MAT1041 · Week 7 · - Subspaces of Vector Spaces

Span and linear independence

You learn to test the subspace spanned by a set of vectors, whether the set spans the whole space, and whether the vectors are linearly independent, using a homogeneous system or a determinant, and to apply the same method in Rⁿ, P₂ and M₂₂.

linear combinationspanspanning setspans Vlinearly independentlinearly dependenttrivial solutiondependence relationcoefficient vector

We defined a subspace inside a set; now we do the opposite: starting from a few vectors, we build the subspace they generate. This idea has two faces. Span answers “where can I get to with these vectors?”, linear independence answers “is any of these vectors unnecessary?”. Combined, they give next week’s topic: the basis. The computing side is familiar: it is all Gaussian elimination.

Golden rules:

Write the vectors as columns of a matrix.
A span question = consistency of a nonhomogeneous system. An independence question = the homogeneous system having only the trivial solution.
$n$ vectors in $\mathbb{R}^n$ : the determinant alone answers both.
Turn polynomials and matrices into coefficient vectors first.

Span

Let $S = \{\mathbf{v}_1, \dots, \mathbf{v}_k\}$ be in a space $V$ . The span of $S$ is the set of all its linear combinations:

$\operatorname{span}(S) = \{c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k : c_i \in \mathbb{R}\}.$

$\operatorname{span}(S)$ is always a subspace of $V$ (the sum and multiples of two combinations are again combinations) and it is the smallest subspace containing $S$ . If $\operatorname{span}(S) = V$ we say ” $S$ spans $V$ ”.

Geometric intuition ( $\mathbb{R}^3$ ):

$S$	span
a single nonzero vector	a line through the origin
two non-parallel vectors	a plane through the origin
three vectors not in the same plane	$\mathbb{R}^3$

Whether a set spans the space

Example 1. Does $S = \{(1, 2, 3), (2, 1, 0), (4, 5, 6)\}$ span $\mathbb{R}^3$ ? Reduce $[A \mid \mathbf{b}]$ for an arbitrary $\mathbf{b} = (b_1, b_2, b_3)$ :

$\left[\begin{array}{ccc|c} 1 & 2 & 4 & b_1 \\ 2 & 1 & 5 & b_2 \\ 3 & 0 & 6 & b_3 \end{array}\right] \xrightarrow[R_3 \leftarrow R_3 - 3R_1]{R_2 \leftarrow R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & 2 & 4 & b_1 \\ 0 & -3 & -3 & b_2 - 2b_1 \\ 0 & -6 & -6 & b_3 - 3b_1 \end{array}\right] \xrightarrow{R_3 \leftarrow R_3 - 2R_2} \left[\begin{array}{ccc|c} 1 & 2 & 4 & b_1 \\ 0 & -3 & -3 & b_2 - 2b_1 \\ 0 & 0 & 0 & b_1 - 2b_2 + b_3 \end{array}\right]$

The right side of the last row is $b_3 - 3b_1 - 2(b_2 - 2b_1) = b_1 - 2b_2 + b_3$ . The system is consistent only when $b_1 - 2b_2 + b_3 = 0$ . So $S$ does not span $\mathbb{R}^3$ ; the span is the plane $x - 2y + z = 0$ . For example $(1, 1, 0)$ is not in the span ( $1 - 2 + 0 \neq 0$ ).

Square shortcut. With $n$ vectors in $\mathbb{R}^n$ the matrix is square: $\det \neq 0 \iff$ a unique solution for every $\mathbf{b}$ $\iff$ $S$ spans. In Example 1 $\det = 0$ , in Q2 $\det = 2$ .

PITFALL · spanning with too few vectors. You can’t span $\mathbb{R}^3$ with two vectors: a $3 \times 2$ matrix has at most 2 pivots, the third row becomes zero and a consistency condition appears. To span $\mathbb{R}^n$ you need at least $n$ vectors.

