Vector Space of Real Numbers

Lessons

Senior 4 · Mathematics

Vector Space of Real Numbers

5 lessons

Lesson 2 of 5

Subspaces of ℝ²

25 min

A space inside a space

Take the line $y = 2 x$ and treat each of its points as a vector from the origin. Add two of them: $(1, 2) + (3, 6) = (4, 8)$ , and $8 = 2 \times 4$ , so the sum is on the line again. Multiply one by a scalar: $- 5 (1, 2) = (- 5, - 10)$ , still on the line. Whatever you do with the two operations, you never leave the line. The line is a vector space in its own right, sitting inside $\mathbb{R}^2$ .

Now shift the line up to $y = 2 x + 2$ . The points $(0, 2)$ and $(1, 4)$ are on it, but their sum $(1, 6)$ is not, since $2 \times 1 + 2 = 4 \ne 6$ . The shifted line is not closed under the operations. This lesson makes the difference precise.

Definition

Subspace: a subset $W$ of a vector space $V$ is a subspace of $V$ when $W$ is non-empty and is itself a vector space under the same addition and scalar multiplication as $V$ .

Notation: we write $W \le V$ or simply say " $W$ is a subspace of $V$ ".

Example: the line $W = \{ (x, y) : y = 2x \} = \{ (t, 2t) : t \in \mathbb{R} \}$ is a subspace of $\mathbb{R}^2$ .

Important

Theorem (subspace test). A subset $W \subseteq \mathbb{R}^2$ is a subspace of $\mathbb{R}^2$ if and only if

$\vec{0} \in W$ ;
$\vec{u}, \vec{v} \in W \Rightarrow \vec{u} + \vec{v} \in W$ (closed under addition);
$\vec{u} \in W,\ k \in \mathbb{R} \Rightarrow k\vec{u} \in W$ (closed under scalar multiplication).

Why three checks are enough. Conditions 2 and 3 are rules A1 and M1 for $W$ . Rules A2, A3, M2, M3, M4 and M5 are equations that hold for all vectors of $\mathbb{R}^2$ , so they hold in particular for the vectors of $W$ . Condition 1 gives A4, and condition 3 with $k = - 1$ puts $-\vec{u}$ in $W$ , which gives A5.

Conditions 2 and 3 can be merged into one: $a\vec{u} + b\vec{v} \in W$ for all $\vec{u}, \vec{v} \in W$ and all $a, b \in \mathbb{R}$ .

Lines through the origin, and lines that miss it

The picture shows why the origin matters. On the blue line $y = 2 x$ every sum and every multiple stays on the line. On the orange line $y = 2 x + 2$ the vector $\vec{0}$ is missing, and the sum of the position vectors of $P$ and $Q$ lands off the line. A line, a pair of lines or a region of the plane can fail in different ways, so test each condition:

Subset of $\mathbb{R}^2$	$\vec{0} \in W$ ?	Closed under $+$ ?	Closed under $k\cdot$ ?	Subspace?
$\{ \vec{0} \}$	yes	yes	yes	yes (the zero subspace)
line $y = 2 x$	yes	yes	yes	yes
line $y = 2 x + 2$	no	no	no	no
first quadrant $x \ge 0,\ y \ge 0$	yes	yes	no: $- 1 (1, 1) = (- 1, - 1)$	no
both axes $x y = 0$	yes	no: $(1, 0) + (0, 1) = (1, 1)$	yes	no
parabola $y = x^{2}$	yes	no: $(1, 1) + (1, 1) = (2, 2)$	no: $2 (1, 1) = (2, 2)$	no
the whole plane $\mathbb{R}^2$	yes	yes	yes	yes

Worked example 1: proving that a line is a subspace

Show that $W = \{ (x, y) \in \mathbb{R}^2 : 3x - 2y = 0 \}$ is a subspace of $\mathbb{R}^2$ .

Solution.

1. $3 (0) - 2 (0) = 0$ , so $\vec{0} \in W$ .

2. Let $(x_1, y_1), (x_2, y_2) \in W$ , so $3 x_{1} - 2 y_{1} = 0$ and $3 x_{2} - 2 y_{2} = 0$ . Their sum is $(x_1 + x_2,\ y_1 + y_2)$ and

3 (x_{1} + x_{2}) - 2 (y_{1} + y_{2}) = (3 x_{1} - 2 y_{1}) + (3 x_{2} - 2 y_{2}) = 0 + 0 = 0,

so the sum is in $W$ .

