State the condition for vectors v_1, , v_k to be linearly independent. [2 marks]

SEAB Original-style practice: Determine whether the vectors pmatrix 1 \\ 2 \\ 1 pmatrix, pmatrix 2 \\ 1 \\ 0 pmatrix and pmatrix 4 \\ 5 \\ 1 pmatrix are linearly independent.

Vectors are linearly independent if the only solution of av_1 + bv_2 + cv_3 = 0 is a = b = c = 0. A quick test in three dimensions is the determinant of the matrix with the vectors as columns: a non-zero determinant means independent. $pmatrix 1 & 2 & 4 \\ 2 & 1 & 5 \\ 1 & 0 & 1 pmatrix = 1(11 - 50) - 2(21 - 51) + 4(20 - 11).$ $= 1(1) - 2(-3) + 4(-1) = 1 + 6 - 4 = 3 ≠ 0.$ Since the determinant is non-zero the vectors are linearly independent. (Note v_3 = 2v_1 + v_2 would have given dependence, but here the combination does not close.) Markers reward the independence criterion, computing the determinant, and concluding independence from the non-zero value.

SingaporeFurther MathsSyllabus dot point

What is a vector space, and how do the ideas of linear independence, basis, dimension and rank organise it?

Use the concepts of vector spaces and subspaces, linear independence, spanning sets, basis, dimension and the rank of a matrix

A focused answer to the H2 Further Mathematics outcome on linear spaces. Vector spaces and subspaces, linear independence, spanning sets, basis and dimension, the column and null spaces of a matrix, and the rank-nullity relationship.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

SEAB wants you to work with the abstract structure of linear algebra: vector spaces and subspaces, linear independence and spanning, basis and dimension, and the rank of a matrix together with the rank-nullity relationship. These ideas give the language to say exactly how big a solution set is and which vectors are genuinely "new".

The answer

Vector spaces and subspaces

A vector space is a set of objects (vectors) closed under addition and scalar multiplication, obeying the usual algebraic rules. A subspace is a subset that is itself a vector space: it must contain the zero vector and be closed under addition and scalar multiplication. Lines and planes through the origin are subspaces of three-dimensional space; a line not through the origin is not.

Linear independence and spanning

A set of vectors is linearly independent if no non-trivial combination gives the zero vector, that is

a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_k\mathbf{v}_k = \mathbf{0} \;\Rightarrow\; a_1 = a_2 = \cdots = a_k = 0.

A set spans a space if every vector in the space is some linear combination of them. Independence says "no redundancy"; spanning says "enough to build everything".

Basis and dimension

A basis is a linearly independent spanning set: the minimal building blocks that generate the space without redundancy. Every basis of a given space has the same number of vectors, and this number is the dimension. For example the standard basis $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ shows three-dimensional space has dimension $3$ .

Rank of a matrix

The rank of a matrix is the number of linearly independent columns (equivalently independent rows), found by reducing to row-echelon form and counting the non-zero rows. It is the dimension of the column space, the subspace spanned by the columns.

The null space and rank-nullity

The null space of $\mathbf{A}$ is the set of all $\mathbf{x}$ with $\mathbf{A}\mathbf{x} = \mathbf{0}$ ; its dimension is the nullity. The rank-nullity theorem connects them:

\operatorname{rank}(\mathbf{A}) + \operatorname{nullity}(\mathbf{A}) = n,

where $n$ is the number of columns. A high rank leaves a small null space (few free parameters in the homogeneous solution), and vice versa.

Worked example

Find the rank and nullity of $\mathbf{A} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 0 \end{pmatrix}$ , and describe the solution set of $\mathbf{A}\mathbf{x} = \mathbf{0}$ .

Step 1: Row reduce

$R_2 \to R_2 - 2R_1$ gives a zero row $\begin{pmatrix} 0 & 0 & 0 \end{pmatrix}$ . $R_3 \to R_3 - R_1$ gives $\begin{pmatrix} 0 & -1 & -3 \end{pmatrix}$ . The reduced rows are $\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$ and $\begin{pmatrix} 0 & -1 & -3 \end{pmatrix}$ , with one zero row.

