Subspaces of a vector space – Intersection and Sum of Subspaces (Senior 6 Mathematics) | Ganzaa

Intersection and Sum of Subspaces

Lessons

Senior 6 · Mathematics

Intersection and Sum of Subspaces

3 lessons

Lesson 1 of 3

Subspaces of a vector space

30 min

Smaller vector spaces inside bigger ones

In S4 (Vector Space of Real Numbers) and S5 (Vector Space $\mathbb{R}^3$ ) you added vectors, multiplied them by scalars, and described them with a basis. Inside $\mathbb{R}^3$ some sets of vectors behave like small vector spaces of their own: a plane through the origin, or a line through the origin. Adding two vectors of the plane, or stretching one, never takes you out of it. This unit studies such subspaces, and then asks what happens when two of them meet (their intersection) or are combined (their sum).

Definition

Vector space (real): a set $V$ with an addition and a multiplication by real scalars that satisfy the usual rules (commutative and associative addition, a zero vector $\mathbf{0}$ , opposites, $1\mathbf{v} = \mathbf{v}$ , and the distributive laws). Examples: $\mathbb{R}^n$ ; the set $M_2(\mathbb{R})$ of $2 \times 2$ real matrices; the set of polynomials of degree at most 2.

Subspace: a subset $W$ of a vector space $V$ that is itself a vector space with the same addition and scalar multiplication. We write $W \le V$ .

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

Important

Subspace test. A subset $W$ of a vector space $V$ is a subspace if and only if

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

Conditions 2 and 3 can be combined: $\alpha\mathbf{u} + \beta\mathbf{v} \in W$ for all $\mathbf{u}, \mathbf{v} \in W$ and all scalars $\alpha, \beta$ .

Reason: the other rules (commutativity, distributive laws and so on) hold in $W$ because they hold in $V$ . The opposite of $\mathbf{u}$ is $(-1)\mathbf{u}$ , which is in $W$ by condition 3.

The line y = 2x (blue) is a subspace of ℝ²: it contains O, and sums and multiples of its vectors stay on it. The line y = 2x + 2 (red) is not: it misses O, and p + q falls off it.

Worked example 1: testing subsets of $\mathbb{R}^2$

Which of these are subspaces of $\mathbb{R}^2$ ? (a) $A = \{(x, y) : y = 2x\}$ (b) $B = \{(x, y) : y = 2x + 2\}$ (c) $Q = \{(x, y) : x \ge 0,\ y \ge 0\}$ (d) $X = \{(x, y) : xy = 0\}$ , the union of the two axes.

Solution

Yes. $(0, 0) \in A$ . If $(a, 2 a)$ and $(b, 2 b)$ are in $A$ , then $\alpha(a, 2a) + \beta(b, 2b) = (\alpha a + \beta b,\ 2(\alpha a + \beta b)) \in A$ .
No: $0 \ne 2(0) + 2$ , so $(0, 0) \notin B$ .
No: $(1, 1) \in Q$ but $(-1)(1, 1) = (-1, -1) \notin Q$ . The first quadrant is closed under addition, not under multiplication by negative scalars.
No: $(1, 0) \in X$ and $(0, 1) \in X$ , but their sum $(1, 1) \notin X$ .

Warning

Check the zero vector first. It is the quickest test, and it rules out every line or plane that misses the origin. But passing it is not enough: $Q$ and $X$ above both contain $(0, 0)$ and still fail.

Span, basis and dimension (reminder)

The span of vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ is the set of all their linear combinations, $\operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} = \{\alpha_1\mathbf{v}_1 + \cdots + \alpha_k\mathbf{v}_k\}$ . A span is always a subspace, since a combination of combinations is again a combination. A basis of a subspace $W$ is a linearly independent set that spans $W$ , and the dimension $\dim W$ is the number of vectors in any basis.

subspaces of $\mathbb{R}^3$	dimension
$\{\mathbf{0}\}$ , the zero subspace	0
a line through the origin, $\operatorname{span}\{\mathbf{v}\}$ with $\mathbf{v} \ne \mathbf{0}$	1
a plane through the origin, $a x + b y + c z = 0$	2
$\mathbb{R}^3$ itself	3

If $W \le V$ , then $\dim W \le \dim V$ , with equality only when $W = V$ .

Worked example 2: a basis from equations

(a) Show that $W = \{(x, y, z) : x - 2y + z = 0\}$ is a subspace of $\mathbb{R}^3$ , and find a basis and its dimension. (b) Do the same for $L = \{(x, y, z) : x + y + z = 0,\ x - z = 0\}$ .

