Linear combinations and linear dependence – Vector Space of Real Numbers (Senior 4 Mathematics) | Ganzaa

Vector Space of Real Numbers

Lessons

Senior 4 · Mathematics

Vector Space of Real Numbers

5 lessons

Lesson 3 of 5

Linear combinations and linear dependence

25 min

Mixing to order

An agro-dealer stocks two fertiliser blends. A bag of blend A supplies $2\,\mathrm{kg}$ of nitrogen and $1\,\mathrm{kg}$ of phosphorus; a bag of blend B supplies $3\,\mathrm{kg}$ of nitrogen and $2\,\mathrm{kg}$ of phosphorus (illustrative figures). A cooperative's maize plot needs exactly $14\,\mathrm{kg}$ of nitrogen and $9\,\mathrm{kg}$ of phosphorus. How many bags of each?

Write each bag as a vector (nitrogen, phosphorus). We want numbers $a$ and $b$ with

a(2, 1) + b(3, 2) = (14, 9) \iff \begin{cases} 2a + 3b = 14 \\ a + 2b = 9 \end{cases}

From the second equation $a = 9 - 2 b$ ; then $2 (9 - 2 b) + 3 b = 14$ gives $b = 4$ and $a = 1$ . One bag of A and four of B: $(2, 1) + 4 (3, 2) = (14, 9)$ . The target is a linear combination of the two blends. Could every target be reached this way, and when is a third blend useless? Those are the questions of this lesson.

Definition

Linear combination: a vector $\vec{w}$ is a linear combination of $\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n$ when there are real numbers $a_1, a_2, \ldots, a_n$ (the coefficients) such that

\vec{w} = a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots + a_n\vec{v}_n .

Span: the set of all linear combinations of $\vec{v}_1, \ldots, \vec{v}_n$ is their span, written $\operatorname{span}\{ \vec{v}_1, \ldots, \vec{v}_n \}$ . If the span is all of $\mathbb{R}^2$ , the vectors span (or generate) $\mathbb{R}^2$ .

Example: $(14, 9) = 1 (2, 1) + 4 (3, 2)$ , so $(14, 9) \in \operatorname{span}\{ (2, 1), (3, 2) \}$ .

Seeing a combination

Let $\vec{u} = (2, 1)$ and $\vec{v} = (-1, 1)$ . To reach $\vec{w} = (1, 5)$ , solve $a (2, 1) + b (- 1, 1) = (1, 5)$ : $2 a - b = 1$ and $a + b = 5$ . Adding gives $3 a = 6$ , so $a = 2$ and $b = 3$ . In the diagram you walk two steps of $\vec{u}$ , then three steps of $\vec{v}$ . The dashed lines form a slanted grid built from $\vec{u}$ and $\vec{v}$ ; in that grid $\vec{w}$ has "coordinates" $2$ and $3$ , an idea the next lesson develops.

The span is always a subspace. It contains $\vec{0} = 0\vec{v}_1 + \cdots + 0\vec{v}_n$ ; the sum of two combinations is a combination (add the coefficients); and $k$ times a combination is a combination (multiply the coefficients by $k$ ). In $\mathbb{R}^2$ the span of one non-zero vector is a line through $O$ , and the span of two non-parallel vectors is the whole plane.

Worked example 1: is it in the span?

(a) Write $\vec{w} = (5, 3)$ as a linear combination of $\vec{u} = (1, 2)$ and $\vec{v} = (3, -1)$ . (b) Is $(4, 5)$ in $\operatorname{span}\{ (2, -3) \}$ ?

Solution.

(a) $a (1, 2) + b (3, - 1) = (5, 3)$ gives $a + 3 b = 5$ and $2 a - b = 3$ . From the second, $b = 2 a - 3$ ; then $a + 6 a - 9 = 5$ , so $a = 2$ and $b = 1$ . Hence $\vec{w} = 2\vec{u} + \vec{v}$ . Check: $(2, 4) + (3, - 1) = (5, 3)$ . ✓

(b) We need $(4, 5) = t (2, - 3)$ : the first components give $t = 2$ , the second give $t = -\frac{5}{3}$ . No single $t$ works, so $(4, 5)$ is not in the span; it is not on the line through $O$ in the direction $(2, - 3)$ .

Definition

Linear dependence and independence: vectors $\vec{v}_1, \ldots, \vec{v}_n$ are linearly dependent if there are scalars $a_1, \ldots, a_n$ , not all zero, with

a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots + a_n\vec{v}_n = \vec{0}.

