Linear-Algebra-Done-Right · Unit 9 · Video 1 · Interactive Practice

The Dual Basis Reads Coefficients and T′ Reverses Arrows

IKey Formulas

Formula	Name	What it says
$V' = \mathcal{L}(V, \mathbf{F})$ , $\;\dim V' = \dim V$	Dual space	The linear functionals on $V$ ; for finite-dimensional $V$ , $\dim \mathcal{L}(V, \mathbf{F}) = (\dim V)(1)$
$\varphi_j(v_k) = \begin{cases} 1 & k = j \\ 0 & k \neq j \end{cases}$	Dual basis	One functional per basis vector, fixed by the linear map lemma; $\varphi_1, \dots, \varphi_n$ is a basis of $V'$
$v = \varphi_1(v)\,v_1 + \cdots + \varphi_n(v)\,v_n$	Coefficient reader	$\varphi_j(v)$ is the coefficient of $v_j$ when $v$ is written in the basis
$T'(\varphi) = \varphi \circ T$ , $\;(ST)' = T'S'$	Dual map	$T \colon V \to W$ gives $T' \colon W' \to V'$ ; the order of a product reverses

Key Insight: $\varphi_j$ depends on the whole basis, not only on $v_j$ : with $v_1 = (1, 0)$ and $v_2 = (1, 1)$ , the functional $\varphi_1(x, y) = x - y$ is not the first coordinate, because it must also vanish on $v_2$ .

IIVisualization 1 — The Dual Basis Reads Coefficients

For a basis $v_1, v_2$ of $\mathbf{R}^2$ , the dual basis returns the two coefficients of $v$ : $v = \varphi_1(v)\,v_1 + \varphi_2(v)\,v_2$ .

💡 Challenge: keep $v_1 = (1, 0)$ and move only $v_2$ until $\varphi_1$ becomes the first-coordinate functional $(x, y) \mapsto x$ .

IIIVisualization 2 — A Functional on $\mathcal{P}_2(\mathbf{R})$ Is Three Numbers

A linear functional on $\mathcal{P}_2(\mathbf{R})$ is fixed by its values on $1$ , $x$ , $x^2$ — those values are its coordinates in the dual basis.

φ a₀ 0 a₁ 0 a₂ 1

IVVisualization 3 — T′ Reverses the Arrow

For $T(x, y) = (x + 2y,\ 3x,\ x - y)$ , each $\varphi \in (\mathbf{R}^3)'$ becomes $T'(\varphi) = \varphi \circ T \in (\mathbf{R}^2)'$ .

c₁ 1 c₂ 1 c₃ 1 v = (x, y): x 1 y 2

VQuiz Questions

Problem 1 · Dual Basis in R²

Given: the basis $v_1 = (1, 1)$ , $v_2 = (1, 2)$ of $\mathbf{R}^2$ with dual basis $\varphi_1, \varphi_2$ — find $\varphi_1(x, y)$ .

✅ Correct!

\varphi_1(1, 1) = 2 - 1 = 1

and

\varphi_1(1, 2) = 2 - 2 = 0

, exactly the defining values.

❌ Not the first coordinate. The first-coordinate functional gives

1

v_2 = (1, 2)

, but

\varphi_1

must vanish there.

❌ That is $\varphi_2$ .

-x + y

takes the value

0

v_1

and

1

v_2

— it reads the coefficient of

v_2

❌ That belongs to another basis.

x - y

\varphi_1

for

(1, 0), (1, 1)

; here it gives

\varphi_1(1, 1) = 0

, not

1

❌ Not quite. Solve

(x, y) = c_1(1, 1) + c_2(1, 2)

for

c_1

Show solution

By the coefficient-reader theorem, $\varphi_1(x, y)$ is the coefficient $c_1$ in

(x, y) = c_1(1, 1) + c_2(1, 2) = (c_1 + c_2,\ c_1 + 2c_2).

Subtracting the first coordinate equation from the second gives $c_2 = y - x$ , and then $c_1 = x - c_2 = 2x - y$ . So

\varphi_1(x, y) = 2x - y, \qquad \varphi_2(x, y) = -x + y.

Check: $\varphi_1(1, 1) = 1$ , $\varphi_1(1, 2) = 0$ , $\varphi_2(1, 1) = 0$ , $\varphi_2(1, 2) = 1$ .

Problem 2 · Which Map Is a Linear Functional?

Given: four maps defined on $\mathcal{P}_2(\mathbf{R})$ — find the one that is a linear functional, then its value at $p(x) = 1 + 3x - x^2$ .

Which is a linear functional on $\mathcal{P}_2(\mathbf{R})$ ?

What is $\varphi(1 + 3x - x^2)$ for that functional?

✅ Correct! Differentiation, evaluation and integration are linear, so

2p'(1) - \int_0^1 p

is linear, and here it equals

2 - \tfrac{13}{6} = -\tfrac{1}{6}

❌ Not linear. Doubling

p

multiplies

p(1)\,p(2)

4

, not

2

❌ Not linear. A linear map sends

0

0

, but this one sends the zero polynomial to

1

❌ Linear, but not a functional.

p \mapsto p'

takes values in

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

, not in

\mathbf{R}

; a functional must return a scalar.

❌ Check the definition. A linear functional is a linear map into the scalars

\mathbf{F}

❌ Check the sign. The integral is subtracted:

2p'(1) - \int_0^1 p

❌ Recompute the two pieces.

p'(x) = 3 - 2x

gives

p'(1) = 1

;

\int_0^1 (1 + 3x - x^2)\,dx = 1 + \tfrac{3}{2} - \tfrac{1}{3}

Show solution

(a) fails homogeneity: $\varphi(2p) = 4\,p(1)\,p(2)$ . (b) fails $\varphi(0) = 0$ . (d) is linear but lands in $\mathcal{P}_2(\mathbf{R})$ , so it is not a map into $\mathbf{R}$ . (c) is a linear combination of $p \mapsto p'(1)$ and $p \mapsto \int_0^1 p$ , both linear functionals, so it is one.

