Rank and nullity; rank-nullity theorem

At a Glance

• Frequency: 7 sub-parts across 6 of 13 years (2017, 2019, 2020, 2023, 2024, 2025)
• Priority tier: T2
• Marks (count): 5 (1), 10 (4), 15 (2)
• Average solve time: ~7 min
• Difficulty mix: easy 6, medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Rank–nullity appears in 6 of the last 13 years — always in Section A — and is one of the most reliably easy atoms in the corpus. Six of the seven questions are rated easy. The core skill is: given a linear map, find bases for its range and kernel by row-reducing the matrix, then quote the rank–nullity theorem as a check. The 2020 question (left-multiplication $T(A)=BA$) adds a small twist by working in $M_2(\mathbb R)$ rather than $\mathbb R^n$, and the 2017 question requires genuine image/kernel bases (not just dimensions). Master the row-reduction routine and the theorem, and every question in the corpus is straightforward.

Minimum Theory

Linear map notation. For a linear map $T:V\to W$ (or equivalently an $m\times n$ matrix $A$):

• The kernel (null space) is $\ker T=\{v\in V:Tv=0\}$, a subspace of $V$.
• The range (image, column space) is $\operatorname{Im}T=\{Tv:v\in V\}$, a subspace of $W$.
• Rank $=\dim\operatorname{Im}T=$ number of non-zero rows in REF of $A=$ number of pivot columns.
• Nullity $=\dim\ker T=$ number of free variables $=n-\operatorname{rank}A$.

Rank–nullity theorem. $\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim V.$ For an $m\times n$ matrix: rank$+$nullity$=n$.

Finding bases in practice.

1. Write the matrix $A$ of $T$.
2. Row-reduce to RREF.
3. Range basis: use the original columns of $A$ at pivot positions (column-space basis).
4. Kernel basis: from RREF, express pivot variables in terms of free variables; set each free variable to 1 in turn (others 0) to get a basis vector.
5. Verify: rank + nullity = $n$; check $T(\text{kernel vector})=0$.

Left-multiplication operator. For $T:M_n(\mathbb R)\to M_n(\mathbb R)$ defined by $T(A)=BA$: find $\ker T$ by solving $BA=O$ componentwise. The formula $\operatorname{rank}(T)=n\cdot\operatorname{rank}(B)$ holds on $M_n$.

Question Archetypes

Two patterns cover every rank–nullity question in the corpus.

range-kernel-bases (4 question(s); 2017, 2020-Q1b, 2020-Q3b, 2024, 2025)

Recognition Cues

• “Find a basis and dimension of the image of $A$ and of the kernel $A$.”
• “Find the rank and nullity of $T$; find a matrix that maps to the null matrix.”
• “Find the range, rank, kernel and nullity of $T:\mathbb R^4\to\mathbb R^3$.”

Solution Template

1. Write the standard-basis matrix $A$ of $T$ (columns = images of $e_1,\ldots,e_n$).
2. Row-reduce to RREF; identify pivot and free columns.
3. Range: span of original columns at pivot positions; $\dim=\operatorname{rank}$.
4. Kernel: parametrise free variables; read off basis vectors; $\dim=\operatorname{nullity}$.
5. State rank + nullity = $n$ as verification.

Worked Example(s)

2017 Paper 1, 2017-P1-Q3a (15 marks)

$A=\begin{pmatrix}1&2&3&1\\ 1&3&5&-2\\ 3&8&13&-3\end{pmatrix}:\mathbb R^4\to\mathbb R^3$. Find bases and dimensions of image and kernel.

Row-reduce $A$ to RREF: $A\;\longrightarrow\;\begin{pmatrix}1&0&-1&7\\ 0&1&2&-3\\ 0&0&0&0\end{pmatrix}.$ Pivots in columns 1, 2. $\operatorname{rank}(A)=2$.

Image basis (original columns 1 and 2): $\operatorname{Im}(A)=\operatorname{span}\!\left\{\begin{pmatrix}1\\1\\3\end{pmatrix},\,\begin{pmatrix}2\\3\\8\end{pmatrix}\right\},\quad\dim=2.$

Kernel basis (free variables $x_3=s$, $x_4=t$; from RREF: $x_1=s-7t$, $x_2=-2s+3t$): $\ker(A)=\operatorname{span}\!\left\{(1,-2,1,0)^T,\,(-7,3,0,1)^T\right\},\quad\dim=2.$

$\boxed{\;\dim\operatorname{Im}=2,\;\dim\ker=2;\quad\operatorname{rank}+\operatorname{nullity}=4=\dim\mathbb R^4.\;\checkmark}$

2025 Paper 1, 2025-P1-Q1b (10 marks, compulsory)

$T:\mathbb R^4\to\mathbb R^3$, $T(x,y,z,w)=(x-w,\,y+z,\,z-w)$. Find range, rank, kernel, nullity.

Matrix (columns = $Te_1,Te_2,Te_3,Te_4$): $A=\begin{pmatrix}1&0&0&-1\\ 0&1&1&0\\ 0&0&1&-1\end{pmatrix}.$ Three pivot rows → $\operatorname{rank}=3$, range $=\mathbb R^3$ (onto).