Linear independence

For $S = \{\mathbf{v}_1, \dots, \mathbf{v}_k\}$ , if the equation

$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \dots + c_k\mathbf{v}_k = \mathbf{0}$

has only the solution $c_1 = \dots = c_k = 0$ (the trivial solution), $S$ is linearly independent. If there is a nontrivial solution, it is dependent and that solution is a dependence relation.

Equivalent definition: $S$ is dependent $\iff$ at least one of the vectors is a linear combination of the others. In a dependent set there is an “unnecessary” vector: removing it does not change the span.

The test

Write the vectors as columns of a matrix $A$ .
Solve $A\mathbf{c} = \mathbf{0}$ (REF is enough).
No free variable $\Rightarrow$ independent. A free variable $\Rightarrow$ dependent; give the free variable 1 and read off the relation.

Example 2. $\mathbf{v}_1 = (1, 0, 2)$ , $\mathbf{v}_2 = (2, 1, 1)$ , $\mathbf{v}_3 = (0, 1, -3)$ .

$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 1 & -3 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 - 2R_1} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & -3 & -3 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 + 3R_2} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$

$c_3$ is free. $c_3 = 1$ : $c_2 = -1$ , $c_1 = -2c_2 = 2$ . The relation: $2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}$ . Check: $(2 - 2 + 0, 0 - 1 + 1, 4 - 1 - 3) = (0, 0, 0)$ . Dependent, and $\mathbf{v}_3 = \mathbf{v}_2 - 2\mathbf{v}_1$ .

Cases that decide without computation

Case	Result
The set contains $\mathbf{0}$	dependent ( $1 \cdot \mathbf{0} = \mathbf{0}$ )
Two vectors, one a multiple of the other	dependent
Two vectors, not multiples	independent
More than $n$ vectors in $\mathbb{R}^n$	dependent
A subset of an independent set	independent
Adding a vector to a dependent set	stays dependent

Polynomials and matrices

In $P_2$ match the polynomial $a + bx + cx^2$ with the vector $(a, b, c)$ , in $M_{2,2}$ match $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $(a, b, c, d)$ . Since linear combinations are preserved exactly in the coefficients, span and independence questions turn into questions in $\mathbb{R}^3$ or $\mathbb{R}^4$ .

Example 3. Does $\{1 + x, x + x^2, 1 + x^2\}$ span $P_2$ ? The coefficient vectors are $(1, 1, 0)$ , $(0, 1, 1)$ , $(1, 0, 1)$ ; the determinant is $2 \neq 0$ . It is both independent and spanning. The set in Q9 is different because it contains $1 - x^2$ : there the determinant is 0.

TIP · fix the order. When converting to coefficient vectors, use the same order for every polynomial (constant, $x$ , $x^2$ ). If you write $x^2$ first in one polynomial, the columns get mixed up.

Span and independence together

	Independent	Dependent
Spans	a basis (next week)	there are extra vectors, remove them
Does not span	a vector is missing, add one	both problems

In $\mathbb{R}^n$ , whether exactly $n$ vectors span and whether they are independent is the same question: $\det \neq 0$ .

Common mistakes

Wrong	Right
Writing the vectors as rows and setting up $[A \mid \mathbf{b}]$	Vectors are columns, $\mathbf{b}$ is the last column
Deciding “doesn’t span” without finding a single example $\mathbf{b}$	Write the consistency condition or use $\det$
Putting $\mathbf{b}$ on the right in the independence test	Independence is a homogeneous system: right side $\mathbf{0}$
Thinking three vectors are independent because two are	The third may be a combination of the first two
Confusing spanning with independence	$\{(1,2),(3,4),(5,6)\}$ spans but is dependent
Saying “no solution” when $\det = 0$	In a homogeneous system $\det = 0$ means infinitely many solutions = dependent
Writing polynomial coefficients in a different order	Always (constant, $x$ , $x^2$ )

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

Are $\mathbf{u} = (1, 2)$ and $\mathbf{v} = (-3, -6)$ linearly independent?

