Linear-Algebra-Done-Right · Unit 28 · Video 1 · Interactive Practice

Alternating Multilinear Forms: Dependent Lists Give Zero

IKey Formulas

Formula	Name	What it says
$v \mapsto \beta(u_1, \dots, u_{k-1}, v, u_{k+1}, \dots, u_m)$ is linear	$m$ -linear form, $\beta \in V^{(m)}$	Linear in each slot, the others held fixed
$v_j = v_k,\ j \ne k \ \Rightarrow\ \alpha(v_1, \dots, v_m) = 0$	Alternating, $\alpha \in V_{\text{alt}}^{(m)}$	Zero on a repeated input
$v_1, \dots, v_m$ dependent $\Rightarrow \alpha(v_1, \dots, v_m) = 0$	Dependent lists give zero	Linear dependence lemma + one slot expanded
$m > \dim V \ \Rightarrow\ V_{\text{alt}}^{(m)} = \{0\}$	Too many slots	Every list of length $m$ is dependent

Key Insight: linear in each slot is not linear on $V^m$ : $\beta(2v_1, \dots, 2v_m) = 2^m \beta(v_1, \dots, v_m)$ , one factor of $2$ per slot.

IIExpanding One Slot

If $v_3 = a v_1 + b v_2$ , linearity in slot $3$ leaves only terms with a repeated input.

a 2 b 3 c 0

IIIThree Vectors in a Plane

In a space of dimension $2$ , any three vectors are dependent, so every alternating $3$ -linear form is $0$ .

IVQuiz Questions

Problem 1 · Doubling Every Input

Given: $\beta \in V^{(4)}$ and vectors $v_1, v_2, v_3, v_4 \in V$ with $\beta(v_1, v_2, v_3, v_4) = 1$ . Find $\beta(2v_1, 2v_2, 2v_3, 2v_4)$ .

✅ Correct! Each of the four slots contributes its own factor of

2

, so the value is

2^4 = 16

❌ Not quite. That would be true for a linear map on

V^4

. A

4

-linear form is linear in each slot separately, so each slot gives a factor of

2

❌ Not quite. Pull the

2

out of one slot at a time, using linearity in that slot. How many slots are there?

Show solution

Linearity in slot $1$ (the other slots held fixed) gives $\beta(2v_1, 2v_2, 2v_3, 2v_4) = 2\,\beta(v_1, 2v_2, 2v_3, 2v_4)$ . Repeating in slots $2$ , $3$ and $4$ :

\beta(2v_1, 2v_2, 2v_3, 2v_4) = 2^4\,\beta(v_1, v_2, v_3, v_4) = 16.

So a nonzero $4$ -linear form is not a linear map on $V^4$ : a linear map would only double.

Problem 2 · Which One Is Bilinear?

Given: four functions $\beta \colon \mathbf{R}^2 \times \mathbf{R}^2 \to \mathbf{R}$ , with $x = (x_1, x_2)$ and $y = (y_1, y_2)$ . Which one is a $2$ -linear form on $\mathbf{R}^2$ ?

✅ Correct! With

y

fixed,

x \mapsto y_2 x_1

is linear; with

x

fixed,

y \mapsto x_1 y_2

is linear.

❌ Not quite.

x_1 + y_1

is linear on

\mathbf{R}^4

, but with

y

fixed,

x \mapsto x_1 + y_1

is not linear: it sends

0

y_1

❌ Not quite. With

y

fixed,

x \mapsto x_1 y_1 + 1

sends

0

1

, so it is not linear in slot

1

❌ Not quite. Doubling

x

multiplies

(x_1 y_1)^2

4

, not

2

, so the function is not linear in slot

1

Show solution

A $2$ -linear form must be linear in $x$ for each fixed $y$ , and linear in $y$ for each fixed $x$ . A linear map sends $0$ to $0$ and scales by $\lambda$ when its input is scaled by $\lambda$ .

$x_1 + y_1$ and $x_1 y_1 + 1$ are nonzero at $x = 0$ for suitable $y$ : not linear in slot $1$ .
$(x_1 y_1)^2$ scales by $\lambda^2$ when $x$ is scaled by $\lambda$ : not linear in slot $1$ .
$x_1 y_2$ : for fixed $y$ it is $y_2$ times the coordinate functional $x \mapsto x_1$ ; for fixed $x$ it is $x_1$ times $y \mapsto y_2$ . Both are linear.

So only $\beta(x, y) = x_1 y_2$ is in $(\mathbf{R}^2)^{(2)}$ .

Problem 3 · Expand the Last Slot

Given: $\alpha \in V_{\text{alt}}^{(3)}$ and $u, v, w \in V$ with $\alpha(u, v, w) = 5$ . Find $\alpha(u, v, 2u - 3v + 4w)$ .

✅ Correct! The

u

and

v

terms repeat an input and vanish; only

4\,\alpha(u, v, w) = 20

survives.

❌ Not quite.

(2 - 3 + 4) \cdot 5 = 15

treats every term as

\alpha(u, v, w)

. The terms

\alpha(u, v, u)

and

\alpha(u, v, v)

have a repeated input.

❌ Not quite. The list

u, v, 2u - 3v + 4w

need not be dependent: the

4w

part is not a combination of

u

and

v

in general.

❌ Not quite. Expand slot

3

by linearity and use the fact that a repeated input gives

0

Show solution

Linearity in slot $3$ :

\alpha(u, v, 2u - 3v + 4w) = 2\,\alpha(u, v, u) - 3\,\alpha(u, v, v) + 4\,\alpha(u, v, w).

$\alpha(u, v, u)$ has $u$ in slots $1$ and $3$ , and $\alpha(u, v, v)$ has $v$ in slots $2$ and $3$ , so both are $0$ because $\alpha$ is alternating. Hence the value is $4 \cdot 5 = 20$ .

Problem 4 · Reading a Nonzero Value

Given: $\alpha \in V_{\text{alt}}^{(3)}$ and $u, v, w \in V$ with $\alpha(u, v, w) \ne 0$ . Which conclusion must hold?

✅ Correct! A dependent list would give

0

, so a nonzero value certifies independence.

❌ Not quite. An independent list of length

3

only forces

\dim V \ge 3

. For example,

\alpha

could be an alternating

3

-linear form on

\mathbf{R}^4

❌ Not quite. Nothing forces

\dim V = 3

, so three independent vectors need not span

V

❌ Not quite. Expanding slot

3

\alpha(u, v, u + w) = \alpha(u, v, u) + \alpha(u, v, w) = \alpha(u, v, w) \ne 0

❌ Not quite. Read the theorem "dependent ⇒ value

0

" in the other direction.

Show solution

If $u, v, w$ were linearly dependent, the theorem of this video would give $\alpha(u, v, w) = 0$ . The value is not $0$ , so $u, v, w$ is linearly independent.

That forces $\dim V \ge 3$ (an independent list is no longer than a spanning list), but not $\dim V = 3$ , so the list need not span $V$ . And $\alpha(u, v, u + w) = \alpha(u, v, u) + \alpha(u, v, w) = \alpha(u, v, w) \ne 0$ .

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