Given a [linear map](/page/Linear%20Map) $A: \mathbb{R}^n \to \mathbb{R}^m$ represented by a matrix, the transpose $A^\top$ satisfies $(Ax) \cdot y = x \cdot (A^\top y)$ for all $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. This simple identity — the ability to "move" the operator from one side of the inner product to the other — is the foundation of duality in linear algebra: the rank of $A$ equals the rank of $A^\top$, the null space of $A^\top$ is the orthogonal complement of the column space of $A$, and the solvability of $Ax = b$ reduces to a condition on $b$ relative to $\ker(A^\top)$. The adjoint of an operator is the infinite-dimensional generalisation of this idea. It is the central construction linking an operator to its dual, and it provides the language in which spectral theory, the [Fredholm alternative](/page/The%20Fredholm%20Alternative), and the theory of [self-adjoint operators](/page/Self-Adjoint%20Operators) are formulated.

This page develops the adjoint in two settings: first for bounded operators between [Banach spaces](/page/Banach%20Space) (the Banach adjoint, which acts on dual spaces), and then for bounded operators between [Hilbert spaces](/page/Hilbert%20Space) (the Hilbert adjoint, which acts on the spaces themselves). The distinction matters: the Banach adjoint always exists by pure duality, while the Hilbert adjoint relies on the [Riesz Representation Theorem](/theorems/221) to identify a Hilbert space with its dual. The Hilbert adjoint is the one that appears most often in PDE theory and spectral theory, and we develop its properties in detail. We also introduce self-adjoint operators and establish their most basic properties — that their eigenvalues are real and their spectral radius equals their operator norm — as preparation for the [separate page](/page/Self-Adjoint%20Operators) devoted to the deeper theory.

## Motivation

[motivation] ### The Finite-Dimensional Template Consider a linear map $A: \mathbb{R}^n \to \mathbb{R}^m$ and the question: for which $b \in \mathbb{R}^m$ does $Ax = b$ have a solution? The answer is classical: $b$ must lie in the column space of $A$, which is the orthogonal complement of $\ker(A^\top)$. That is, $Ax = b$ is solvable if and only if $y \cdot b = 0$ for every $y$ with $A^\top y = 0$. This characterisation depends entirely on the relationship between $A$ and its transpose $A^\top$, defined by the identity \begin{align*} (Ax) \cdot y = x \cdot (A^\top y) \quad \text{for all } x \in \mathbb{R}^n, \, y \in \mathbb{R}^m. \end{align*} The transpose gives us a "dual" operator that moves in the reverse direction, and the interplay between $A$ and $A^\top$ — their kernels, their ranges, the dimensional relationships — is the backbone of finite-dimensional linear algebra. ### The Problem in Infinite Dimensions When we move to operators $T: X \to Y$ between infinite-dimensional Banach spaces, we lose the matrix representation, and with it the concrete recipe for computing a transpose. More fundamentally, a general Banach space has no inner product, so the identity $(Ax) \cdot y = x \cdot (A^\top y)$ does not even make sense. We need a substitute for the inner product pairing. The natural candidate is the *duality pairing* between a space and its dual: for $f \in X^*$ and $x \in X$, the evaluation $f(x)$ plays the role of the inner product. This leads to the Banach adjoint, which operates on dual spaces. ### The Hilbert Space Advantage In Hilbert spaces, we have both the duality pairing and the inner product, and the [Riesz Representation Theorem](/theorems/221) provides a canonical isometric isomorphism between a Hilbert space and its dual. This identification allows us to "internalise" the adjoint: instead of an operator $T^*: K^* \to H^*$ between dual spaces, we obtain an operator $T^*: K \to H$ between the original spaces, satisfying exactly the same identity as the finite-dimensional transpose: \begin{align*} (Tx, y)_K = (x, T^*y)_H. \end{align*} The Hilbert adjoint is far more powerful than its Banach counterpart because it lives in the same spaces as the original operator. Self-adjointness ($T = T^*$) becomes a meaningful condition, eigenvalue theory acquires a geometric flavour through orthogonal decompositions, and the kernel-range duality relations become statements about orthogonal complements rather than annihilators. [/motivation]

## The Banach Space Adjoint

Before introducing the Hilbert adjoint, we define the more general Banach adjoint. The construction requires no inner product — only the duality pairing between a Banach space and its [continuous](/page/Continuity) dual.

The idea is simple: if $T: X \to Y$ is bounded and linear, and $f \in Y^*$ is a functional on $Y$, then the composition $f \circ T$ is a functional on $X$. This "pullback" operation defines an operator from $Y^*$ to $X^*$.