Solution

$(0, 0, 0)$ satisfies the equation. If $\mathbf{u}$ and $\mathbf{v}$ satisfy it, then so does $\alpha\mathbf{u} + \beta\mathbf{v}$ , because the left side is linear: $(\alpha u_1 + \beta v_1) - 2(\alpha u_2 + \beta v_2) + (\alpha u_3 + \beta v_3) = \alpha \cdot 0 + \beta \cdot 0 = 0$ . Solve for $x$ : $x = 2 y - z$ , with $y$ and $z$ free: $(2y - z,\ y,\ z) = y(2, 1, 0) + z(-1, 0, 1) .$ The vectors $(2, 1, 0)$ and $(- 1, 0, 1)$ span $W$ and are independent (neither is a multiple of the other), so they form a basis and $\dim W = 2$ : a plane through $O$ .
From $x - z = 0$ , $z = x$ ; then $y = - x - z = - 2 x$ . So $L = \{x(1, -2, 1)\} = \operatorname{span}\{(1, -2, 1)\}$ , a line through $O$ , and $\dim L = 1$ .

The pattern: one independent linear equation in $\mathbb{R}^3$ leaves 2 free variables (a plane); two independent equations leave 1 (a line).

Worked example 3: a subspace of matrices

Show that the symmetric matrices $S = \left\{\begin{pmatrix} a & b \\ b & d \end{pmatrix}\right\}$ form a subspace of $M_2(\mathbb{R})$ , and find its dimension.

Solution The zero matrix is symmetric, and a combination of symmetric matrices is symmetric, since $(\alpha A + \beta B)^T = \alpha A^T + \beta B^T = \alpha A + \beta B$ . Every element can be written

a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},

and these three matrices are independent, so $\dim S = 3$ , while $\dim M_2(\mathbb{R}) = 4$ .

Tip

A set described by linear, homogeneous equations (every term contains a variable to the first power, and the right-hand side is 0) is always a subspace. A constant term (like $+ 2$ ), a power (like $x^{2}$ ), a product (like $x y$ ) or an inequality (like $x \ge 0$ ) is a warning sign: test it.

3Blue1Brown: Linear combinations, span, and basis vectors, Chapter 2 of Essence of Linear Algebra (about 10 min). Watch how the span of two vectors in space is a plane through the origin, and the span of one vector a line through it: the subspaces of the table above.

✦

Practice

Questions 6 and 7 ask for reasons.

Which of these are subspaces of $\mathbb{R}^3$ ? (a) $\{(x, y, z) : z = 0\}$ (b) $\{(x, y, z) : z = 1\}$ (c) $\{(x, y, z) : x = 2y,\ z = 3y\}$ (d) $\{(x, y, z) : x^2 = y^2\}$ (e) $\{(t, t^2, 0) : t \in \mathbb{R}\}$
Find a basis and the dimension of $W = \{(x, y, z, t) \in \mathbb{R}^4 : x + y = 0,\ z - 2t = 0\}$ .
Show that $W = \{(a, b, a + b) : a, b \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^3$ . Find a basis, and one equation that describes $W$ .
Is $(3, 1, 4)$ in $\operatorname{span}\{(1, 0, 1), (1, 1, 2)\}$ ?
Show that the $2 \times 2$ matrices with trace 0 ( $a + d = 0$ ) form a subspace of $M_2(\mathbb{R})$ , and find a basis.
Explain why a subspace of $\mathbb{R}^3$ of dimension 3 must be $\mathbb{R}^3$ itself.
For which values of $k$ is $\{(x, y, z) : x + 2y - z = k\}$ a subspace of $\mathbb{R}^3$ ? Explain.

Show answersHide answers

(a) yes (b) no: $\mathbf{0}$ is missing (c) yes: $\operatorname{span}\{(2, 1, 3)\}$ (d) no: $(1, 1, 0) + (1, - 1, 0) = (2, 0, 0)$ , and $4 \ne 0$ (e) no: $2 (1, 1, 0) = (2, 2, 0)$ , but $2 \ne 2^2$
$(- y, y, 2 t, t) = y (- 1, 1, 0, 0) + t (0, 0, 2, 1)$ ; basis $\{(-1, 1, 0, 0), (0, 0, 2, 1)\}$ , dimension 2.
$W = \operatorname{span}\{(1, 0, 1), (0, 1, 1)\}$ , a span, hence a subspace; basis $\{(1, 0, 1), (0, 1, 1)\}$ ; equation $z = x + y$ .
Yes: $2 (1, 0, 1) + 1 (1, 1, 2) = (3, 1, 4)$ .
Zero has trace 0, and $\operatorname{tr}(\alpha A + \beta B) = \alpha\operatorname{tr}A + \beta\operatorname{tr}B = 0$ . Basis $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ ; dimension 3.
A basis of the subspace is 3 independent vectors in $\mathbb{R}^3$ , and any 3 independent vectors in $\mathbb{R}^3$ span $\mathbb{R}^3$ .
Only $k = 0$ : for $k \ne 0$ the zero vector does not satisfy the equation.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed values), the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S6 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Linear combinations, span, and basis vectors | Chapter 2, Essence of linear algebra, 3Blue1Brown, embedded from YouTube.

1Smaller vector spaces inside bigger ones

2Span, basis and dimension (reminder)

Smaller vector spaces inside bigger ones

Span, basis and dimension (reminder)