Bases and dimension; coordinates in a basis

At a Glance

• Frequency: 7 sub-parts across 6 of 13 years (2014, 2015, 2018, 2022, 2024, 2025)
• Priority tier: T2
• Marks (count): 10 (4), 12 (1), 15 (1), 13? — 10(4), 12(1), 15(1)
• Average solve time: ~6 min
• Difficulty mix: easy 5, medium 1, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Bases and dimension appear in 6 of the last 13 years, always in Section A. The questions are overwhelmingly easy to medium: five of seven are rated easy, and most carry 10 marks. The four recurring tasks are: (1) find a basis and dimension of a given subspace or span; (2) express one set of vectors in the coordinates of another basis; (3) extend an independent set to a basis of $\mathbb{R}^n$ by adding standard basis vectors; and (4) prove the theoretical result that $n$ linearly independent vectors in a dimension-$n$ space form a basis. Mastering row reduction as the universal tool covers the first three; the proof question has a fixed two-paragraph argument.

Minimum Theory

Basis. A basis of a vector space $V$ is a set $\mathcal{B}=\{v_1,\ldots,v_n\}\subset V$ that is (i) linearly independent and (ii) spans $V$. The dimension $\dim V=n$ is the cardinality of any basis; it is well-defined (all bases of a finite-dimensional space have the same size).

Coordinates. If $\mathcal{B}=\{v_1,\ldots,v_n\}$ is a basis of $V$, every $v\in V$ can be written uniquely as $v=c_1v_1+\cdots+c_nv_n$; the column $(c_1,\ldots,c_n)^T$ is the coordinate vector of $v$ with respect to $\mathcal{B}$.

Finding a basis of a span. Given spanning vectors $u_1,\ldots,u_k$: form the matrix with these as rows, row-reduce to echelon form, and discard zero rows. The non-zero rows form a basis; equivalently, the original vectors at the pivot positions are a basis (column-space version).

Subspace from equations. A subspace defined by $m$ homogeneous linear equations in $n$ variables has dimension $n-r$ where $r$ is the rank of the coefficient matrix.

Dimension formula. For subspaces $V,W$ of a finite-dimensional space: $\dim(V+W)=\dim V+\dim W-\dim(V\cap W).$

Extending to a basis. Any linearly independent set $\{v_1,\ldots,v_k\}$ in $\mathbb{R}^n$ can be extended to a basis of $\mathbb{R}^n$ by adjoining $n-k$ standard basis vectors $e_i$ chosen so that the resulting $n\times n$ matrix is non-singular (nonzero determinant).

Key theorem (10-mark proof target). Any $n$ linearly independent vectors in a vector space $V$ with $\dim V=n$ form a basis. Proof: let $v\in V$; the $n+1$ vectors $\{v,v_1,\ldots,v_n\}$ are linearly dependent (dimension bound), so $\alpha v+\sum\alpha_iv_i=0$ for scalars not all zero. If $\alpha=0$, independence of $\{v_i\}$ forces all $\alpha_i=0$ — contradiction. So $\alpha\ne0$ and $v\in\operatorname{span}\{v_i\}$.

Question Archetypes

Four patterns cover every basis question in the corpus.

find-basis-dimension (3 question(s); 2014, 2015, 2022)

Recognition Cues

• “Find a basis and dimension of (i) $V$ (ii) $W$ (iii) $V\cap W$” where subspaces are given by equations.
• “Find the dimension of the subspace spanned by {…}” followed by “find a basis.”

Solution Template

Subspace defined by equations (2014 type):

1. For each subspace, identify the free variables (dimension = $n$ - number of independent constraints).
2. Write the general element parametrically in terms of the free variables; each “free direction” vector is a basis element.
3. For the intersection $V\cap W$: combine all constraints from both subspaces; solve; count free variables.

Span of given vectors (2015, 2022 types):

1. Form the matrix with the vectors as rows (or columns).
2. Row-reduce; count non-zero rows = dimension.
3. For a basis: either take the non-zero rows from the reduced form, or take the original vectors corresponding to pivot positions.

Worked Example(s)

2014 Paper 1, 2014-P1-Q2a (15 marks)

$V=\{(a,b,c,d)\in\mathbb R^4: b-2c+d=0\}$, $W=\{(a,b,c,d): a=d,\,b=2c\}$. Find bases and dimensions of $V$, $W$, $V\cap W$.

$V$ (one constraint, $d=2c-b$ free in $a,b,c$): $(a,b,c,d)=a(1,0,0,0)+b(0,1,0,-1)+c(0,0,1,2).$ Basis: $\{(1,0,0,0),\,(0,1,0,-1),\,(0,0,1,2)\}$, $\dim V=3$.

