Linear-Algebra-Done-Right · Unit 3 · Video 2 · Interactive Practice

Linear Independence and the Linear Dependence Lemma

IKey Formulas

Formula	Name	Meaning
$a_1v_1 + \cdots + a_mv_m = 0 \implies a_1 = \cdots = a_m = 0$	Linearly independent	Equivalently, each vector in $\operatorname{span}(v_1, \dots, v_m)$ has only one representation; the empty list $(\,)$ is independent
$a_1v_1 + \cdots + a_mv_m = 0$ with some $a_k \neq 0$	Linearly dependent	For example $2(2, 3, 1) + 3(1, -1, 2) - (7, 3, 8) = 0$ ; any list containing $0$ is dependent
$v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$	Linear dependence lemma	Some $k \in \{1, \dots, m\}$ satisfies this when the list is dependent: take the largest $k$ with $a_k \neq 0$
$\operatorname{span}(v_1, \dots, v_{k-1}, v_{k+1}, \dots, v_m) = \operatorname{span}(v_1, \dots, v_m)$	Removal step	Holds for every $k$ with $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$

Key Insight: Subtracting two representations $v = \sum a_kv_k = \sum c_kv_k$ gives $0 = \sum (a_k - c_k)v_k$ , so uniqueness of the representation of $0$ forces uniqueness for every vector in the span.

IIUnique Coefficients in $\mathbf{R}^2$

For an independent list $v_1, v_2$ , each $v$ in the span has exactly one pair $a_1, a_2$ ; a dependent list gives many.

IIIThe Linear Dependence Lemma, One $k$ at a Time

Testing $k = 1, 2, \dots$ in turn finds the smallest $k$ with $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$ .

List in R³

💡 Challenge: in the list $(2, 3, 1), (1, -1, 2), (7, 3, c)$ , find every value of $c$ that makes the list dependent, then predict the coefficients of $v_3$ before changing $c$ .

IVQuiz Questions

Problem 1 · Two Representations

Given: vectors $v_1, v_2, v_3 \in V$ and a vector $v$ with $v = v_1 + 2v_2 + v_3$ and also $v = 3v_1 + v_2$ . Which combination equal to $0$ with scalars not all zero does this produce?

✅ Correct! Subtracting the two representations gives

(1 - 3)v_1 + (2 - 1)v_2 + (1 - 0)v_3 = 0

, and the scalar

-2

is not zero, so the list is dependent.

❌ Not quite. Adding the two equations gives

2v

on the left, not

0

. Subtract them instead: the left side becomes

v - v = 0

❌ Close, but check the sign. The coefficient of

v_1

1 - 3 = -2

, not

2

❌ Not quite. The two choices of scalars differ (

1 \neq 3

), so representations are not unique, and a list with non-unique representations is linearly dependent.

Show solution

Subtract the second equation from the first:

0 = v - v = (v_1 + 2v_2 + v_3) - (3v_1 + v_2 + 0v_3) = (1 - 3)v_1 + (2 - 1)v_2 + (1 - 0)v_3.

So $-2v_1 + v_2 + v_3 = 0$ with scalars $-2, 1, 1$ , not all zero. The list $v_1, v_2, v_3$ is therefore linearly dependent: an independent list gives each vector in its span only one representation. (Multiplying by $-1$ gives the equally valid relation $2v_1 - v_2 - v_3 = 0$ .)

Problem 2 · Which Statement Is True?

Exactly one of these statements holds for every vector space $V$ . Which one?

✅ Correct!

1v_1 + 0v_2 + (-1)v_1 = 0

has the nonzero scalar

1

, whatever

v_1

and

v_2

are.

❌ Not quite. A list of length one is independent exactly when its vector is nonzero:

1 \cdot 0 = 0

uses the nonzero scalar

1

❌ Not quite. The pairwise test works only for lists of length two. In

\mathbf{F}^3

, no vector in

(2, 3, 1), (1, -1, 2), (7, 3, 8)

is a multiple of another, yet

2(2, 3, 1) + 3(1, -1, 2) - (7, 3, 8) = 0

❌ Not quite. The true fact runs the other way: removing vectors from an independent list keeps it independent. Removing

(7, 3, 8)

from the dependent list

(2, 3, 1), (1, -1, 2), (7, 3, 8)

leaves an independent list.

Show solution

(c) is true. The combination $1v_1 + 0v_2 + (-1)v_1$ equals $0$ and its first scalar is $1 \neq 0$ , so $v_1, v_2, v_1$ is dependent. Equivalently, the third vector $v_1$ is a linear combination of the others.

(a) is false: $1 \cdot 0 = 0$ with scalar $1 \neq 0$ ; every list containing $0$ is dependent.

