined as $i(I + u)(I - u)^{-1}$ (where $I$ is the identity operator), is equal to the range of the operator $I - u$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. If the kernel of the operator $I - u$ is trivial (where $I$ is the identity operator), i.e., $\ker(I - u) = \{0\}$, then the range of $I - u$ is dense in $H$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint linear partial map on $H$. Then the unitary operator resulting from the Cayley transform of $T$, when viewed as a linear partial map, is equal to the Cayley transform of $T$ defined as a partial map.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear map on $H$. Let $U$ be the unitary operator obtained by the Cayley transform of $T$. Then the kernel of the operator $I - U$ is trivial, i.e., $\ker(I - U) = \{0\}$.

For any partial linear map $Q$ on a complex Hilbert space $H$ and any non-zero scalar $a \in \mathbb{C} \setminus \{0\}$, the range of the scaled operator $a Q$ is equal to the range of $Q$, i.e., $\operatorname{range}(a Q) = \operatorname{range}(Q)$.

Let $H$ be a complex Hilbert space and $Q$ be a partial linear map on $H$. If $Q$ is injective (that is, its kernel is trivial, $\ker Q = \{0\}$), then for any $y$ in the domain of its inverse $Q^{-1}$, applying $Q$ to $Q^{-1}(y)$ returns $y$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear map on $H$. Let $u$ be the Cayley transform of $T$. Then the domain of the inverse Cayley transform of $u$ 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 $I - u$ is injective (i.e., its kernel is trivial), where $I$ denotes the identity operator on $H$. For any vector $a \in H$, the inverse Cayley transform $\Phi(u) = i(I + u)(I - u)^{-1}$ satisfies: $$\Phi(u) ((I - u) a) = i(I + u) a$$ 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 $T$. Then the inverse Cayley transform of $u$ recovers the original operator $T$: $$\text{inverseCayleyPMap}(u) = T$$ where the inverse Cayley transform is defined as $i(I + u)(I - u)^{-1}$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. 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 operator. 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 operator $T = i(I + u)(I - u)^{-1}$ on the domain $\text{ran}(I - u)$, is symmetric. That is, for all $x, y$ in the domain of $T$, the identity $\langle Tx, y \rangle = \langle x, Ty \rangle$ holds.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator such that the operator $I - u$ is injective (i.e., its kernel is $\{0\}$), where $I$ denotes the identity operator. Let $C^{-1}(u) = i(I + u)(I - u)^{-1}$ be the inverse Cayley transform of $u$. For any vector $a$ in the domain of $I - u$, let $x = (I - u)a$ be the corresponding element in the domain of $C^{-1}(u)$. Then the following identity holds: $$C^{-1}(u)x + ix = 2ia$$ where $i$ is the imaginary unit.

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

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

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. If the kernel of the operator $I - u$ is trivial (i.e., $1$ is not an eigenvalue of $u$), then the range of the partial linear map $i(I + u)(I - u)^{-1} + iI$ is the entire space $H$, where $I$ denotes 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. Suppose that the kernel of the operator $I - u$ is trivial (i.e., $\ker(I - u) = \{0\}$), where $I$ denotes the identity operator. Let $C(u) = i(I + u)(I - u)^{-1}$ be the partial inverse Cayley transform of $u$. Then the range of the operator $C(u) - iI$ is the entire space $H$.

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

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. Suppose that the kernel of $I - u$ is trivial (i.e., $\ker(I - u) = \{0\}$) and that the range of $I - u$ is dense in $H$. Let $T = i(I + u)(I - u)^{-1}$ be the partial inverse Cayley transform of $u$. Then for any $x \in H$, applying the Cayley transform $C(T) = I - 2i(T + iI)^{-1}$ to $x$ yields $u(x)$.

Let $H$ be a complex Hilbert space and $u: H \to H$ be a unitary operator. Suppose that the kernel of the operator $I - u$ is trivial, i.e., $\ker(I - u) = \{0\}$, and that the range of $I - u$ is dense in $H$. Let $T = i(I + u)(I - u)^{-1}$ be the inverse Cayley transform of $u$. Then the Cayley transform of $T$, defined as $C(T) = I - 2i(T + iI)^{-1}$ (where $I$ is the identity operator and $i$ is the imaginary unit), is equal to the original operator $u$.

Let $H$ be a complex Hilbert space and $u : H \to H$ be a unitary operator. If the kernel of the operator $I - u$ 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. Suppose $u$ satisfies the conditions of `CayleyUnitaryData`, which include that $1$ is not a fixed vector of $u$ (i.e., $\ker(I - u) = \{0\}$) and that the range of the operator $I - u$ is dense in $H$. Then the inverse Cayley transform of $u$, defined as the partial linear map $i(I + u)(I - u)^{-1}$, is a self-adjoint operator.