$W$ (two constraints, free in $c,d$): $(a,b,c,d)=c(0,2,1,0)+d(1,0,0,1).$ Basis: $\{(0,2,1,0),\,(1,0,0,1)\}$, $\dim W=2$.

$V\cap W$ (all three constraints combined): $b=2c$ substituted into $b-2c+d=0$ gives $d=0$, then $a=d=0$. Free: $c$. $(0,2c,c,0)=c(0,2,1,0).$ $\boxed{\;\text{Basis of }V\cap W:\;\{(0,2,1,0)\},\;\dim(V\cap W)=1.\;}$

Check (dimension formula): $\dim V+\dim W=5$, $\dim(V\cap W)=1$, so $\dim(V+W)=4=\dim\mathbb R^4$ ✓.

2022 Paper 1, 2022-P1-Q3a-ii (10 marks)

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

Subtract the first equation from the second: $x+2z=0\Rightarrow x=-2z$. Back into equation 1: $y=x-z=-3z$. So $(x,y,z)=z(-2,-3,1)$. $\boxed{\;\text{Basis: }\{(-2,-3,1)\},\;\dim P=1.\;}$

Common Traps

• For $V\cap W$: don’t just intersect the basis vectors — combine all defining equations and solve from scratch.
• When using the row-reduction approach to find a basis of a span: the non-zero rows after reduction form a basis for the row space, which equals the original span. These are not the same as the original vectors unless you track pivot rows.
• The dimension formula $\dim(V+W)=\dim V+\dim W-\dim(V\cap W)$ provides a useful sanity check.

coordinates-in-basis (1 question(s); 2018)

Recognition Cues

• “Express $e_1,e_2,\ldots$ as linear combinations of $\alpha_1,\alpha_2,\ldots$”
• Equivalently: “find the coordinates of [standard basis vectors] in the basis $\{\alpha_i\}$.”

Solution Template

1. Form the matrix $M$ with $\alpha_1,\alpha_2,\ldots$ as columns.
2. Invert $M$ (using the $2\times2$ formula $M^{-1}=\frac{1}{\det M}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ for $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, or by augmented-matrix row reduction for larger matrices).
3. The columns of $M^{-1}$ give the coordinates of $e_1,e_2,\ldots$ in the $\alpha$-basis.
4. Verify by substituting back.

Worked Example(s)

2018 Paper 1, 2018-P1-Q1b (10 marks, compulsory)

Express $e_1=(1,0)$ and $e_2=(0,1)$ as linear combinations of $\alpha_1=(2,-1)$ and $\alpha_2=(1,3)$.

$M=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$, $\det M=7$. $M^{-1}=\frac{1}{7}\begin{pmatrix}3&-1\\1&2\end{pmatrix}.$ Reading off columns: $e_1=\tfrac{3}{7}\alpha_1+\tfrac{1}{7}\alpha_2$, $e_2=-\tfrac{1}{7}\alpha_1+\tfrac{2}{7}\alpha_2$. $\boxed{\;e_1=\tfrac{3}{7}\alpha_1+\tfrac{1}{7}\alpha_2,\qquad e_2=-\tfrac{1}{7}\alpha_1+\tfrac{2}{7}\alpha_2.\;}$

Verification: $\tfrac{3}{7}(2,-1)+\tfrac{1}{7}(1,3)=(7/7,0/7)=(1,0)=e_1$ ✓.

Common Traps

• Put $\alpha_1,\alpha_2$ as columns of $M$ (not rows); then columns of $M^{-1}$ give coordinates.
• The $\det M=7$ factor appears in every coefficient — keep as exact fractions.

extend-to-basis (2 question(s); 2024, 2025)

Recognition Cues

• “Find a basis and dimension of $H$, and extend the basis of $H$ to a basis of $\mathbb{R}^n$.”
• “Can the set $\{v_1,v_2,v_3\}$ be extended to a basis of $\mathbb{R}^4$?”

Solution Template

1. Row-reduce the spanning vectors to find a basis of the subspace and its dimension $k$.
2. To extend: try adjoining standard basis vectors $e_{k+1},\ldots,e_n$ (or whatever is needed) and verify independence by computing the $n\times n$ determinant.
3. The right choice of $e_i$ to adjoin: use the columns not already “covered” by pivots in the row reduction of $H$.

