Linear-Algebra-Done-Right · Unit 15 · Video 3 · Interactive Practice

Linear Functionals and the Riesz Representation Theorem

IKey Formulas

Formula	Name	What it says
$\varphi(u) = \langle u, v \rangle$ for every $u \in V$	Riesz representation theorem	$V$ finite-dimensional, $\varphi \in V'$ : exactly one such $v$ exists
$v = \overline{\varphi(e_1)}\, e_1 + \cdots + \overline{\varphi(e_n)}\, e_n$	Formula (R)	$e_1, \dots, e_n$ any orthonormal basis of $V$
$\langle u, \lambda e_k \rangle = \overline{\lambda}\, \langle u, e_k \rangle$	Second slot is conjugate linear	Why the bars appear in (R)
$\int_{-1}^{1} p(t)\cos(\pi t)\,dt = \int_{-1}^{1} p(t)\,\tfrac{15}{2\pi^2}(1 - 3t^2)\,dt$	The example	Holds for every $p \in \mathcal{P}_2(\mathbf{R})$

Key Insight: The right side of (R) seems to depend on the orthonormal basis, but uniqueness says $v$ is determined by $\varphi$ alone — every orthonormal basis gives the same sum.

IIVisualization 1 — Formula (R) Ignores the Basis

For $\varphi(x, y) = 2x + y$ on $\mathbf{R}^2$ , every orthonormal basis builds the same vector $v = (2, 1)$ .

Angle of

e_1

(θ) 30°

IIIVisualization 2 — Why Formula (R) Needs the Bar

On $\mathbf{C}^2$ , $\varphi(z_1, z_2) = \lambda z_1 + 2z_2$ is represented by $v = (\overline{\lambda}, 2)$ , not by $(\lambda, 2)$ .

💡 Challenge: find every $\lambda$ for which $(\lambda, 2)$ also represents $\varphi$ , so that dropping the bar does no harm.

IVVisualization 3 — Same Integral, Different Function

Against every $p \in \mathcal{P}_2(\mathbf{R})$ , the parabola $q$ and $\cos(\pi t)$ have the same integral — beyond $\mathcal{P}_2(\mathbf{R})$ they need not.

Degree-4 part

c\,t^2(t^2 - 1)

c

0 Representing vector

💡 The representing vector depends on the space: on $\mathcal{P}_5(\mathbf{R})$ the same $\varphi$ is represented by a polynomial of degree four, not by $q$ .

VQuiz Questions

Problem 1 · Formula (R) in a Rotated Basis

Given: $\varphi(x, y) = x + 3y$ on $\mathbf{R}^2$ with the Euclidean inner product, and the orthonormal basis $e_1 = \tfrac{1}{\sqrt2}(1, 1)$ , $e_2 = \tfrac{1}{\sqrt2}(1, -1)$ . Find $v = \varphi(e_1)e_1 + \varphi(e_2)e_2$ .

✅ Correct! The rotated basis gives back

(1, 3)

, the coefficient vector of

\varphi

— as uniqueness demands.

❌ Not quite.

(2\sqrt2, -\sqrt2)

lists the numbers

\varphi(e_1), \varphi(e_2)

; formula (R) still has to multiply them by

e_1

and

e_2

and add.

❌ Close, but… that is what

(1, 1)

and

(1, -1)

give before normalizing. Formula (R) needs an orthonormal basis; with unnormalized vectors each term is off by the factor

\|e_k\|^2 = 2

❌ Not quite. Compute

\varphi(e_1) = \tfrac{4}{\sqrt2}

and

\varphi(e_2) = -\tfrac{2}{\sqrt2}

, then form

\varphi(e_1)e_1 + \varphi(e_2)e_2

Show solution

Everything is real, so the conjugates in (R) do nothing.

\varphi(e_1) = \tfrac{1}{\sqrt2}(1 + 3) = 2\sqrt2, \qquad \varphi(e_2) = \tfrac{1}{\sqrt2}(1 - 3) = -\sqrt2.

v = 2\sqrt2 \cdot \tfrac{1}{\sqrt2}(1, 1) - \sqrt2 \cdot \tfrac{1}{\sqrt2}(1, -1) = (2, 2) - (1, -1) = (1, 3).

