Basis and dimension – 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 4 of 5

Basis and dimension

25 min

One point, two addresses

The point $(1, 4)$ is "1 along the $x$ -axis and 4 up". But in the diagram the same arrow $\vec{w}$ is also "2 steps of $\vec{u} = (1, 1)$ and 1 step of $\vec{v} = (-1, 2)$ ". Both descriptions are correct, and both pin down $\vec{w}$ exactly. A surveyor whose base line runs diagonally across a field, or a robot arm whose joints move in slanted directions, works naturally with the second kind of address.

What made both descriptions work? Each pair of directions does two jobs: it reaches every vector of the plane (it spans), and it does so in only one way (it is independent). A set that does both jobs is a basis.

Definition

Basis: a set $B = \{ \vec{v}_1, \ldots, \vec{v}_n \}$ of vectors in a vector space $V$ is a basis of $V$ when

$B$ is linearly independent, and
$B$ spans $V$ .

Standard basis of $\mathbb{R}^2$ : $\vec{e}_1 = (1, 0)$ , $\vec{e}_2 = (0, 1)$ , since $(x, y) = x\vec{e}_1 + y\vec{e}_2$ for every $(x, y)$ , and $a\vec{e}_1 + b\vec{e}_2 = (a, b) = \vec{0}$ only when $a = b = 0$ .

Example: $\{ (1, 1), (-1, 2) \}$ is a basis of $\mathbb{R}^2$ : $1(2) - 1(-1) = 3 \ne 0$ , so the vectors are independent, and two independent vectors of $\mathbb{R}^2$ span it (previous lesson).

Important

Theorem (unique coordinates). If $B = \{ \vec{u}, \vec{v} \}$ is a basis of $\mathbb{R}^2$ , every vector $\vec{w}$ can be written as $\vec{w} = a\vec{u} + b\vec{v}$ in exactly one way.

Proof. A way exists because $B$ spans. Suppose $\vec{w} = a\vec{u} + b\vec{v} = a'\vec{u} + b'\vec{v}$ . Subtracting, $(a - a')\vec{u} + (b - b')\vec{v} = \vec{0}$ . Independence forces $a - a^{'} = 0$ and $b - b^{'} = 0$ . $\blacksquare$

The unique numbers $(a, b)$ are the coordinates of $\vec{w}$ relative to $B$ , written $[\vec{w}]_B = (a, b)$ . The order of the basis vectors matters. In the diagram, $[\vec{w}]_B = (2, 1)$ for $B = \{ (1, 1), (-1, 2) \}$ , while the standard coordinates are $(1, 4)$ .

Worked example 1: coordinates relative to a basis

Show that $B = \{ (2, 1), (1, 3) \}$ is a basis of $\mathbb{R}^2$ , and find the coordinates of $\vec{w} = (7, 11)$ relative to $B$ .

Solution. $2(3) - 1(1) = 5 \ne 0$ : the vectors are independent, so they form a basis. Solve $a (2, 1) + b (1, 3) = (7, 11)$ :

\begin{cases} 2a + b = 7 \\ a + 3b = 11 \end{cases}

From the first, $b = 7 - 2 a$ ; then $a + 21 - 6 a = 11$ , so $a = 2$ and $b = 3$ . Check: $2(2, 1) + 3(1, 3) = (4 + 3,\ 2 + 9) = (7, 11)$ . ✓ Hence $[\vec{w}]_B = (2, 3)$ .

Worked example 2: from coordinates back to the vector

With the same basis $B = \{ (2, 1), (1, 3) \}$ , the vector $\vec{z}$ has $[\vec{z}]_B = (-1, 4)$ . Find $\vec{z}$ .

Solution. The coordinates are the coefficients: $\vec{z} = -1(2, 1) + 4(1, 3) = (-2 + 4,\ -1 + 12) = (2, 11)$ .

Definition

Dimension: if a vector space $V$ has a basis with $n$ vectors, then every basis of $V$ has $n$ vectors, and $n$ is the dimension of $V$ , written $\dim V = n$ . The zero space $\{ \vec{0} \}$ has the empty set as basis and dimension $0$ .

Example: $\dim \mathbb{R}^2 = 2$ , since the standard basis has two vectors. A line $\{ t\vec{u} : t \in \mathbb{R} \}$ with $\vec{u} \ne \vec{0}$ has basis $\{ \vec{u} \}$ and dimension $1$ .

Why every basis of $\mathbb{R}^2$ has two vectors

One vector spans at most a line, so it cannot span $\mathbb{R}^2$ . Three or more vectors of $\mathbb{R}^2$ are always dependent (previous lesson), so they cannot be a basis. That leaves exactly two. It also gives two shortcuts for $\mathbb{R}^2$ :

any two linearly independent vectors form a basis (they automatically span);
any two vectors that span $\mathbb{R}^2$ form a basis (they are automatically independent).

Subspace of $\mathbb{R}^2$	A basis	Dimension
$\{ \vec{0} \}$	$\varnothing$ (no vectors)	$0$
line $a x + b y = 0$ , not both $a, b$ zero	${(b, - a)}$	$1$
$\mathbb{R}^2$	$\{ \vec{e}_1, \vec{e}_2 \}$ , or any two independent vectors	$2$

The name "dimension" agrees with geometry: a point has dimension 0, a line 1 and a plane 2.

Worked example 3: basis and dimension of a subspace

Find a basis and the dimension of $W = \{ (x, y) \in \mathbb{R}^2 : 4x + 6y = 0 \}$ . Is $\{ (3, -2), (-6, 4) \}$ a basis of $W$ ?