Otherwise they are linearly independent: the only way to make $\vec{0}$ is the trivial combination $a_1 = a_2 = \cdots = a_n = 0$ .

Example: $(2, - 3)$ and $(- 4, 6)$ are dependent, because $2(2, -3) + 1(-4, 6) = \vec{0}$ . In words: one of them is a multiple of the other.

Important

Theorem (two vectors in $\mathbb{R}^2$ ). For $\vec{u} = (a, b)$ and $\vec{v} = (c, d)$ , the following are equivalent:

$\vec{u}$ and $\vec{v}$ are linearly dependent;
one of them is a scalar multiple of the other (they are parallel, or one is $\vec{0}$ );
$a d - b c = 0$ .

Proof of the test. Look for $x, y$ with $x\vec{u} + y\vec{v} = \vec{0}$ , that is $a x + c y = 0$ and $b x + d y = 0$ . Multiply the first equation by $d$ , the second by $c$ and subtract: $(a d - b c) x = 0$ . In the same way $(a d - b c) y = 0$ . If $ad - bc \ne 0$ , then $x = y = 0$ is forced: independent. If $a d - b c = 0$ , the pair $(x, y) = (d, - b)$ satisfies both equations (check: $a d - c b = 0$ and $b d - d b = 0$ ), and so does $(x, y) = (c, - a)$ . One of these pairs is non-zero unless $a = b = c = d = 0$ , and in that case both vectors are $\vec{0}$ . Either way the vectors are dependent. $\blacksquare$

The number $a d - b c$ is the determinant of the pair; you will study it in the next unit.

Worked example 2: testing pairs

Decide whether each pair is linearly dependent or independent. (a) $(2, - 3)$ , $(- 4, 6)$ (b) $(1, 2)$ , $(3, 1)$ (c) Find every $k$ for which $(k, 4)$ and $(1, k)$ are dependent.

Solution.

(a) $a d - b c = 2 (6) - (- 3) (- 4) = 12 - 12 = 0$ : dependent. Indeed $(- 4, 6) = - 2 (2, - 3)$ .

(b) $1(1) - 2(3) = -5 \ne 0$ : independent.

(c) $k \cdot k - 4 \cdot 1 = k^2 - 4 = 0 \iff k = 2$ or $k = - 2$ . Check $k = 2$ : $(2, 4) = 2 (1, 2)$ ; $k = - 2$ : $(- 2, 4) = - 2 (1, - 2)$ . ✓

Important

Theorem (three vectors in $\mathbb{R}^2$ ). Any three vectors $\vec{u}, \vec{v}, \vec{w}$ of $\mathbb{R}^2$ are linearly dependent.

Proof. If $\vec{u}$ and $\vec{v}$ are dependent, a non-trivial relation $x\vec{u} + y\vec{v} = \vec{0}$ already exists; add $0\vec{w}$ . If they are independent, then $ad - bc \ne 0$ and the system $x\vec{u} + y\vec{v} = \vec{w}$ can be solved (eliminate as in the proof above: $(a d - b c) x$ and $(a d - b c) y$ come out as known numbers, and you divide by $a d - b c$ ). Then $x\vec{u} + y\vec{v} - 1\vec{w} = \vec{0}$ , and the coefficient $- 1$ is not zero. $\blacksquare$

Worked example 3: finding the dependence relation

Show that $\vec{u} = (1, 1)$ , $\vec{v} = (2, -1)$ and $\vec{w} = (4, 7)$ are linearly dependent by finding a relation between them.

Solution. $\vec{u}$ and $\vec{v}$ are independent ( $1(-1) - 1(2) = -3 \ne 0$ ), so write $\vec{w} = x\vec{u} + y\vec{v}$ : $x + 2 y = 4$ and $x - y = 7$ . Subtracting, $3 y = - 3$ , so $y = - 1$ and $x = 6$ . Check: $6 (1, 1) - (2, - 1) = (4, 7)$ . ✓ The relation is

6\vec{u} - \vec{v} - \vec{w} = \vec{0},

with coefficients $6, - 1, - 1$ , not all zero.

Warning

Independence belongs to the whole set. $(1, 0)$ and $(0, 1)$ are independent, but add $(1, 1)$ and the set of three is dependent: $(1, 0) + (0, 1) - (1, 1) = \vec{0}$ . And any set that contains $\vec{0}$ is dependent, because $1 \cdot \vec{0} = \vec{0}$ is a non-trivial relation. Never call a single vector of a set "the dependent one" without saying which relation you mean.