3. For $k \in \mathbb{R}$ , $k (x_{1}, y_{1}) = (k x_{1}, k y_{1})$ and $3(kx_1) - 2(ky_1) = k(3x_1 - 2y_1) = k \times 0 = 0$ , so $k(x_1, y_1) \in W$ .

All three conditions hold, so $W$ is a subspace. Geometrically it is the line $y = \frac{3}{2}x$ through the origin.

Worked example 2: deciding, with reasons

Which of these subsets of $\mathbb{R}^2$ are subspaces? (a) $A = \{ (x, y) : x + y = 1 \}$ (b) $B = \{ (t, -4t) : t \in \mathbb{R} \}$ (c) $C = \{ (x, y) : x \in \mathbb{Z},\ y \in \mathbb{Z} \}$

Solution.

(a) $0 + 0 \ne 1$ , so $\vec{0} \notin A$ : not a subspace.

(b) $t = 0$ gives $\vec{0} \in B$ . $(s, -4s) + (t, -4t) = (s + t,\ -4(s + t)) \in B$ and $k(t, -4t) = (kt,\ -4(kt)) \in B$ . So $B$ is a subspace: the line $y = - 4 x$ .

(c) $C$ contains $\vec{0}$ and is closed under addition, because a sum of integers is an integer. But $(1, 0) \in C$ and $\frac{1}{2}(1, 0) = \left(\frac{1}{2}, 0\right) \notin C$ . Scalars are all real numbers, so $C$ is not a subspace.

All the subspaces of $\mathbb{R}^2$

Suppose $W$ is a subspace that contains a non-zero vector $\vec{u}$ . By condition 3 it contains every multiple $t\vec{u}$ , that is, the whole line through $O$ in the direction of $\vec{u}$ . If $W$ also contains a vector $\vec{v}$ that is not on that line, then (as you will prove in the next two lessons) every vector of the plane can be written as $a\vec{u} + b\vec{v}$ , so $W = \mathbb{R}^2$ . This leaves only three possibilities:

Subspace of $\mathbb{R}^2$	Picture
$\{ \vec{0} \}$	the origin alone
$\{ t\vec{u} : t \in \mathbb{R} \}$ , with $\vec{u} \ne \vec{0}$	a line through the origin
$\mathbb{R}^2$	the whole plane

A line through the origin with equation $a x + b y = 0$ is the set of multiples of the direction vector $(b, - a)$ : for $3 x - 2 y = 0$ that is $(- 2, - 3)$ , or more neatly its multiple $(2, 3)$ .

Worked example 3: intersection and union

Let $W_1 = \{ (x, y) : y = x \}$ and $W_2 = \{ (x, y) : y = -x \}$ . Find $W_1 \cap W_2$ and decide whether $W_1 \cup W_2$ is a subspace.

Solution. A vector in both satisfies $y = x$ and $y = - x$ , so $x = - x$ , $x = 0$ and $y = 0$ : $W_1 \cap W_2 = \{ \vec{0} \}$ , a subspace.

The union contains $(1, 1) \in W_1$ and $(1, -1) \in W_2$ , but their sum $(2, 0)$ satisfies neither $y = x$ nor $y = - x$ . So $W_1 \cup W_2$ is not closed under addition and is not a subspace.

In general, the intersection of two subspaces is always a subspace (each condition holds in both, so it holds in the intersection), while a union of two different lines through $O$ never is.

Warning

"It contains $\vec{0}$ " does not finish the job. Condition 1 is a quick way to say no: if $\vec{0} \notin W$ , stop. But the axes $x y = 0$ and the parabola $y = x^{2}$ both contain $\vec{0}$ and are still not subspaces. To say yes you must prove all three conditions for general vectors, not for one or two examples.

Tip

For a set given by an equation, the test usually comes down to one question: is the equation linear and homogeneous (every term has exactly one variable to the first power, and there is no constant)? $3 x - 2 y = 0$ is; $x + y = 1$ , $y = x^{2}$ and $x y = 0$ are not.

Dr. Trefor Bazett: Subspaces are the natural subsets of linear algebra (about 6 min). Watch for the three conditions of the subspace test and for the examples that fail one of them.

✦

Practice

Give a reason for every yes and a counter-example for every no.

