Adjoint of a linear map
Consider finite-dimensional inner product spaces $V$ and $W$ and a linear map $T : V \to W$. The adjoint of $T$ is the linear map $T^{\ast} : W \to V$ such that
\[\ip{T v}{w} = \ip{v}{T^{\ast} w} \quad \text{for all } v \in V \text{ and } w \in W.\]Consider Hilbert spaces $H$ and $K$ and a bounded linear operator $T : H \to K$. The adjoint of $T$ is the bounded linear operator $T^{\ast} : K \to H$ such that
\[\ip{T x}{y} = \ip{x}{T^{\ast} y} \quad \text{for all } x \in H \text{ and } y \in K.\]Properties
- The adjoint depends on the inner products, not only on the map. For example, see the adjoint of $x \mapsto A x$ with a weighted inner product.
- Relative to orthonormal bases of $V$ and $W$, the matrix of $T^{\ast}$ is the conjugate transpose of the matrix of $T$.
- $(S T)^{\ast} = T^{\ast} S^{\ast}$, and $(T^{\ast})^{\ast} = T$.
- A linear map $T : V \to V$ such that $T^{\ast} = T$ is called self-adjoint.
Notes
Adjoints of unbounded operators
The graduate definition applies only to bounded operators, but differential operators are not bounded. For example, consider the derivative $D f = f^{\prime}$ on functions on $[0, 1]$, with the inner product $\ip{f}{g} = \int_0^1 f g \, \mathrm{d}x$. The functions $f_n(x) = \sin(n \pi x)$ all have the norm $\norm{f_n} = 1/\sqrt{2}$, but their derivatives $D f_n(x) = n \pi \cos(n \pi x)$ have the norms $\norm{D f_n} = n \pi / \sqrt{2}$, which grow without bound as $n$ increases. So no constant $c$ satisfies $\norm{D f} \leq c \norm{f}$ for every $f$, and $D$ is not bounded.
The derivative $D$ is also defined only on some functions, such as the differentiable functions that satisfy given boundary conditions. The set of functions on which $D$ is defined is called the domain of $D$, written $\dom D$. Restricting the domain does not make $D$ bounded, because every $f_n$ satisfies the boundary conditions $f_n(0) = f_n(1) = 0$. Instead, the definition of the adjoint is extended to unbounded operators: when $\dom D$ is dense, as it is here, the adjoint $D^{\ast}$ is defined by the same equation,
\[\ip{D f}{g} = \ip{f}{D^{\ast} g} \quad \text{for all } f \in \dom D,\]but $\dom D^{\ast}$ contains only the functions $g$ for which some function $D^{\ast} g$ satisfies this equation. So the boundary conditions in $\dom D$ determine $\dom D^{\ast}$, because the boundary conditions decide which boundary terms vanish in integration by parts:
\[\int_0^1 f^{\prime} g \, \mathrm{d}x = f(1) g(1) - f(0) g(0) - \int_0^1 f g^{\prime} \, \mathrm{d}x.\]If $\dom D$ requires $f(0) = f(1) = 0$, then $D^{\ast} g = -g^{\prime}$ for all differentiable $g$. If $\dom D$ has no boundary conditions, then $D^{\ast} g = -g^{\prime}$ only for the differentiable $g$ such that $g(0) = g(1) = 0$.
Other meanings of “adjoint”
“Adjoint” has an unrelated meaning for matrices: the adjugate (or classical adjoint) $\operatorname{adj} A$ is the transpose of the matrix of cofactors of $A$, which satisfies $A \operatorname{adj} A = (\det A) I$. The dual map $T^{\top} : W^{\ast} \to V^{\ast}$, defined by $(T^{\top} \omega)(v) = \omega(T v)$ for covectors $\omega$, is also sometimes written $T^{\ast}$, but the dual map needs no inner products.