Dual Spaces and the Riesz Representation
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You should be comfortable with the definition of an inner product, the concept of a linear functional, and the Gram–Schmidt process for constructing orthonormal bases.
Why this matters
The Riesz Representation Theorem provides the crucial bridge between abstract functionals and concrete vectors.
In quantum mechanics and classical mechanics, we often distinguish between a state vector and the measurement of an observable. While these are conceptually distinct, the Riesz theorem tells us that in finite-dimensional spaces, they are mathematically isomorphic. This allows us to treat "row vectors" (covectors) and "column vectors" (vectors) as interchangeable objects, provided we respect the conjugation rules. This identification is fundamental to understanding the adjoint operator and the spectral theorem, which we will encounter in subsequent lectures.
The Intuition: Measuring Direction
Imagine a vector space as a landscape. A vector points to a specific location or direction. A linear functional, however, is a rule that assigns a number to every vector in the space. Think of it as a "measurement device" that reads off a value based on the input vector. In two dimensions, if you have a vector \(\mathbf{v}\), a functional \(\varphi\) might measure the projection of \(\mathbf{v}\) onto some fixed direction. The Riesz Representation Theorem states that this "measurement direction" is not an abstract concept; it is actually a vector in the same space. Specifically, the functional \(\varphi\) is completely determined by a unique vector \(\mathbf{r}_\varphi\), such that measuring \(\mathbf{v}\) is equivalent to taking the inner product of \(\mathbf{v}\) with \(\mathbf{r}_\varphi\). This means the dual space (the space of all measurements) is structurally identical to the original space, though with a subtle twist regarding complex conjugation.
Formal Definition and Uniqueness
Let \( V \) be a finite-dimensional inner-product space over the field \( \mathbb{F} \) (where \( \mathbb{F} \) is either \( \mathbb{R} \) or \( \mathbb{C} \)). We define the inner product \( \langle \cdot , \cdot \rangle \) to be linear in its first argument and conjugate-linear in its second. The dual space \( V^{*} \) consists of all linear maps \( \varphi : V \to \mathbb{F} \).
The Riesz Representation Theorem asserts that for every \( \varphi \in V^{*} \), there exists a unique vector \( \mathbf{r}_\varphi \in V \) such that:
The vector \( \mathbf{r}_\varphi \) is called the Riesz representative of \( \varphi \). The uniqueness follows directly from the non-degeneracy of the inner product: if two vectors \( \mathbf{a} \) and \( \mathbf{b} \) both represented \( \varphi \), then \( \langle \mathbf{x}, \mathbf{a} - \mathbf{b} \rangle = 0 \) for all \( \mathbf{x} \). Setting \( \mathbf{x} = \mathbf{a} - \mathbf{b} \) implies \( \|\mathbf{a} - \mathbf{b}\|^2 = 0 \), hence \( \mathbf{a} = \mathbf{b} \).
Constructing the Representative
To find the specific vector \( \mathbf{r}_\varphi \), we utilize an orthonormal basis. Let \( \{ \mathbf{e}_1, \dots, \mathbf{e}_n \} \) be an orthonormal basis for \( V \). Any vector \( \mathbf{x} \in V \) can be written as \( \mathbf{x} = \sum_{j=1}^{n} c_j \mathbf{e}_j \), where \( c_j = \langle \mathbf{x}, \mathbf{e}_j \rangle \).
Since \( \varphi \) is linear, we have \( \varphi(\mathbf{x}) = \sum_{j=1}^{n} c_j \varphi(\mathbf{e}_j) \). We seek \( \mathbf{r}_\varphi = \sum_{i=1}^{n} r_i \mathbf{e}_i \) such that \( \langle \mathbf{x}, \mathbf{r}_\varphi \rangle = \varphi(\mathbf{x}) \). Using the conjugate-linearity in the second slot:
Comparing this with \( \varphi(\mathbf{x}) = \sum_{j=1}^{n} c_j \varphi(\mathbf{e}_j) \), we see that we must have \( \overline{r_j} = \varphi(\mathbf{e}_j) \), or equivalently \( r_j = \overline{\varphi(\mathbf{e}_j)} \). Thus, the Riesz representative is constructed as:
Conjugate-Linearity of the Map
A critical property of the Riesz map \( \Phi : V^{*} \to V \), defined by \( \Phi(\varphi) = \mathbf{r}_\varphi \), is that it is not linear, but conjugate-linear. Let \( \varphi, \psi \in V^{*} \) and \( \alpha \in \mathbb{F} \). Consider the functional \( \alpha \varphi + \psi \). Its representative is \( \mathbf{r}_{\alpha \varphi + \psi} \).