Check: $\langle (x, y), (1, 3) \rangle = x + 3y = \varphi(x, y)$ . The standard basis gives the same $v$ immediately: $\varphi(1, 0) = 1$ , $\varphi(0, 1) = 3$ .

Problem 2 · The Conjugate Trap

Given: $\varphi(z_1, z_2) = (2 + i)z_1 - 3i\,z_2$ on $\mathbf{C}^2$ with the Euclidean inner product $\langle z, w \rangle = z_1\overline{w_1} + z_2\overline{w_2}$ . Find the unique $v \in \mathbf{C}^2$ with $\varphi(z) = \langle z, v \rangle$ for every $z$ .

✅ Correct!

\langle z, (2 - i, 3i) \rangle = z_1\overline{(2 - i)} + z_2\overline{3i} = (2 + i)z_1 - 3i\,z_2

❌ Close, but… this drops the bars of (R). Then

\langle z, v \rangle = (2 - i)z_1 + 3i\,z_2

— the second slot conjugates the coefficients, giving a different functional.

❌ Not quite. Every coordinate gets conjugated, including the purely imaginary one:

\overline{-3i} = 3i

❌ Not quite. Conjugation flips the sign of the imaginary part only:

\overline{2 + i} = 2 - i

, not

-2 - i

Show solution

Use (R) with the standard basis: $\varphi(e_1) = 2 + i$ and $\varphi(e_2) = -3i$ , so

v = \overline{(2 + i)}\,e_1 + \overline{(-3i)}\,e_2 = (2 - i,\ 3i).

Check:

\langle z, v \rangle = z_1\overline{(2 - i)} + z_2\overline{(3i)} = (2 + i)z_1 - 3i\,z_2 = \varphi(z).

The bar in (R) undoes the bar that the second slot applies.

Problem 3 · Representing Evaluation at Zero

Given: $\mathcal{P}_2(\mathbf{R})$ with $\langle p, q \rangle = \int_{-1}^{1} p\,q$ and orthonormal basis $e_1 = \sqrt{\tfrac12}$ , $e_2 = \sqrt{\tfrac32}\,x$ , $e_3 = \sqrt{\tfrac{45}{8}}\left(x^2 - \tfrac13\right)$ . Find the $q \in \mathcal{P}_2(\mathbf{R})$ with $p(0) = \int_{-1}^{1} p\,q$ for every $p \in \mathcal{P}_2(\mathbf{R})$ .

✅ Correct!

\int_{-1}^1 q = 1

\int_{-1}^1 xq = 0

\int_{-1}^1 x^2 q = 0

— exactly

p(0)

for

p = 1, x, x^2

❌ Close, but… you used

\varphi(e_3)

correctly but multiplied it by

x^2

alone. Formula (R) multiplies by all of

e_3

, whose

-\tfrac13

contributes the constant

\tfrac58

❌ Not quite.

e_3(0) = \sqrt{\tfrac{45}{8}}\cdot\left(-\tfrac13\right) \neq 0

: the third basis element is not zero at the origin, so its term survives.

❌ Not quite.

\int_{-1}^{1} p \cdot 1

is the integral of

p

, not

p(0)

: for

p = x^2

it gives

\tfrac23

instead of

0

Show solution

Let $\varphi(p) = p(0)$ , a linear functional. Over $\mathbf{R}$ , formula (R) reads $q = \varphi(e_1)e_1 + \varphi(e_2)e_2 + \varphi(e_3)e_3$ .

\varphi(e_1) = \sqrt{\tfrac12}, \qquad \varphi(e_2) = 0, \qquad \varphi(e_3) = \sqrt{\tfrac{45}{8}}\left(-\tfrac13\right).

q = \tfrac12 + \tfrac{45}{8}\left(-\tfrac13\right)\left(x^2 - \tfrac13\right) = \tfrac12 - \tfrac{15}{8}x^2 + \tfrac{5}{8} = \tfrac98 - \tfrac{15}{8}x^2.

Check on $1, x, x^2$ : $\int_{-1}^1 q = \tfrac94 - \tfrac54 = 1 = p(0)$ for $p = 1$ ; $\int_{-1}^1 xq = 0$ (odd integrand); $\int_{-1}^1 x^2 q = \tfrac34 - \tfrac34 = 0$ . By linearity it holds for every $p$ .