[definition:Banach Adjoint] Let $X$ and $Y$ be Banach spaces and let $T \in \mathcal{L}(X, Y)$. The **Banach adjoint** (also called the **dual operator** or **transpose**) of $T$ is the operator \begin{align*} T^*: Y^* &\to X^* \\ f &\mapsto f \circ T, \end{align*} so that $(T^*f)(x) = f(Tx)$ for all $x \in X$ and $f \in Y^*$. [/definition]

The Banach adjoint is automatically bounded with $\|T^*\|_{\mathcal{L}(Y^*, X^*)} = \|T\|_{\mathcal{L}(X, Y)}$. The bound $\|T^*\| \le \|T\|$ is immediate: $|(T^*f)(x)| = |f(Tx)| \le \|f\|_{Y^*} \|Tx\|_Y \le \|f\|_{Y^*} \|T\| \|x\|_X$. The reverse inequality $\|T\| \le \|T^*\|$ follows from the Hahn–Banach theorem: for each $x \in X$ with $Tx \neq 0$, there exists $f \in Y^*$ with $\|f\|_{Y^*} = 1$ and $f(Tx) = \|Tx\|_Y$, giving $\|Tx\|_Y = (T^*f)(x) \le \|T^*f\|_{X^*} \|x\|_X \le \|T^*\| \|x\|_X$.

[example:The Matrix Transpose As Banach Adjoint] Let $X = \mathbb{R}^n$ and $Y = \mathbb{R}^m$ with the Euclidean norms. Every linear map $A: \mathbb{R}^n \to \mathbb{R}^m$ is represented by an $m \times n$ matrix. The dual $(\mathbb{R}^m)^*$ is canonically identified with $\mathbb{R}^m$ via $f \leftrightarrow y$ where $f(z) = y \cdot z$. Under this identification, the Banach adjoint $A^*: (\mathbb{R}^m)^* \to (\mathbb{R}^n)^*$ corresponds to the matrix transpose $A^\top: \mathbb{R}^m \to \mathbb{R}^n$. Indeed, for $y \in \mathbb{R}^m$ and $x \in \mathbb{R}^n$: \begin{align*} (A^*f_y)(x) = f_y(Ax) = y \cdot (Ax) = (A^\top y) \cdot x = f_{A^\top y}(x), \end{align*} so $A^* f_y = f_{A^\top y}$, which under the identification $f_z \leftrightarrow z$ gives $A^* y = A^\top y$. [/example]

The Banach adjoint has a significant limitation: it maps between dual spaces, not between the original spaces. If $T: X \to Y$, then $T^*: Y^* \to X^*$. In general, $X^*$ and $X$ are different objects — there is no canonical way to identify them. This asymmetry makes the Banach adjoint less suitable for spectral theory, where we want to compare an operator with its adjoint on the *same* space. In Hilbert spaces, the Riesz Representation Theorem resolves this asymmetry.

## The Hilbert Adjoint

The defining identity of the matrix transpose — $(Ax) \cdot y = x \cdot (A^\top y)$ — makes sense in Hilbert spaces because the inner product provides the pairing. The question is whether, given $T \in \mathcal{L}(H, K)$, there always exists an operator $T^* \in \mathcal{L}(K, H)$ satisfying $(Tx, y)_K = (x, T^*y)_H$. This is not a formal tautology: the existence of $T^*$ requires producing, for each $y \in K$, a specific element of $H$ that represents the functional $x \mapsto (Tx, y)_K$. The [Riesz Representation Theorem](/theorems/221) is precisely the tool that supplies this element. The resulting adjoint has the same operator norm as $T$, which is a stronger statement than mere existence — it says the adjoint operation preserves the metric structure of the operator space.

[quotetheorem:550]

The norm identity $\|T^*\| = \|T\|$ has an important consequence that goes beyond the individual operator. Combined with the fact that $(T^*)^* = T$ (which follows from the symmetry of the defining identity), it implies the $C^*$-identity: $\|T^*T\|_{\mathcal{L}(H)} = \|T\|_{\mathcal{L}(H,K)}^2$. To see this, note that $\|T^*T\| \le \|T^*\| \|T\| = \|T\|^2$ by submultiplicativity. For the reverse, $\|Tx\|_K^2 = (Tx, Tx)_K = (x, T^*Tx)_H \le \|x\|_H \|T^*Tx\|_H \le \|x\|_H \|T^*T\| \|x\|_H$, giving $\|T\|^2 \le \|T^*T\|$. This identity characterises the algebras of bounded operators on Hilbert spaces and is the starting point of $C^*$-algebra theory.

### The Relationship Between the Two Adjoints

If $H$ and $K$ are Hilbert spaces, $T \in \mathcal{L}(H, K)$ has both a Banach adjoint $T^*_B: K^* \to H^*$ and a Hilbert adjoint $T^*_H: K \to H$. These are related by the Riesz isomorphisms $\Phi_H: H^* \to H$ and $\Phi_K: K^* \to K$ (which send a functional to its Riesz representer):

\begin{align*} T^*_H = \Phi_H \circ T^*_B \circ \Phi_K^{-1}. \end{align*}

In other words, the Hilbert adjoint is obtained from the Banach adjoint by "translating" from dual spaces back to the original spaces via the Riesz identification. Because this identification is isometric, the norm equality $\|T^*_H\| = \|T\|$ is consistent with the Banach adjoint norm equality $\|T^*_B\| = \|T\|$.

### Algebraic Properties of the Hilbert Adjoint