Tip

For two vectors of $\mathbb{R}^2$ , write them side by side, $\begin{matrix} a & c \\ b & d \end{matrix}$ , and cross-multiply: $a d - b c$ . Zero means dependent (parallel); anything else means independent.

Dr. Trefor Bazett: Linear dependence and independence, geometrically (about 8 min). Watch for how a dependent set always has a vector that lies in the span of the others.

✦

Practice

Show your working; then open the answers.

Write $(7, 1)$ as a linear combination of $(1, 2)$ and $(3, - 1)$ .
Is $(6, - 9)$ in $\operatorname{span}\{ (-2, 3) \}$ ? Is $(6, 9)$ ?
Show that $(1, 1)$ and $(1, - 1)$ span $\mathbb{R}^2$ by writing a general vector $(x, y)$ as a combination of them.
Dependent or independent? (a) $(3, - 1), (- 6, 2)$ (b) $(4, 5), (5, 4)$ (c) $(0, 0), (1, 7)$ (d) $\left(\frac{1}{2}, 2\right), (1, 4)$
Find every value of $k$ for which $(k, 3)$ and $(12, k)$ are linearly dependent.
Find a linear relation between $\vec{u} = (2, 1)$ , $\vec{v} = (1, 3)$ and $\vec{w} = (4, -3)$ .
Blend A supplies $(4, 1)$ kg and blend B supplies $(2, 2)$ kg of (nitrogen, phosphorus) per bag. How many bags of each give exactly $(22, 10)$ kg?
Prove that if $\vec{u}$ and $\vec{v}$ are linearly independent, then so are $\vec{u} + \vec{v}$ and $\vec{u} - \vec{v}$ .

Show answersHide answers

$a + 3 b = 7$ , $2 a - b = 1$ : $b = 2 a - 1$ , $7 a - 3 = 7$ , $a = \frac{10}{7}$ , $b = \frac{13}{7}$ . Check: $\frac{10}{7} + \frac{39}{7} = 7$ and $\frac{20}{7} - \frac{13}{7} = 1$ . ✓
$(6, - 9) = - 3 (- 2, 3)$ : yes. $(6, 9) = t (- 2, 3)$ needs $t = - 3$ and $t = 3$ : no.
$a + b = x$ and $a - b = y$ give $a = \frac{x + y}{2}$ , $b = \frac{x - y}{2}$ , which exist for every $x, y$ .
(a) $3 (2) - (- 1) (- 6) = 0$ : dependent. (b) $16 - 25 = - 9$ : independent. (c) dependent: the set contains $\vec{0}$ . (d) $\frac{1}{2}(4) - 2(1) = 0$ : dependent; $(1, 4) = 2\left(\frac{1}{2}, 2\right)$ .
$k^{2} - 36 = 0$ , so $k = 6$ or $k = - 6$ .
$2 a + b = 4$ , $a + 3 b = - 3$ give $a = 3$ , $b = - 2$ : $\vec{w} = 3\vec{u} - 2\vec{v}$ , so $3\vec{u} - 2\vec{v} - \vec{w} = \vec{0}$ .
$4 a + 2 b = 22$ , $a + 2 b = 10$ : subtracting, $3 a = 12$ , $a = 4$ , $b = 3$ . Four bags of A, three of B.
If $a(\vec{u} + \vec{v}) + b(\vec{u} - \vec{v}) = \vec{0}$ , then $(a + b)\vec{u} + (a - b)\vec{v} = \vec{0}$ . Independence gives $a + b = 0$ and $a - b = 0$ , so $a = b = 0$ .

Quick check

Key points

A linear combination is $a_1\vec{v}_1 + \cdots + a_n\vec{v}_n$ ; finding the coefficients means solving a linear system.
The span of a set is the subspace of all its combinations.
Dependent: some non-trivial combination is $\vec{0}$ . For $(a, b)$ and $(c, d)$ : dependent $\iff ad - bc = 0$ .
In $\mathbb{R}^2$ , two independent vectors span the plane, and any three vectors are dependent.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples (the fertiliser figures are illustrative), the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Linear Dependence and Independence - Geometrically, Dr. Trefor Bazett, embedded from YouTube.

1Mixing to order

2Seeing a combination

Mixing to order

Seeing a combination