Is it a subspace of $\mathbb{R}^2$ ? (a) $\{ (x, y) : y = -3x \}$ (b) $\{ (x, y) : x - y = 2 \}$ (c) $\{ (x, y) : x = 0 \}$ (d) $\{ (x, y) : y = x^2 \}$
Prove that $W = \{ (x, y) : 5x + y = 0 \}$ is a subspace of $\mathbb{R}^2$ , and give a direction vector of the line.
For which value of $k$ is $\{ (x, y) : 2x + 3y = k \}$ a subspace? Justify.
Show that the upper half-plane $H = \{ (x, y) : y \ge 0 \}$ is closed under addition but is not a subspace.
Show that $S = \{ (x, y) : x^2 = y^2 \}$ is not a subspace. Describe $S$ geometrically.
Let $W_{1}$ be the line $y = x$ and $W_{2}$ the line $y = 3 x$ . Find $W_1 \cap W_2$ and show that $W_1 \cup W_2$ is not a subspace.
Find the smallest subspace of $\mathbb{R}^2$ that contains $\vec{u} = (2, -1)$ . Give it as a set of multiples and as an equation.
$W$ is a subspace of $\mathbb{R}^2$ that contains $(1, 2)$ and $(3, 1)$ . Show that $W$ contains $(5, 0)$ .

Show answersHide answers

(a) Yes: a line through $O$ with a linear homogeneous equation. (b) No: $0 - 0 \ne 2$ , so $\vec{0} \notin W$ . (c) Yes: the $y$ -axis, $1 x + 0 y = 0$ . (d) No: $(1, 1)$ is on it but $2 (1, 1) = (2, 2)$ is not, since $2^2 = 4 \ne 2$ .
$5 (0) + 0 = 0$ ; if $5 x_{1} + y_{1} = 0$ and $5 x_{2} + y_{2} = 0$ then $5 (x_{1} + x_{2}) + (y_{1} + y_{2}) = 0$ ; and $5 (k x_{1}) + k y_{1} = k (5 x_{1} + y_{1}) = 0$ . Direction vector $(1, - 5)$ .
$k = 0$ . If $k \ne 0$ , then $2(0) + 3(0) = 0 \ne k$ and $\vec{0}$ is missing; if $k = 0$ it is a line through $O$ with a linear homogeneous equation.
If $y_1 \ge 0$ and $y_2 \ge 0$ then $y_1 + y_2 \ge 0$ , so $H$ is closed under addition. But $(0, 1) \in H$ and $-1(0, 1) = (0, -1) \notin H$ .
$x^2 = y^2 \iff y = x \text{ or } y = -x$ : $S$ is the pair of lines $y = x$ and $y = - x$ . $(1, 1) + (1, - 1) = (2, 0)$ , and $2^2 \ne 0^2$ , so $S$ is not closed under addition.
$x = 3 x$ gives $x = 0$ , so $W_1 \cap W_2 = \{ \vec{0} \}$ . $(1, 1) + (1, 3) = (2, 4)$ : $4 \ne 2$ and $4 \ne 6$ , so the sum is in neither line.
$\{ t(2, -1) : t \in \mathbb{R} \}$ , the line $x + 2 y = 0$ . Any subspace containing $\vec{u}$ contains all its multiples, and this line is already a subspace.
$2(3, 1) - (1, 2) = (6 - 1,\ 2 - 2) = (5, 0)$ , and $W$ is closed under scalar multiplication and addition, so $(5, 0) \in W$ .

Quick check

The line y = 4x − 1 is a subspace of ℝ².

Key points

$W$ is a subspace when $\vec{0} \in W$ and $W$ is closed under addition and under multiplication by every real scalar.
$\vec{0} \notin W$ is a quick "no"; a "yes" needs a general proof.
The subspaces of $\mathbb{R}^2$ are $\{ \vec{0} \}$ , the lines through the origin, and $\mathbb{R}^2$ itself.
Intersections of subspaces are subspaces; unions usually are not.

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.
Video: Subspaces are the Natural Subsets of Linear Algebra, Dr. Trefor Bazett, embedded from YouTube.

1A space inside a space

2Lines through the origin, and lines that miss it

3All the subspaces of R2 \mathbb{R}^2 R2

A space inside a space

Lines through the origin, and lines that miss it

All the subspaces of $\mathbb{R}^2$