Solution. $4x + 6y = 0 \iff y = -\frac{2}{3}x$ . Put $x = 3 t$ : then $y = - 2 t$ , and

W = \{ (3t, -2t) : t \in \mathbb{R} \} = \{ t(3, -2) : t \in \mathbb{R} \}.

${(3, - 2)}$ spans $W$ and is independent (a single non-zero vector), so it is a basis and $\dim W = 1$ .

$\{ (3, -2), (-6, 4) \}$ spans $W$ too, but $(- 6, 4) = - 2 (3, - 2)$ , so the set is dependent: not a basis. A basis of a 1-dimensional space has exactly one vector.

Warning

A basis is not unique, but the dimension is. $\mathbb{R}^2$ has infinitely many bases, and the coordinates of a vector change when the basis changes. What never changes is the number of vectors in a basis. Also note: $\{ (1, 0), (0, 1), (1, 1) \}$ spans $\mathbb{R}^2$ but is not a basis, because it is dependent; a basis has no spare vectors.

Tip

In $\mathbb{R}^2$ , "is it a basis?" is one line of arithmetic: there must be exactly two vectors, and $ad - bc \ne 0$ .

Dr. Trefor Bazett: The dimension of a subspace, definition and first examples (about 5 min). Watch for why every basis of a space has the same number of vectors.

✦

Practice

Answer each part with a reason.

Which are bases of $\mathbb{R}^2$ ? (a) $\{ (1, 3), (2, 6) \}$ (b) $\{ (1, 3), (2, 5) \}$ (c) ${(4, 0)}$ (d) $\{ (1, 0), (0, 1), (2, 3) \}$
$B = \{ (1, 1), (1, -1) \}$ . Find $[\vec{w}]_B$ for $\vec{w} = (6, 2)$ .
$B = \{ (3, 1), (-2, 1) \}$ and $[\vec{w}]_B = (2, -1)$ . Find $\vec{w}$ .
Find a basis and the dimension of $W = \{ (x, y) : x = 5y \}$ .
Find every value of $k$ for which $\{ (k, 2), (3, k + 1) \}$ is not a basis of $\mathbb{R}^2$ .
Show that $B = \{ (2, 1), (1, 3) \}$ is a basis and find the coordinates of $(1, 8)$ relative to it.
Explain why the line $W = \{ (x, y) : 2x - y = 0 \}$ cannot contain two linearly independent vectors.
The polynomials $a + b x$ form a 2-dimensional vector space. Show that $\{ 1 + x,\ 1 - x \}$ is a basis and write $5 + 3 x$ in this basis.

Show answersHide answers

(a) No: $1 (6) - 3 (2) = 0$ , dependent. (b) Yes: $1(5) - 3(2) = -1 \ne 0$ . (c) No: one vector spans only a line. (d) No: three vectors of $\mathbb{R}^2$ are dependent.
$a + b = 6$ , $a - b = 2$ : $a = 4$ , $b = 2$ . $[\vec{w}]_B = (4, 2)$ .
$\vec{w} = 2(3, 1) - (-2, 1) = (8, 1)$ .
$W = \{ (5t, t) \}$ : basis ${(5, 1)}$ , $\dim W = 1$ .
$k(k + 1) - 2(3) = 0 \iff k^2 + k - 6 = 0 \iff (k + 3)(k - 2) = 0$ : $k = 2$ or $k = - 3$ .
$2(3) - 1(1) = 5 \ne 0$ : a basis. $2 a + b = 1$ , $a + 3 b = 8$ : $b = 1 - 2 a$ , $a + 3 - 6 a = 8$ , $a = - 1$ , $b = 3$ . Coordinates $(- 1, 3)$ ; check $(-2 + 3,\ -1 + 9) = (1, 8)$ . ✓
Two independent vectors in $W$ would span $\mathbb{R}^2$ , and since $W$ is a subspace it would contain their span, so $W = \mathbb{R}^2$ . But $(1, 0) \notin W$ . (Equivalently: $\dim W = 1$ .)
$a (1 + x) + b (1 - x) = (a + b) + (a - b) x$ . Setting this equal to $0$ gives $a + b = 0$ and $a - b = 0$ , so $a = b = 0$ : independent, and two independent vectors in a 2-dimensional space form a basis. For $5 + 3 x$ : $a + b = 5$ , $a - b = 3$ , so $a = 4$ , $b = 1$ : $5 + 3 x = 4 (1 + x) + 1 (1 - x)$ .

Quick check

Which set is a basis of ℝ²?

Key points

A basis is a linearly independent set that spans the space.
Relative to a basis, every vector has unique coordinates $[\vec{w}]_B$ , found by solving a linear system.
All bases of a space have the same number of vectors, its dimension: $\dim \mathbb{R}^2 = 2$ , a line through $O$ has dimension 1, $\{ \vec{0} \}$ has dimension 0.
In $\mathbb{R}^2$ : two vectors with $ad - bc \ne 0$ form a basis.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, 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.
History: Hermann Günter Grassmann, J. J. O'Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews.
Etching: Hermann Grassmann etching, unknown author, Wikimedia Commons, public domain.
Video: The Dimension of a Subspace: Definition + First Examples, Dr. Trefor Bazett, embedded from YouTube.

1One point, two addresses

2Why every basis of R2 \mathbb{R}^2 R2 has two vectors

One point, two addresses

Why every basis of $\mathbb{R}^2$ has two vectors