Worked Example(s)

2024 Paper 1, 2024-P1-Q1a (10 marks, compulsory)

$H=\operatorname{span}\{v_1,v_2,v_3\}$ with $v_1=(1,-2,5,-3)$, $v_2=(2,3,1,-4)$, $v_3=(3,8,-3,-5)$. Find a basis of $H$; extend to a basis of $\mathbb{R}^4$.

Basis of $H$: Row-reduce; $v_3-3v_1-2\cdot(\text{row}_2\text{ step})=0$ (third row zeroes out). Pivots in columns 1 and 2, so $\dim H=2$, basis $\{v_1,v_2\}$.

Extension: Adjoin $e_3=(0,0,1,0)$ and $e_4=(0,0,0,1)$. Compute: $\det\begin{pmatrix}1&-2&5&-3\\2&3&1&-4\\0&0&1&0\\0&0&0&1\end{pmatrix}=7\ne0.$ $\boxed{\;\text{Basis of }\mathbb{R}^4:\;\{v_1,\,v_2,\,(0,0,1,0),\,(0,0,0,1)\}.\;}$

2025 Paper 1, 2025-P1-Q1a (10 marks, compulsory)

Can $\{(0,0,0,3),(1,1,0,0),(0,1,-1,0)\}$ be extended to a basis of $\mathbb{R}^4$? Justify; exhibit one such extension.

Independence check: The three vectors have leading non-zero entries in distinct columns (4, 1, 2); thus rank $=3$ — they are linearly independent, so extension is possible.

Extension: Adjoin $e_1=(1,0,0,0)$ and check the $4\times4$ determinant $=3\ne0$. $\boxed{\;\{(0,0,0,3),(1,1,0,0),(0,1,-1,0),(1,0,0,0)\}\text{ is a basis of }\mathbb R^4.\;}$

Common Traps

• A set that is linearly dependent cannot be extended to a basis (any extension will still include the dependence). Always check independence first.
• The $e_i$ vectors to adjoin are not always $e_3,e_4$; use the columns not already pinned by pivots from the given vectors.

basis-theory-proof (1 question(s); 2022)

Recognition Cues

• “Prove that any $n$ linearly independent vectors in a vector space of dimension $n$ constitute a basis.”

Solution Template

1. State what needs to be shown: the set already spans $V$ (independence is given).
2. Take any $v\in V$; the set $\{v,v_1,\ldots,v_n\}$ has $n+1$ vectors in a dimension-$n$ space, so it is linearly dependent (standard theorem).
3. Write the dependence relation $\alpha v+\sum\alpha_iv_i=0$; show $\alpha\ne0$ (if $\alpha=0$, independence of $\{v_i\}$ forces all $\alpha_i=0$ — contradicts “not all zero”).
4. Conclude $v\in\operatorname{span}\{v_i\}$.

Worked Example(s)

2022 Paper 1, 2022-P1-Q1a (10 marks, compulsory)

Prove that any $n$ linearly independent vectors in a vector space $V$ with $\dim V=n$ form a basis.

Let $S=\{v_1,\ldots,v_n\}$ be linearly independent in $V$ with $\dim V=n$.

Step 1 (S spans V). Take any $v\in V$. The set $S'=\{v,v_1,\ldots,v_n\}$ contains $n+1$ vectors. Since $\dim V=n$, any $n+1$ vectors are linearly dependent: there exist scalars not all zero with $\alpha v+\sum\alpha_iv_i=0$.

If $\alpha=0$ then $\sum\alpha_iv_i=0$; since $\{v_i\}$ is linearly independent, all $\alpha_i=0$ — contradiction. So $\alpha\ne0$, and $v=-\sum(\alpha_i/\alpha)v_i\in\operatorname{span}(S)$.

Step 2 (Conclusion). $S$ is linearly independent and spans $V$, so $S$ is a basis. $\blacksquare$

Common Traps

• The proof relies on a cited prior theorem: any $n+1$ vectors in $\dim V=n$ are linearly dependent. State this explicitly before using it.
• The sub-argument ”$\alpha=0$ is impossible” is the pivot of the proof — don’t skip it.

Marks-Aware Writing

10-mark questions (2018, 2022-Q1a, 2022-Q3a-ii, 2024, 2025): One to two displayed equations, the basis as a boxed set, and the dimension stated. For the proof, four to five labelled sentences are sufficient; don’t ramble.

12–15-mark questions (2014, 2015): Three-part structure (find basis/dim of $V$, $W$, $V\cap W$) deserves three clearly labelled parts. Verify each basis element satisfies the defining equations. Check the dimension formula as a final sanity test.

Practice Set