Linear combinations, basis and dimension – Vector Space ℝ³ of Real Numbers (Senior 5 Mathematics) | Ganzaa

Vector Space ℝ³ of Real Numbers

Lessons

Senior 5 · Mathematics

Vector Space ℝ³ of Real Numbers

4 lessons

Lesson 2 of 4

Linear combinations, basis and dimension

40 min

Building every vector from a few

Look closely at a phone screen and each pixel turns out to be three tiny lights: red, green and blue. Every colour you see is made by choosing how strongly to light each of the three. Space works the same way: every vector of $\mathbb{R}^3$ is $x i + y j + z k$ , a mixture of three building blocks. This lesson asks which sets of vectors can serve as building blocks, how many are needed, and how to describe a vector using building blocks other than $i, j, k$ .

Definition

A linear combination of vectors $v_1, v_2, \dots, v_n$ is any vector $\lambda_1v_1 + \lambda_2v_2 + \dots + \lambda_nv_n$ with real scalars $\lambda_1, \dots, \lambda_n$ . The set of all their linear combinations is their span, written $\operatorname{span}\{v_1, \dots, v_n\}$ or $\langle v_1, \dots, v_n \rangle$ .

Example: $(3, 2, 5) = 3 (1, 0, 1) + 2 (0, 1, 1)$ , so $(3, 2, 5) \in \operatorname{span}\{(1, 0, 1), (0, 1, 1)\}$ .

Worked example 1: is it a linear combination?

Let $u = (1, 2, 3)$ and $v = (0, 1, 2)$ . Is $w = (1, 4, 7)$ a linear combination of $u$ and $v$ ? Is $w^{'} = (1, 4, 6)$ ?

Solution. Solve $\lambda(1, 2, 3) + \mu(0, 1, 2) = (1, 4, 7)$ , that is $\lambda = 1$ , $2\lambda + \mu = 4$ , $3\lambda + 2\mu = 7$ . The first two give $\lambda = 1$ , $\mu = 2$ ; the third checks: $3 + 4 = 7$ . ✓ So $w = u + 2 v$ .

For $w^{'}$ , the same first two equations force $\lambda = 1$ , $\mu = 2$ , but then $3\lambda + 2\mu = 7 \ne 6$ . No scalars work: $w^{'}$ is not in $\operatorname{span}\{u, v\}$ .

Definition

A subspace of $\mathbb{R}^3$ is a non-empty subset $W$ that is closed under the two operations: if $u, v \in W$ and $\lambda \in \mathbb{R}$ , then $u + v \in W$ and $\lambda u \in W$ . Every subspace contains $0$ (take $\lambda = 0$ ), and is itself a vector space.

Examples: ${0}$ ; any line through $O$ ; any plane through $O$ ; $\mathbb{R}^3$ itself. A span is always a subspace. The plane $x + y + z = 1$ is not a subspace, because it does not contain $(0, 0, 0)$ . (In S6 you will test subspaces in more detail and combine them.)

The span of two vectors

Two non-parallel vectors $u$ and $v$ span a plane through the origin. Every vector of that plane is $\lambda u + \mu v$ for exactly one pair $(\lambda, \mu)$ ; a vector sticking out of the plane, like $p$ in the figure, is not a combination of $u$ and $v$ at all. A third building block is needed to reach it.

Definition

Vectors $v_1, \dots, v_n$ are linearly independent if

\lambda_1v_1 + \lambda_2v_2 + \dots + \lambda_nv_n = 0 \quad \Longrightarrow \quad \lambda_1 = \lambda_2 = \dots = \lambda_n = 0 .

Otherwise they are linearly dependent: some combination with scalars not all 0 gives $0$ , and then one of the vectors is a combination of the others. Two vectors are dependent exactly when they are parallel (or one is 0).

Important

Test for three vectors of $\mathbb{R}^3$ . $u, v, w$ are linearly independent if and only if the determinant with $u, v, w$ as its columns is not 0:

\det(u, v, w) = \begin{vmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{vmatrix} \ne 0 .

Reason. $\lambda u + \mu v + \nu w = 0$ is a system of three equations whose coefficient matrix has columns $u, v, w$ and whose right side is 0. It always has the solution $(0, 0, 0)$ , and by unit 6 it has only that solution exactly when the determinant is not 0. Geometrically, $\det = 0$ means the three vectors lie in one plane through $O$ .

Worked example 2: a dependence relation

Are $u = (1, 2, - 1)$ , $v = (0, 1, 3)$ and $w = (2, 5, 1)$ linearly independent?

Solution.

\begin{vmatrix} 1 & 0 & 2 \\ 2 & 1 & 5 \\ -1 & 3 & 1 \end{vmatrix} = 1(1 - 15) - 0 + 2(6 + 1) = -14 + 14 = 0 .

They are dependent. To find the relation, try $w = \lambda u + \mu v$ : the first components give $\lambda = 2$ , the second $4 + \mu = 5$ , so $\mu = 1$ ; the third checks: $- 2 + 3 = 1$ . ✓ So $2 u + v - w = 0$ .

Definition

A basis of a vector space (or subspace) $V$ is a set of vectors that is linearly independent and spans $V$ . Every basis of $V$ has the same number of vectors, the dimension of $V$ , written $\dim V$ .

$\{i, j, k\}$ is the standard basis of $\mathbb{R}^3$ , so $\dim \mathbb{R}^3 = 3$ .
A line through $O$ has dimension 1; a plane through $O$ has dimension 2; $\dim\{0\} = 0$ .
In $\mathbb{R}^3$ , any 3 linearly independent vectors form a basis, and any 4 vectors are linearly dependent.

