Physlib.Mathematics.InnerProductSpace.Adjoint

For a map $f : E \to F$ between two inner product spaces $E$ and $F$ over a field $\mathbb{k}$, the adjoint $f^\dagger : F \to E$ is defined as the unique map satisfying the property $\langle f(x), y \rangle = \langle x, f^\dagger(y) \rangle$ for all $x \in E$ and $y \in F$. If such a map exists (i.e., if $f$ has an adjoint), the definition returns that map; otherwise, it returns the zero map.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If $f: E \to F$ is a map and $f': F \to E$ is its adjoint (meaning they satisfy the `HasAdjoint` property), then for any $x \in E$ and $y \in F$, the inner product satisfies $\langle x, f'(y) \rangle = \langle f(x), y \rangle$.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). Suppose $E$ and $F$ are complete with respect to their respective norms. For any continuous linear map $f : E \to F$, there exists an adjoint map $f^\dagger : F \to E$ such that $\langle f(x), y \rangle = \langle x, f^\dagger(y) \rangle$ for all $x \in E$ and $y \in F$. Specifically, this adjoint is given by the composition: $$f^\dagger = \text{fromL2} \circ (\text{toL2} \circ f \circ \text{fromL2})^\dagger \circ \text{toL2}$$ where $\text{toL2}$ and $\text{fromL2}$ are the continuous linear identity maps between the spaces equipped with their original norms and the same spaces equipped with the $L_2$ norm $\|x\|_2 = \sqrt{\text{Re} \langle x, x \rangle}$, and $(\cdot)^\dagger$ denotes the standard Hilbert space adjoint.

Let $E$ and $F$ be complete generalized inner product spaces over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). For any continuous linear map $f : E \to F$, the adjoint $f^\dagger : F \to E$ is equal to the composition of maps: $$f^\dagger = \text{fromL2} \circ (\text{toL2} \circ f \circ \text{fromL2})^\dagger \circ \text{toL2}$$ where $\text{toL2}$ and $\text{fromL2}$ are the continuous linear identity maps between the spaces equipped with their original norms and the same spaces equipped with the $L_2$ norm $\|x\|_2 = \sqrt{\text{Re} \langle x, x \rangle}$, and the inner $(\cdot)^\dagger$ denotes the standard Hilbert space adjoint of the map between these $L_2$ spaces.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If $f' : F \to E$ is an adjoint of a map $f : E \to F$ (meaning they satisfy the adjoint relation with respect to the inner products on $E$ and $F$), then the image of the zero vector under $f'$ is zero, i.e., $f'(0) = 0$.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If a map $f: E \to F$ has an adjoint $f': F \to E$ (meaning $\langle f(x), y \rangle = \langle x, f'(y) \rangle$ for all $x \in E$ and $y \in F$), then the adjoint operator applied to $f$, denoted $f^\dagger$, is equal to $f'$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. Let $f : E \to F$ be a linear map. If $g' : F \to E$ is an adjoint of $f$ and $g' = f'$, then $f'$ is also an adjoint of $f$.

Let $E$ be a generalized inner product space over a field $\mathbb{k}$. The identity map $\text{id}_E : E \to E$ is its own adjoint. That is, the identity function has an adjoint, and that adjoint is also the identity function.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. The zero map $f : E \to F$, defined by $f(x) = 0$ for all $x \in E$, has an adjoint $g : F \to E$ which is also the zero map, defined by $g(y) = 0$ for all $y \in F$.

Let $E, F$, and $G$ be generalized inner product spaces over a field $\mathbb{k}$. If a map $f: F \to G$ has an adjoint $f': G \to F$ and a map $g: E \to F$ has an adjoint $g': F \to E$, then the composition $f \circ g$ has an adjoint given by the composition of the adjoints in reverse order, $g' \circ f'$.

Let $E, F$, and $G$ be generalized inner product spaces over a field $\mathbb{k}$. If a map $f: E \to F$ has an adjoint $f': F \to E$ and a map $g: E \to G$ has an adjoint $g': G \to E$, then the map $E \to F \times G$ defined by $x \mapsto (f(x), g(x))$ has an adjoint $F \times G \to E$ defined by $(y, z) \mapsto f'(y) + g'(z)$. Here, the product space $F \times G$ is equipped with the standard product inner product $\langle (y_1, z_1), (y_2, z_2) \rangle = \langle y_1, y_2 \rangle_F + \langle z_1, z_2 \rangle_G$.

Let $E, F$, and $G$ be generalized inner product spaces over a field $\mathbb{k}$. If a map $f: E \to F \times G$ has an adjoint $f': F \times G \to E$, then the map from $E$ to $F$ defined by taking the first component of $f$, $x \mapsto (f(x))_1$, has an adjoint given by $y \mapsto f'(y, 0)$.

Let $E, F,$ and $G$ be generalized inner product spaces over a field $\mathbb{k}$. If a map $f: E \to F \times G$ has an adjoint $f': F \times G \to E$, then the map $x \mapsto (f(x))_2$ (the projection of $f$ onto the second component) has an adjoint given by $z \mapsto f'(0, z)$.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If $f: E \to F$ is a map that has an adjoint $f': F \to E$, then the map defined by $x \mapsto -f(x)$ has an adjoint defined by $y \mapsto -f'(y)$.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If $f, g: E \to F$ are maps that have adjoints $f', g': F \to E$ respectively, then the map $f + g$ (defined by $x \mapsto f(x) + g(x)$) has an adjoint given by $f' + g'$ (defined by $y \mapsto f'(y) + g'(y)$).

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$. If the maps $f, g: E \to F$ have adjoints $f', g': F \to E$ respectively, then the map $f - g$ (defined by $x \mapsto f(x) - g(x)$) has an adjoint $f' - g'$ (defined by $y \mapsto f'(y) - g'(y)$).

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). If a map $f: E \to F$ has an adjoint $f': F \to E$, then for any scalar $c \in \mathbb{k}$, the map $x \mapsto c \cdot f(x)$ has an adjoint given by $y \mapsto \bar{c} \cdot f'(y)$, where $\bar{c}$ denotes the complex conjugate of $c$.

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$ (typically $\mathbb{R}$ or $\mathbb{C}$). Let $f: E \to \mathbb{k}$ be a linear map and $f': \mathbb{k} \to E$ be its adjoint, such that $\langle f(x), s \rangle_{\mathbb{k}} = \langle x, f'(s) \rangle_E$ for all $x \in E$ and $s \in \mathbb{k}$. For any vector $v \in F$, the map $g: E \to F$ defined by $g(x) = f(x)v$ has an adjoint $g': F \to E$ given by $g'(y) = f'(\overline{\langle y, v \rangle})$, where $\overline{\langle y, v \rangle}$ denotes the complex conjugate of the inner product in $F$.