(b) is false: no two of $(2, 3, 1), (1, -1, 2), (7, 3, 8)$ are multiples of each other (compare first coordinates and then second coordinates), yet $2(2, 3, 1) + 3(1, -1, 2) - (7, 3, 8) = (0, 0, 0)$ .

(d) is false: the same dependent list loses its dependence when $(7, 3, 8)$ is removed, since $(2, 3, 1)$ and $(1, -1, 2)$ are not multiples of each other.

Problem 3 · When Is the List Dependent?

Given: the list $(1, 1, 2), (1, -1, 0), (3, 1, c)$ in $\mathbf{R}^3$ . Find the value of $c$ for which the list is linearly dependent.

✅ Correct!

(3, 1, 4) = 2(1, 1, 2) + 1(1, -1, 0)

, so

2(1, 1, 2) + (1, -1, 0) - (3, 1, 4) = 0

❌ Close, but… the scalars are swapped. From

x + y = 3

and

x - y = 1

, the scalar on

(1, 1, 2)

x = 2

and the scalar on

(1, -1, 0)

y = 1

❌ Not quite. The first coordinate alone does not fix the scalars:

x + y = 3

must be solved together with

x - y = 1

❌ Not quite. Write

(3, 1, c) = x(1, 1, 2) + y(1, -1, 0)

, solve the first two coordinates for

x

and

y

, then read off the third.

Show solution

Suppose $a_1(1, 1, 2) + a_2(1, -1, 0) + a_3(3, 1, c) = 0$ with $a_1, a_2, a_3$ not all zero.

If $a_3 = 0$ , the first two coordinates give $a_1 + a_2 = 0$ and $a_1 - a_2 = 0$ ; adding gives $2a_1 = 0$ , so $a_1 = a_2 = 0$ , a contradiction. So $a_3 \neq 0$ , and dividing by it writes $(3, 1, c) = x(1, 1, 2) + y(1, -1, 0)$ .

x + y = 3, \qquad x - y = 1 \implies 2x = 4 \implies x = 2,\ y = 1.

The third coordinate is then $c = 2x + 0y = 4$ . Conversely, for $c = 4$ :

2(1, 1, 2) + 1(1, -1, 0) - (3, 1, 4) = (2 + 1 - 3,\ 2 - 1 - 1,\ 4 + 0 - 4) = (0, 0, 0).

So the list is dependent exactly when $c = 4$ .

Problem 4 · The Lemma in $\mathcal{P}(\mathbf{R})$

Given: the dependent list $1 + z,\ 1 - z,\ z,\ z^2$ in $\mathcal{P}(\mathbf{R})$ . Find the smallest $k$ with $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$ , and write that $v_k$ as a combination of the earlier vectors.

What is the smallest k?

How is that vector a combination of the earlier ones?

✅ Correct!

k = 3

is the smallest index that works, and removing

z

leaves

\operatorname{span}(1 + z, 1 - z, z^2) = \operatorname{span}(1 + z, 1 - z, z, z^2)

❌ Check your k.

k = 1

needs

1 + z = 0

, which fails since

1 + z

is not the zero polynomial.

❌ Not quite.

1 - z = a(1 + z)

would need

a = 1

from the constant terms and

a = -1

from the

z

terms.

❌ Not quite. Every combination of

1 + z, 1 - z, z

has degree at most

1

, so

z^2

is not in their span; a smaller

k

works.

❌ Check the combination. Expand the right side: it must equal

z

, with constant term

0

and

z

-coefficient

1

❌ Close, but…

(1 + z) - (1 - z) = 2z

, not

z

: divide both scalars by

2

Show solution

$k = 1$ : $1 + z \neq 0$ , so $v_1 \notin \operatorname{span}(\,) = \{0\}$ .

$k = 2$ : if $1 - z = a(1 + z)$ , comparing constant terms gives $a = 1$ and comparing $z$ -coefficients gives $a = -1$ , a contradiction.

$k = 3$ : solve $z = a(1 + z) + b(1 - z) = (a + b) + (a - b)z$ :

a + b = 0, \qquad a - b = 1 \implies a = \tfrac{1}{2},\ b = -\tfrac{1}{2}.

So $z = \tfrac{1}{2}(1 + z) - \tfrac{1}{2}(1 - z)$ , and the smallest $k$ is $3$ . Check: $\tfrac{1}{2} + \tfrac{1}{2}z - \tfrac{1}{2} + \tfrac{1}{2}z = z$ . By the second part of the lemma, removing $z$ leaves the span unchanged.

The other options expand to $2z$ , to $1$ , and to $-z$ .

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Linear Independence and the Linear Dependence Lemma

IKey Formulas

IIUnique Coefficients in R2\mathbf{R}^2R2

IIIThe Linear Dependence Lemma, One kkk at a Time

IVQuiz Questions

IIUnique Coefficients in $\mathbf{R}^2$

IIIThe Linear Dependence Lemma, One $k$ at a Time