Show answer

$\mathbf{v} = -3\mathbf{u}$ . Two vectors are dependent if one is a multiple of the other: $3\mathbf{u} + \mathbf{v} = \mathbf{0}$ is a nontrivial relation.
Geometrically the two are on the same line.

Question 2medium

Does $S = \{(1, 1, 0), (0, 1, 1), (1, 0, 1)\}$ span $\mathbb{R}^3$ ?

Show answer

Yes

Make the vectors columns: $\det\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} = 1(1 - 0) - 0 + 1(1 - 0) = 2 \neq 0$ .
The square matrix is invertible, so $A\mathbf{c} = \mathbf{b}$ can be solved for every $\mathbf{b}$ . It spans $\mathbb{R}^3$ (and is independent).

Question 3medium

Describe $\operatorname{span}\{(1, 2, 3), (2, 1, 0), (4, 5, 6)\}$ as a geometric object in $\mathbb{R}^3$ .

Show answer

The plane

x - 2y + z = 0

$\det = 0$ , so the three are dependent and do not span $\mathbb{R}^3$ . Elimination on $[A \mid \mathbf{b}]$ gives the consistency condition $b_1 - 2b_2 + b_3 = 0$ .
Check: $(1, 2, 3)$ : $1 - 4 + 3 = 0$ ; $(2, 1, 0)$ : $2 - 2 = 0$ ; $(4, 5, 6)$ : $4 - 10 + 6 = 0$ .
The first two vectors are not parallel, so the span is a plane through the origin.

Question 4medium

Show that $\mathbf{v}_1 = (1, -1, 2)$ , $\mathbf{v}_2 = (2, 1, 1)$ , $\mathbf{v}_3 = (0, 3, -3)$ are linearly dependent and find a dependence relation.

Show answer

2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}

Solve the system $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}$ . The column matrix is $\begin{bmatrix} 1 & 2 & 0 \\ -1 & 1 & 3 \\ 2 & 1 & -3 \end{bmatrix}$ . $R_2 \leftarrow R_2 + R_1$ : $(0, 3, 3)$ . $R_3 \leftarrow R_3 - 2R_1$ : $(0, -3, -3)$ . $R_3 \leftarrow R_3 + R_2$ : a zero row. $c_3 = t$ is free: $c_2 = -t$ , $c_1 = -2c_2 = 2t$ . $t = 1$ : $2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = (2 - 2 + 0, -2 - 1 + 3, 4 - 1 - 3) = \mathbf{0}$ .

Question 5hard

For which values of $k$ are $(1, k, 0)$ , $(0, 1, k)$ , $(k, 0, 1)$ linearly dependent?

Show answer

k = -1

The determinant of the matrix written with these as rows (the transpose does not change the determinant): $\begin{vmatrix} 1 & k & 0 \\ 0 & 1 & k \\ k & 0 & 1 \end{vmatrix} = 1(1 - 0) - k(0 - k^2) + 0 = 1 + k^3$ .
Dependent $\iff 1 + k^3 = 0 \iff k = -1$ (the only real root).

Question 6easy

True or false: Any set of four vectors in $\mathbb{R}^3$ is linearly dependent.

Show answer

True

$c_1\mathbf{v}_1 + \dots + c_4\mathbf{v}_4 = \mathbf{0}$ : a homogeneous system with 3 equations and 4 unknowns. There are more unknowns, at least one free variable, so there is a nontrivial solution.
General rule: more than $n$ vectors in $\mathbb{R}^n$ are always dependent.

Question 7medium

Is $\{1 + x,\ x + x^2,\ 1 + x^2\}$ linearly independent in $P_2$ ?

Show answer

Yes

The coefficient vectors, in the order $(1, x, x^2)$ : $(1, 1, 0)$ , $(0, 1, 1)$ , $(1, 0, 1)$ .
The determinant of the column matrix is $2 \neq 0$ , so $c_1 = c_2 = c_3 = 0$ is the only solution. Independent.

Question 8medium

Are $A_1 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ , $A_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ , $A_3 = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}$ linearly independent in $M_{2,2}$ ?