Definition

If $B = (b_{1}, b_{2}, b_{3})$ is a basis of $\mathbb{R}^3$ , every vector $x$ can be written in exactly one way as $x = \lambda_1b_1 + \lambda_2b_2 + \lambda_3b_3$ . The numbers $(\lambda_1, \lambda_2, \lambda_3)$ are the coordinates of $x$ relative to $B$ , written $[x]_{B}$ .

Why exactly one way: if also $x = \mu_1b_1 + \mu_2b_2 + \mu_3b_3$ , subtracting gives $(\lambda_1 - \mu_1)b_1 + (\lambda_2 - \mu_2)b_2 + (\lambda_3 - \mu_3)b_3 = 0$ , and independence forces every $\lambda_i = \mu_i$ .

Worked example 3: coordinates in a new basis

Show that $B = \big((1, 1, 0), (0, 1, 1), (1, 0, 1)\big)$ is a basis of $\mathbb{R}^3$ , and find the coordinates of $x = (5, 3, 4)$ relative to $B$ .

Solution. $\begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 1(1 - 0) - 0 + 1(1 - 0) = 2 \ne 0$ , so the three vectors are independent and form a basis. Now solve $a (1, 1, 0) + b (0, 1, 1) + c (1, 0, 1) = (5, 3, 4)$ :

a + c = 5, \qquad a + b = 3, \qquad b + c = 4 .

Adding all three: $2 (a + b + c) = 12$ , so $a + b + c = 6$ . Then $b = 6 - 5 = 1$ , $c = 6 - 3 = 3$ , $a = 6 - 4 = 2$ . So $[x]_{B} = (2, 1, 3)$ . Check: $2 (1, 1, 0) + (0, 1, 1) + 3 (1, 0, 1) = (5, 3, 4)$ . ✓

Worked example 4: a basis of a plane through $O$

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

Solution. Solve for $x$ : $x = 2 y - z$ . So every vector of $W$ is

(2y - z,\ y,\ z) = y(2, 1, 0) + z(-1, 0, 1), \qquad y, z \in \mathbb{R} .

$(2, 1, 0)$ and $(- 1, 0, 1)$ span $W$ and are not parallel, so they are independent: $\{(2, 1, 0), (-1, 0, 1)\}$ is a basis and $\dim W = 2$ , a plane through $O$ .

Warning

A set containing the zero vector is always dependent: $1 \cdot 0 = 0$ uses a non-zero scalar.
Spanning is not enough for a basis. $(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)$ span $\mathbb{R}^3$ but are dependent, so they are not a basis.
Coordinates depend on the order of the basis. Swapping $b_{1}$ and $b_{2}$ swaps the first two coordinates.

Tip

Three vectors in $\mathbb{R}^3$ ? Put them as columns and take the determinant: not 0 means "independent, a basis"; 0 means "dependent, all in one plane through $O$ ".

3Blue1Brown: Linear combinations, span, and basis vectors, Chapter 2 of Essence of Linear Algebra (about 10 min). Near the end, watch what the span of two and of three vectors in space looks like.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Write $(7, 1, 5)$ as a linear combination of $(1, 0, 1)$ and $(2, 1, 0)$ .
Is $(3, 1, 2)$ in $\operatorname{span}\{(1, 0, 1), (2, 1, 0)\}$ ? Explain.
Show that $(1, 1, 0)$ , $(1, 0, 1)$ and $(0, 1, 1)$ form a basis of $\mathbb{R}^3$ .
Find the real value of $k$ for which $(1, k, 0)$ , $(0, 1, k)$ and $(k, 0, 1)$ are linearly dependent.
Find the coordinates of $(2, 3, 7)$ relative to the basis $\big((1, 1, 0), (1, 0, 1), (0, 1, 1)\big)$ .
Find a basis and the dimension of $W = \{(x, y, z) : x + y + z = 0 \text{ and } x - y = 0\}$ .
Explain why the plane $x + 2 y - z = 4$ is not a subspace of $\mathbb{R}^3$ , while $x + 2 y - z = 0$ is.

Show answersHide answers

$a (1, 0, 1) + b (2, 1, 0) = (a + 2 b, b, a)$ , so $b = 1$ , $a = 5$ : $(7, 1, 5) = 5 (1, 0, 1) + (2, 1, 0)$ .
No: the second and third components force $b = 1$ , $a = 2$ , but then $a + 2b = 4 \ne 3$ .
$\begin{vmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = -2 \ne 0$ : three independent vectors in $\mathbb{R}^3$ form a basis.
The determinant is $1 + k^{3}$ , which is 0 only for $k = - 1$ .
$a + b = 2$ , $a + c = 3$ , $b + c = 7$ give $a + b + c = 6$ , so $(a, b, c) = (- 1, 3, 4)$ .
$y = x$ and $z = - 2 x$ , so $W = \{x(1, 1, -2)\}$ : basis $\{(1, 1, -2)\}$ , dimension 1 (a line through $O$ ).
$(0, 0, 0)$ does not satisfy $x + 2 y - z = 4$ , and every subspace contains 0. For $x + 2 y - z = 0$ , the sum of two solutions and any multiple of a solution are again solutions.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every determinant and set of coordinates was checked by computer.
Video: Linear combinations, span, and basis vectors | Chapter 2, Essence of linear algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1Building every vector from a few

2The span of two vectors

Building every vector from a few

The span of two vectors