Kernel: $x-w=0$, $y+z=0$, $z-w=0$. Let $w=t$: $x=t$, $z=t$, $y=-t$. $\ker(T)=\operatorname{span}\{(1,-1,1,1)\},\quad\operatorname{nullity}=1.$ $\boxed{\;\operatorname{rank}=3,\;\operatorname{nullity}=1;\quad\operatorname{Range}(T)=\mathbb R^3,\;\ker(T)=\operatorname{span}\{(1,-1,1,1)\}.\;}$

2020 Paper 1, 2020-P1-Q1b (10 marks, compulsory)

$T:M_2(\mathbb R)\to M_2(\mathbb R)$, $T(A)=BA$ with $B=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}$. Find rank and nullity; exhibit $A\ne O$ with $T(A)=O$.

$\det B=4-4=0$, so $\operatorname{rank}(B)=1$. For $T(A)=BA=O$ with $A=\begin{pmatrix}p&q\\r&s\end{pmatrix}$: $BA=\begin{pmatrix}p-r&q-s\\-4p+4r&-4q+4s\end{pmatrix}=O\;\Rightarrow\;p=r,\;q=s.$ Kernel: $\left\{\begin{pmatrix}p&q\\p&q\end{pmatrix}\right\}=\operatorname{span}\!\left\{\begin{pmatrix}1&0\\1&0\end{pmatrix},\begin{pmatrix}0&1\\0&1\end{pmatrix}\right\}$, nullity $=2$.

$\operatorname{rank}(T)=4-2=2$. A nonzero matrix in the kernel: $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$. $\boxed{\;\operatorname{rank}=2,\;\operatorname{nullity}=2;\quad T\!\begin{pmatrix}1&1\\1&1\end{pmatrix}=O.\;}$

Common Traps

• The range basis uses the original columns at pivot positions, not the RREF columns (which span a different subspace of $\mathbb R^m$).
• Free variables: there are $n-\operatorname{rank}$ of them; each produces one kernel basis vector by setting it to 1 and all others to 0.
• For $T(A)=BA$ on $M_2$: $\operatorname{rank}(T)=2\cdot\operatorname{rank}(B)$, not just $\operatorname{rank}(B)$.

rank-nullity-count (3 question(s); 2019, 2023, 2024)

Recognition Cues

• “Find the rank and nullity of $T$” with no request for explicit bases.
• Often a follow-up to a previous part (where the rank was already computed).
• Or: “find the dimension of the null space” as a standalone 5-mark item.

Solution Template

1. Write and row-reduce the matrix of $T$.
2. Count non-zero rows: rank.
3. Apply rank–nullity: nullity $= n -$ rank.
4. (Optional) State the kernel explicitly as a quick verification.

Worked Example(s)

2023 Paper 1, 2023-P1-Q1b (10 marks, compulsory)

Find rank and nullity of $T:\mathbb R^3\to\mathbb R^3$, $T(x,y,z)=(x+z,\,x+y+2z,\,2x+y+3z)$.

Matrix: $A=\begin{pmatrix}1&0&1\\ 1&1&2\\ 2&1&3\end{pmatrix}$.

$R_2-R_1$, $R_3-2R_1$: $(0,1,1)$ and $(0,1,1)$. Then $R_3-R_2=0$: $\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&0\end{pmatrix}.$ Two pivots: $\operatorname{rank}=2$, $\operatorname{nullity}=3-2=1$. $\boxed{\;\operatorname{rank}(T)=2,\quad\operatorname{nullity}(T)=1.\;}$

Kernel: from RREF, $z$ free; $x=-z$, $y=-z$; kernel $=\operatorname{span}\{(1,1,-1)\}$ (verify: $T(1,1,-1)=(0,0,0)$ ✓).

2019 Paper 1, 2019-P1-Q3c-ii (5 marks)

Find $\dim\ker A$ for the $4\times4$ matrix $A$ from part (i).

From part (i), $\operatorname{rank}(A)=2$. By rank–nullity: $\dim\ker A=4-2=\boxed{2.}$

Common Traps

• The rank–nullity theorem says nullity $=\dim V$ (domain) $-$ rank, not $\dim W$ $-$ rank. For a non-square map, use the domain dimension.
• For follow-up parts: read the rank from the earlier part — don’t recompute.

Marks-Aware Writing

5-mark questions (2019-Q3c-ii): One sentence: state rank from prior part, apply rank–nullity, box the answer.

10-mark questions (2020-Q1b, 2023-Q1b, 2024-Q1b, 2025-Q1b): Full solution with matrix, row reduction, rank statement, nullity from theorem, and kernel basis. Verify with one check ($T(\text{kernel vector})=0$).

15-mark questions (2017-Q3a, 2020-Q3b): Provide both image and kernel bases with full derivation. The 2017 question needs the original-column rule for the image basis stated explicitly. The 2020-Q3b question (abstract field $F^3$) is prose-heavy — state the conditions on $(a,b,c)$ first, then the spanning vector and nullity.

Practice Set