Riesz Representation in Finite Dimensions

Definition Let $V$ be a finite-dimensional inner product space over $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, with the inner product conjugate-linear in its first argument and linear in its second. For every linear functional $\ell \in V^{\vee}$ there is exactly one vector $z \in V$ such that

Existence comes from the orthonormal basis construction below. For uniqueness, suppose $z$ and $w$ both represent $\ell$. Then $\langle z - w, x \rangle = 0$ for every $x$, and the choice $x = z - w$ gives $\lVert z - w \rVert^2 = 0$, so $z = w$ by positive definiteness. Finite dimension and positive definiteness both matter. In infinite dimensions the statement needs more, such as continuity of $\ell$ and completeness of $V$. Without positive definiteness it can fail, as it does for a degenerate form. For the form $s(z, x) = \overline{z_1}\,x_1$ on $\mathbb{C}^2$, with matrix $\operatorname{diag}(1, 0)$, every pairing $s(z, e_2)$ is $0$, so the functional $\ell(x) = x_2$ has no representing vector, and the zero functional is represented by both $0$ and $e_2$.

Definition Let $V$ be an inner product space over $\mathbb{F}$ and $\ell \in V^{\vee}$. A representing vector of $\ell$ is a vector $z \in V$ with $\ell(x) = \langle z, x \rangle$ for every $x \in V$. When $V$ is finite-dimensional it exists and is unique by the Riesz representation theorem, and we write it $z_{\ell}$. The vector sits in the first slot because $x \mapsto \langle z, x \rangle$ is linear, as a functional must be, while $x \mapsto \langle x, z \rangle$ is conjugate-linear over $\mathbb{C}$. On $\mathbb{F}^n$ with the standard inner product, a functional given by a row $r \in \mathbb{F}^{1 \times n}$, so that $\ell(x) = rx$, has representing vector

The functional is zero exactly when $z_{\ell} = 0$. Otherwise its kernel is the set of vectors orthogonal to $z_{\ell}$, and by Cauchy-Schwarz $\lvert \ell(x) \rvert \le \lVert z_{\ell} \rVert \, \lVert x \rVert$, with equality at $x = z_{\ell}$.

Definition Let $e_1, \dots, e_n$ be an orthonormal basis of a finite-dimensional inner product space $V$ over $\mathbb{F}$. Such a basis exists by the Gram-Schmidt process. For $\ell \in V^{\vee}$ the vector

represents $\ell$. To check it, write $x = \sum_k x_k e_k$, so that $\ell(x) = \sum_k x_k \ell(e_k)$ by linearity. The first slot conjugates the coefficients of $z$ back, and orthonormality removes the terms with $j \ne k$, so

By uniqueness, every orthonormal basis gives the same $z$. Over $\mathbb{R}$ the conjugations do nothing. The formula works for every $\ell$ only when the basis is orthonormal. In a basis that is not orthonormal the Gram matrix enters, as shown under the dependence on the chosen inner product.

Definition For a finite-dimensional inner product space $V$ over $\mathbb{F}$, define $J \colon V \to V^{\vee}$ by $J(z)(x) = \langle z, x \rangle$, so that $J(z)$ is the functional represented by $z$. Because a scalar leaves the first slot conjugated, for all $z, w \in V$ and $a \in \mathbb{F}$

The Riesz representation theorem says that $J$ is a bijection. Over $\mathbb{C}$ it is conjugate-linear, and on a nonzero space it is not linear and so not an isomorphism of complex vector spaces. Its inverse $\ell \mapsto z_{\ell}$ is conjugate-linear as well, with $z_{\ell_1 + \ell_2} = z_{\ell_1} + z_{\ell_2}$ and $z_{a\ell} = \overline{a}\,z_{\ell}$. Books that make the first argument linear put the representing vector in the second slot, and the match is conjugate-linear there too. Over $\mathbb{R}$ conjugation is trivial and $J$ is a linear isomorphism, one that depends on the chosen inner product.

Definition A linear functional and the dual space $V^{\vee}$ need no inner product, but the representing vector does. On $\mathbb{F}^n$ let $G \in \mathbb{F}^{n \times n}$ be Hermitian and positive definite, with inner product $\langle x, y \rangle_G = x^{*}Gy$, and let $\ell(x) = rx$ for a row $r$. A vector $z$ represents $\ell$ exactly when $z^{*}G = r$. Taking conjugate transposes and using $G^{*} = G$ gives $Gz = r^{*}$. The matrix $G$ is invertible, because $Gz = 0$ with $z \ne 0$ would give $z^{*}Gz = 0$. Hence

The same formula works in any basis $\mathcal{B} = (b_1, \dots, b_n)$ of an abstract space, with $G$ the Gram matrix of $\mathcal{B}$, the row $r = \begin{bmatrix} \ell(b_1) & \cdots & \ell(b_n) \end{bmatrix}$, and $z$ replaced by its coordinate column $[z]_{\mathcal{B}}$. Changing the inner product changes $G$ and in general moves the representing vector. Changing only the basis can change $G$, $r$ and the coordinates of $z$, but $z$ stays the same vector.