For any \( \mathbf{x} \), \( (\alpha \varphi + \psi)(\mathbf{x}) = \alpha \varphi(\mathbf{x}) + \psi(\mathbf{x}) = \langle \mathbf{x}, \mathbf{r}_\varphi \rangle + \langle \mathbf{x}, \mathbf{r}_\psi \rangle = \langle \mathbf{x}, \mathbf{r}_\varphi + \mathbf{r}_\psi \rangle \). This suggests the additive part is linear. However, consider the scalar multiplication. Let \( \theta = \alpha \varphi \). Then \( \theta(\mathbf{x}) = \alpha \langle \mathbf{x}, \mathbf{r}_\varphi \rangle = \langle \mathbf{x}, \overline{\alpha} \mathbf{r}_\varphi \rangle \). Therefore, \( \mathbf{r}_{\alpha \varphi} = \overline{\alpha} \mathbf{r}_\varphi \).
Combining these, the map satisfies \( \Phi(\alpha \varphi + \psi) = \overline{\alpha} \Phi(\varphi) + \Phi(\psi) \). This conjugate-linearity is essential when translating between the dual space and the primal space, especially in complex Hilbert spaces where the distinction between \( \alpha \) and \( \overline{\alpha} \) is physically significant.
The derivation
Riesz Representation in Finite Dimensions
Let \( V \) be a finite-dimensional inner-product space over the field \( \mathbb{F} \) (either \( \mathbb{R} \) or \( \mathbb{C} \)) with inner product \( \langle \cdot , \cdot \rangle \), taken linear in its first argument and conjugate-linear in its second. Then for every linear functional \( \varphi : V \to \mathbb{F} \) there exists a unique vector \( \mathbf{r}_\varphi \in V \), the Riesz representative, such that \[ \varphi(\mathbf{x}) = \langle \mathbf{x}, \mathbf{r}_\varphi \rangle \qquad \text{for all } \mathbf{x} \in V. \] The assignment \( \varphi \mapsto \mathbf{r}_\varphi \) is a conjugate-linear bijection from the dual space \( V^{*} \) onto \( V \).
A common source of error is assuming the Riesz representative is linear in the functional. If you define \( \varphi(\mathbf{x}) = \langle \mathbf{r}, \mathbf{x} \rangle \) (swapping the arguments), you change the linearity properties. Our convention is linear in the first argument, so the representative involves the conjugate of the functional's values on the basis. Confusing \( \langle \mathbf{x}, \mathbf{r} \rangle \) with \( \langle \mathbf{r}, \mathbf{x} \rangle \) leads to incorrect adjoint relations later. C
Worked example
Problem. Let \( V = \mathbb{C}^2 \) with the standard inner product \( \langle \mathbf{u}, \mathbf{v} \rangle = u_1 \overline{v_1} + u_2 \overline{v_2} \). Define the linear functional \( \varphi : V \to \mathbb{C} \) by \( \varphi(\mathbf{x}) = 2i x_1 - 3 x_2 \) for \( \mathbf{x} = (x_1, x_2)^T \). Find the Riesz representative \( \mathbf{r}_\varphi \).
- Identify the standard orthonormal basis: \( \mathbf{e}_1 = (1, 0)^T \) and \( \mathbf{e}_2 = (0, 1)^T \).
- Evaluate the functional on the basis vectors: \( \varphi(\mathbf{e}_1) = 2i(1) - 3(0) = 2i \) and \( \varphi(\mathbf{e}_2) = 2i(0) - 3(1) = -3 \).
- Apply the Riesz formula \( \mathbf{r}_\varphi = \overline{\varphi(\mathbf{e}_1)} \mathbf{e}_1 + \overline{\varphi(\mathbf{e}_2)} \mathbf{e}_2 \).
- Compute the conjugates: \( \overline{2i} = -2i \) and \( \overline{-3} = -3 \).
- Construct the vector: \( \mathbf{r}_\varphi = -2i \mathbf{e}_1 - 3 \mathbf{e}_2 = \begin{pmatrix} -2i \\ -3 \end{pmatrix} \).
Units check. The functional maps vectors to scalars. The inner product of two vectors in \( \mathbb{C}^2 \) yields a scalar. The components of \( \mathbf{r}_\varphi \) are complex numbers, consistent with the field \( \mathbb{C} \). Verification: \( \langle (1,0)^T, (-2i, -3)^T \rangle = 1 \cdot \overline{-2i} + 0 = 2i = \varphi(\mathbf{e}_1) \). Correct.
Three quick tests
- If \( \varphi(\mathbf{x}) = \langle \mathbf{x}, \mathbf{a} \rangle \), what is \( \mathbf{r}_\varphi \)?
Answer
\( \mathbf{a} \). The representative is the vector in the second slot of the inner product. - Is the map \( \varphi \mapsto \mathbf{r}_\varphi \) linear or conjugate-linear?
Answer
Conjugate-linear. \( \Phi(\alpha \varphi) = \overline{\alpha} \Phi(\varphi) \). - In \( \mathbb{R}^2 \), if \( \varphi(x,y) = 3x + 4y \), find \( \mathbf{r}_\varphi \).
Answer
\( (3, 4)^T \). In real spaces, conjugation has no effect, so the coefficients map directly to the vector components.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.