Linear-Algebra-Done-Right · Unit 30 · Video 1 · Interactive Practice

Why V × W Fails: Bilinear Functionals and dim B(V, W)

IKey Formulas

Formula	Name	Where it comes from
$\dim(V \times W) = \dim V + \dim W$	Product space	Basis $(e_j, 0)$ and $(0, f_k)$ : $m + n$ vectors
$\mathcal{M}(\beta)_{j,k} = \beta(e_j, f_k)$	Matrix of a bilinear functional	Values of $\beta$ on pairs of basis vectors
$\beta_C\big(\sum_j a_j e_j,\ \sum_k b_k f_k\big) = \sum_{j,k} C_{j,k}\, a_j b_k$	Inverse map $C \mapsto \beta_C$	Any $C \in \mathbf{F}^{m,n}$ gives a bilinear functional
$\dim \mathcal{B}(V, W) = (\dim V)(\dim W)$	Dimension of $\mathcal{B}(V, W)$	$\beta \mapsto \mathcal{M}(\beta)$ is an isomorphism onto $\mathbf{F}^{m,n}$

Key Insight: Linearity in each slot separately is what turns the count $m + n$ into $mn$ : a bilinear functional is free on every pair $(e_j, f_k)$ , not just on each basis vector.

IIA Sum Where a Product Is Needed

A basis of $V \times W$ has $m + n$ vectors, while a bilinear functional is fixed by $mn$ values.

dim V = m 2 dim W = n 3

IIIThe Matrix C Determines β_C

With $v = (1, 2)$ and $w = (3, -1, 4)$ , $\beta_C(v, w)$ adds one term $C_{j,k}\, a_j b_k$ per cell.

Inputs

💡 Challenge: change entries of $C$ until $\beta_C(v, w) = 0$ while $C \neq 0$ ; a nonzero bilinear functional can vanish at a pair.

IVQuiz Questions

Problem 1 · Two Dimension Counts

Given: $\dim V = 4$ and $\dim W = 3$ . Find $\dim(V \times W)$ and $\dim \mathcal{B}(V, W)$ .

What is $\dim(V \times W)$ ?

What is $\dim \mathcal{B}(V, W)$ ?

✅ Correct! The product space adds dimensions,

4 + 3 = 7

; bilinear functionals multiply them,

4 \cdot 3 = 12

❌ Check the product space. A basis of

V \times W

is the

4

vectors

(e_j, 0)

together with the

3

vectors

(0, f_k)

❌ Check $\mathcal{B}(V, W)$ . A bilinear functional is determined by one free value

\beta(e_j, f_k)

for each of the

4 \cdot 3

pairs; powers such as

4^3

count functions, not bilinear functionals.

Show solution

With bases $e_1, \dots, e_4$ of $V$ and $f_1, f_2, f_3$ of $W$ , the vectors $(e_1, 0), \dots, (e_4, 0), (0, f_1), (0, f_2), (0, f_3)$ form a basis of $V \times W$ , so

\dim(V \times W) = 4 + 3 = 7.

The map $\beta \mapsto \mathcal{M}(\beta)$ , $\mathcal{M}(\beta)_{j,k} = \beta(e_j, f_k)$ , is an isomorphism $\mathcal{B}(V, W) \to \mathbf{F}^{4,3}$ , so

\dim \mathcal{B}(V, W) = \dim \mathbf{F}^{4,3} = 4 \cdot 3 = 12.

Problem 2 · Which One Is Bilinear?

Given: functions on $\mathbf{R}^2 \times \mathbf{R}^2$ , with $v = (v_1, v_2)$ and $w = (w_1, w_2)$ . Which is a bilinear functional?

✅ Correct! For fixed

w

it is

v \mapsto 2w_2 v_1 - w_1 v_2

, linear in

v

; for fixed

v

it is linear in

w

the same way.

❌ Not quite.

v_1 + w_2

is linear on the pair

(v, w)

, which is a different condition: for fixed

w

v \mapsto v_1 + w_2

sends

0

w_2

, so it is not linear in

v

unless

w_2 = 0

❌ Not quite. A functional linear in

v

must send

v = 0

0

, but

\beta(0, w) = 1

❌ Not quite. For fixed

w

with

w_2 \neq 0

v \mapsto v_1^2 w_2

is not linear: doubling

v

multiplies it by

4

Show solution

Fix $w$ . Then $\beta(v, w) = (2w_2)\,v_1 + (-w_1)\,v_2$ is a linear functional of $v$ with fixed coefficients. Fix $v$ . Then $\beta(v, w) = (-v_2)\,w_1 + (2v_1)\,w_2$ is a linear functional of $w$ . So (d) is bilinear.

The others fail linearity in one slot: (a) and (b) do not send $0$ to $0$ in the first slot (for fixed $w$ with $w_2 \neq 0$ , resp. for every $w$ ), and (c) is quadratic in $v_1$ .

Problem 3 · Evaluate β_C

Given: the standard bases of $\mathbf{F}^2$ and $\mathbf{F}^3$ , and $C = \begin{pmatrix} 1 & 2 & 0 \\ 0 & -1 & 3 \end{pmatrix}$ . Find $\beta_C\big((2, 1),\ (1, 1, -1)\big)$ .

✅ Correct! Row 1 contributes

2 \cdot 3 = 6

and row 2 contributes

1 \cdot (-4) = -4

❌ Close, but…

6

is only the row

j = 1

part; the row

j = 2

terms

C_{2,k}\, a_2 b_k

add

-4

❌ Check the signs.

b_3 = -1

, so

C_{2,3}\, a_2 b_3 = 3 \cdot 1 \cdot (-1) = -3

, and

C_{2,2}\, a_2 b_2 = -1

❌ Not quite. Add

C_{j,k}\, a_j b_k

over all six cells, with

a = (2, 1)

and

b = (1, 1, -1)

Show solution

With $a = (2, 1)$ and $b = (1, 1, -1)$ ,

\beta_C(a, b) = \sum_{j=1}^{2} a_j \sum_{k=1}^{3} C_{j,k}\, b_k = 2\,(1 + 2 + 0) + 1\,(0 - 1 - 3) = 6 - 4 = 2.

Problem 4 · Bilinearity in Both Slots

Given: $\beta \in \mathcal{B}(V, W)$ with $\beta(v, w) = 5$ and $\beta(v, w') = -1$ . Find $\beta(2v,\ 3w - w')$ .

✅ Correct!

\beta(2v, 3w - w') = 2\big(3 \cdot 5 - (-1)\big) = 32

❌ Check the sign. The term is

-\beta(v, w') = -(-1) = +1

❌ Close, but…

\beta

is linear in the first slot too, so the factor

2

2v

comes out.

❌ Not quite. Pull the

2

out of the first slot, then expand the second slot:

\beta(v, 3w - w') = 3\beta(v, w) - \beta(v, w')

Show solution

\beta(2v,\ 3w - w') = 2\,\beta(v,\ 3w - w') = 2\big(3\beta(v, w) - \beta(v, w')\big) = 2\,(15 + 1) = 32.

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