Show answer

No:

A_3 = A_1 + A_2

$A_1 + A_2 = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} = A_3$ , so $A_1 + A_2 - A_3 = O$ .
Systematic way: turn each matrix into the vector $(a_{11}, a_{12}, a_{21}, a_{22})$ ; the $4 \times 3$ matrix has rank 2.

Question 9hard

Is $p(x) = 1 + 4x + x^2$ in $\operatorname{span}\{1 + x,\ x + x^2,\ 1 - x^2\}$ ?

Show answer

$a(1 + x) + b(x + x^2) + c(1 - x^2) = (a + c) + (a + b)x + (b - c)x^2$ .
Equations: $a + c = 1$ , $a + b = 4$ , $b - c = 1$ . Add the 1st and 3rd: $a + b = 2$ . But the 2nd equation says $a + b = 4$ . Inconsistent.
Note: the three polynomials are dependent ( $(1 + x) - (x + x^2) = 1 - x^2$ ), the span is only 2-dimensional.

Question 10hard

Suppose $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ is linearly independent. Is $\{\mathbf{u} + \mathbf{v},\ \mathbf{v} + \mathbf{w},\ \mathbf{u} + \mathbf{w}\}$ linearly independent?

Show answer

Yes

$a(\mathbf{u} + \mathbf{v}) + b(\mathbf{v} + \mathbf{w}) + c(\mathbf{u} + \mathbf{w}) = (a + c)\mathbf{u} + (a + b)\mathbf{v} + (b + c)\mathbf{w} = \mathbf{0}$ .
Because $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are independent, $a + c = 0$ , $a + b = 0$ , $b + c = 0$ . The determinant of the coefficient matrix is $2 \neq 0$ , the only solution is $a = b = c = 0$ .

Question 11medium

Does $\{(1, 2), (3, 4), (5, 6)\}$ span $\mathbb{R}^2$ ? Is it linearly independent?

Show answer

Spans

\mathbb{R}^2

; linearly dependent

$(1, 2)$ and $(3, 4)$ are already independent: $\det\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = -2 \neq 0$ . The two span $\mathbb{R}^2$ , and adding the third does not shrink the span.
But three vectors in $\mathbb{R}^2$ are always dependent. Spanning and being independent are different questions.

Question 12hard

Find $k$ so that $(1, 2, k)$ lies in $\operatorname{span}\{(1, 1, 1), (1, 3, 2)\}$ .

Show answer

k = 3/2

$a(1, 1, 1) + b(1, 3, 2) = (a + b, a + 3b, a + 2b)$ . $a + b = 1$ , $a + 3b = 2$ : $2b = 1$ , $b = \frac12$ , $a = \frac12$ .
Third component: $k = a + 2b = \frac12 + 1 = \frac32$ .

Verification: The span and independence tests were done in node with BigInt fractions using exact RREF, rank and determinants; the dependence relations were found from the null space and recomputed to confirm that the linear combination gives the zero vector; in the parametric independence question det = 1 + k³ was checked at many values of k (verify/MAT1041/07.mjs).

Review card

span(S) = ALL linear combinations of the vectors in S. Always a subspace. w ∈ span(S)? → solve the system [v1 ... vk | w] (vectors as COLUMNS). Is it consistent? Does S span Rⁿ? → is it consistent for every b? If square: det ≠ 0 ⇔ spans. Independent: c1v1 + ... + ckvk = 0 ⇒ only c1 = ... = ck = 0. Test: [v1 ... vk] x = 0 → NO free variable ⇔ independent. Square (n vectors, Rⁿ): det ≠ 0 ⇔ independent ⇔ spans. If dependent, a null space vector = a dependence relation. Quick rules: a set containing the 0 vector → dependent two vectors: one is a multiple of the other ⇔ dependent more than n vectors in Rⁿ → dependent; fewer than n vectors do NOT span Rⁿ P_n, M_mn: turn the polynomial/matrix into a coefficient vector, same test. Dependent ⇔ at least one is a combination of the others.