For $p(x) = 1 + 3x - x^2$ : $p'(x) = 3 - 2x$ , so $p'(1) = 1$ , and

\int_0^1 (1 + 3x - x^2)\,dx = 1 + \tfrac{3}{2} - \tfrac{1}{3} = \tfrac{13}{6}.

\varphi(p) = 2 \cdot 1 - \tfrac{13}{6} = -\tfrac{1}{6}.

Problem 3 · Dual Basis in R³

Given: the basis $v_1 = (1, 0, 0)$ , $v_2 = (1, 1, 0)$ , $v_3 = (1, 1, 1)$ of $\mathbf{R}^3$ with dual basis $\varphi_1, \varphi_2, \varphi_3$ — find $\varphi_2$ and the coefficient $\varphi_1(4, 7, 2)$ .

What is $\varphi_2(x, y, z)$ ?

What is $\varphi_1(4, 7, 2)$ ?

✅ Correct!

(4, 7, 2) = -3\,v_1 + 5\,v_2 + 2\,v_3

, so the dual basis reads off

-3

5

2

❌ Not the second coordinate.

(x, y, z) \mapsto y

gives

1

v_3 = (1, 1, 1)

, but

\varphi_2(v_3)

must be

0

❌ Check the defining values.

\varphi_2

must give

0

v_1

1

v_2

and

0

v_3

❌ That is the first coordinate.

\varphi_1

reads the coefficient of

v_1

, and

v_2

v_3

also contribute to the first coordinate.

❌ Not quite. Solve for the coefficients from the last coordinate upward.

Show solution

Write $(x, y, z) = c_1(1, 0, 0) + c_2(1, 1, 0) + c_3(1, 1, 1) = (c_1 + c_2 + c_3,\ c_2 + c_3,\ c_3)$ .

From the bottom: $c_3 = z$ , then $c_2 = y - z$ , then $c_1 = x - c_2 - c_3 = x - y$ . By the coefficient-reader theorem,

\varphi_1(x, y, z) = x - y, \qquad \varphi_2(x, y, z) = y - z, \qquad \varphi_3(x, y, z) = z.

So $\varphi_1(4, 7, 2) = 4 - 7 = -3$ .

Check: $-3(1, 0, 0) + 5(1, 1, 0) + 2(1, 1, 1) = (4, 7, 2)$ .

Problem 4 · Computing Dual Maps

Given: $T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ with $T(x, y) = (x + y,\ 2x,\ x - 3y)$ and $\varphi \in (\mathbf{R}^3)'$ with $\varphi(u_1, u_2, u_3) = u_1 - u_2 + 2u_3$ ; also $D \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}))$ , $Dp = p'$ , and $\psi(p) = \int_0^1 p$ — find $T'(\varphi)$ and $\big(D'(\psi)\big)(x^3 - 4x)$ .

What is $\big(T'(\varphi)\big)(x, y)$ ?

What is $\big(D'(\psi)\big)(x^3 - 4x)$ ?

✅ Correct!

T'(\varphi) = \varphi \circ T

is the functional

x - 5y

\mathbf{R}^2

, and

\big(D'(\psi)\big)(p) = \int_0^1 p' = p(1) - p(0) = -3

❌ That is $\varphi$ itself.

T'

goes from

(\mathbf{R}^3)'

(\mathbf{R}^2)'

, so

T'(\varphi)

is a functional of

(x, y)

❌ Distribute the coefficient $2$ .

2(x - 3y) = 2x - 6y

, not

2x - 3y

❌ Not quite. Substitute

T(x, y)

into

\varphi

(x + y) - 2x + 2(x - 3y)

❌ That is $\psi(p)$ .

D'(\psi) = \psi \circ D

differentiates first: it integrates

p'

, not

p

❌ Not quite.

\big(D'(\psi)\big)(p) = \psi(Dp) = \int_0^1 p'

Show solution

The map $T$ . By definition $T'(\varphi) = \varphi \circ T$ , so

\big(T'(\varphi)\big)(x, y) = \varphi(x + y,\ 2x,\ x - 3y) = (x + y) - 2x + 2(x - 3y) = x - 5y.

It is a linear functional on $\mathbf{R}^2$ : $T'$ sends $(\mathbf{R}^3)'$ to $(\mathbf{R}^2)'$ .

The map $D$ . $\big(D'(\psi)\big)(p) = (\psi \circ D)(p) = \psi(p') = \int_0^1 p'$ , which by the fundamental theorem of calculus is $p(1) - p(0)$ . For $p(x) = x^3 - 4x$ :

\big(D'(\psi)\big)(p) = (1 - 4) - 0 = -3.

Directly: $p'(x) = 3x^2 - 4$ and $\int_0^1 (3x^2 - 4)\,dx = 1 - 4 = -3$ .

Solved: 0 / 4

The Dual Basis Reads Coefficients and T′ Reverses Arrows

IKey Formulas

IIVisualization 1 — The Dual Basis Reads Coefficients

IIIVisualization 2 — A Functional on P2(R)\mathcal{P}_2(\mathbf{R})P2​(R) Is Three Numbers

IVVisualization 3 — T′ Reverses the Arrow

VQuiz Questions

IIIVisualization 2 — A Functional on $\mathcal{P}_2(\mathbf{R})$ Is Three Numbers