Problem 4 · A Sine Functional on $\mathcal{P}_3(\mathbf{R})$

Given: $\varphi(p) = \int_{-1}^{1} p(t)\sin(\pi t)\,dt$ on $\mathcal{P}_3(\mathbf{R})$ , with $\langle p, q \rangle = \int_{-1}^{1} p\,q$ and the orthonormal basis $e_1 = \sqrt{\tfrac12}$ , $e_2 = \sqrt{\tfrac32}\,x$ , $e_3 = \sqrt{\tfrac{45}{8}}\left(x^2 - \tfrac13\right)$ , $e_4 = \tfrac{\sqrt{14}}{4}(5x^3 - 3x)$ . You may use $\int_{-1}^{1} t\sin(\pi t)\,dt = \tfrac{2}{\pi}$ and $\int_{-1}^{1} t^3\sin(\pi t)\,dt = \tfrac{2}{\pi} - \tfrac{12}{\pi^3}$ . Which $q \in \mathcal{P}_3(\mathbf{R})$ represents $\varphi$ ?

✅ Correct! Only the odd basis elements

e_2

and

e_4

contribute, and

\varphi(e_4) = \tfrac{\sqrt{14}(\pi^2 - 15)}{\pi^3} \neq 0

❌ Close, but… that is the representer on

\mathcal{P}_2(\mathbf{R})

. On

\mathcal{P}_3(\mathbf{R})

formula (R) has a fourth term, and

\varphi(e_4) \neq 0

(5t^3 - 3t)\sin(\pi t)

is even, not odd. Indeed

\langle x^3, \tfrac{3}{\pi}x \rangle = \tfrac{6}{5\pi}

, while

\varphi(x^3) = \tfrac{2}{\pi} - \tfrac{12}{\pi^3}

❌ Not quite.

\sin(\pi x)

is not a polynomial, so it is not an element of

\mathcal{P}_3(\mathbf{R})

— the theorem promises a representer inside the space.

❌ Not quite. This multiplies

\varphi(e_4)

5x^3 - 3x

instead of by

e_4 = \tfrac{\sqrt{14}}{4}(5x^3 - 3x)

. Formula (R) uses the whole basis element, so the two factors

\tfrac{\sqrt{14}}{4}

combine into

\tfrac{14}{16}

Show solution

By (R) over $\mathbf{R}$ , $q = \varphi(e_1)e_1 + \varphi(e_2)e_2 + \varphi(e_3)e_3 + \varphi(e_4)e_4$ .

$\sin(\pi t)$ is odd, so $e_1(t)\sin(\pi t)$ and $e_3(t)\sin(\pi t)$ are odd and integrate to zero over $[-1, 1]$ : $\varphi(e_1) = \varphi(e_3) = 0$ . The other two use the given integrals:

\varphi(e_2) = \sqrt{\tfrac32}\cdot\tfrac{2}{\pi}, \qquad \varphi(e_4) = \tfrac{\sqrt{14}}{4}\left(5\left(\tfrac{2}{\pi} - \tfrac{12}{\pi^3}\right) - 3\cdot\tfrac{2}{\pi}\right) = \tfrac{\sqrt{14}}{4}\left(\tfrac{4}{\pi} - \tfrac{60}{\pi^3}\right).

Multiplying each by its basis element, the square roots combine: $\tfrac32 \cdot \tfrac{2}{\pi} = \tfrac{3}{\pi}$ and $\tfrac{14}{16}\left(\tfrac{4}{\pi} - \tfrac{60}{\pi^3}\right) = \tfrac{7(\pi^2 - 15)}{2\pi^3}$ . Hence

q(x) = \frac{3}{\pi}x + \frac{7(\pi^2 - 15)}{2\pi^3}(5x^3 - 3x).

Since $\pi^2 \neq 15$ , the cubic term is really there: the representer on $\mathcal{P}_3(\mathbf{R})$ differs from the one on $\mathcal{P}_2(\mathbf{R})$ , which is $\tfrac{3}{\pi}x$ .

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