Step 2: Count the rank

There are two non-zero rows, so $\operatorname{rank}(\mathbf{A}) = 2$ .

Step 3: Apply rank-nullity

With $n = 3$ columns, $\operatorname{nullity} = 3 - 2 = 1$ .

Step 4: Solve the homogeneous system

From $-y - 3z = 0$ take $z = t$ , so $y = -3t$ . From $x + 2y + 3z = 0$ : $x - 6t + 3t = 0 \Rightarrow x = 3t$ . The solution is $\mathbf{x} = t\begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$ .

Step 5: Describe the solution set

The null space is a one-dimensional subspace, a line through the origin spanned by $\begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}$ , matching the nullity of $1$ .

Examples in context

Example 1. Degrees of freedom in a design. When a structure or circuit is described by linear constraints, the nullity of the constraint matrix counts the genuine degrees of freedom left. Rank-nullity turns "how many equations" into "how much freedom remains".

Example 2. Independent measurements. A data matrix whose rank is less than the number of variables signals that some measurements are linear combinations of others, the algebraic root of redundancy and the starting point for dimension-reduction methods.

Try this

Q1. State the condition for vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ to be linearly independent. [2 marks]

Cue. The only solution of $\sum a_i\mathbf{v}_i = \mathbf{0}$ is $a_1 = \cdots = a_k = 0$ .

Q2. A $4$ -column matrix has rank $3$ . State its nullity. [1 mark]

Cue. $\operatorname{nullity} = 4 - 3 = 1$ .

Q3. Why is a line not passing through the origin not a subspace? [2 marks]

Cue. It does not contain the zero vector, and it is not closed under scalar multiplication, so it fails the subspace conditions.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksDetermine whether the vectors

\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}

\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}

and

\begin{pmatrix} 4 \\ 5 \\ 1 \end{pmatrix}

are linearly independent.

Show worked answer →

Vectors are linearly independent if the only solution of $a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 = \mathbf{0}$ is $a = b = c = 0$ . A quick test in three dimensions is the determinant of the matrix with the vectors as columns: a non-zero determinant means independent.

\det\begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 5 \\ 1 & 0 & 1 \end{pmatrix} = 1(1\cdot1 - 5\cdot0) - 2(2\cdot1 - 5\cdot1) + 4(2\cdot0 - 1\cdot1).

= 1(1) - 2(-3) + 4(-1) = 1 + 6 - 4 = 3 \neq 0.

Since the determinant is non-zero the vectors are linearly independent. (Note $\mathbf{v}_3 = 2\mathbf{v}_1 + \mathbf{v}_2$ would have given dependence, but here the combination does not close.)

Markers reward the independence criterion, computing the determinant, and concluding independence from the non-zero value.

Original6 marksA

3\times3

matrix

\mathbf{A}

has rank

2

. State the dimension of its column space and, using the rank-nullity theorem, the dimension of its null space. Describe the solution set of

\mathbf{A}\mathbf{x} = \mathbf{0}

Show worked answer →

The rank is the dimension of the column space, so $\dim(\text{column space}) = 2$ .

The rank-nullity theorem states $\operatorname{rank}(\mathbf{A}) + \operatorname{nullity}(\mathbf{A}) = n$ , where $n = 3$ is the number of columns. So $\operatorname{nullity}(\mathbf{A}) = 3 - 2 = 1$ , the dimension of the null space.

The null space is the set of solutions of $\mathbf{A}\mathbf{x} = \mathbf{0}$ . With nullity $1$ it is a one-dimensional subspace, that is a line through the origin: all scalar multiples of a single non-zero vector.

Markers reward identifying rank as the column-space dimension, applying rank-nullity to get nullity $1$ , and describing the solution set as a line through the origin (one-parameter family).