Matrix of a linear transformation

At a Glance

• Frequency: 10 sub-parts across 8 of 13 years (2013, 2015, 2016, 2019, 2021, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (4), 12 (1), 15 (3), 18 (1), 6 (1)
• Average solve time: ~10 min
• Difficulty mix: medium 6, easy 3, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Matrix representations appear in 8 of the last 13 years, almost exclusively in Section A. Seven of the ten questions ask you to find the matrix of a linear map — the most mechanical of all linear algebra tasks. The single non-obvious step is expressing each image in the codomain basis. Every other step is substitution and bookkeeping. The 2016 and 2023 questions add a layer (recovering the map, change of basis), but the underlying procedure is the same.

Minimum Theory

Matrix of $T$ in bases $B$ and $C$. Let $T:V\to W$, domain basis $B=\{b_1,\ldots,b_n\}$, codomain basis $C=\{c_1,\ldots,c_m\}$. The matrix $[T]_{B,C}$ is the $m\times n$ matrix whose $j$-th column is the $C$-coordinate vector of $T(b_j)$: $[T]_{B,C}=\bigl[[T(b_1)]_C\;\big|\;[T(b_2)]_C\;\big|\;\cdots\;\big|\;[T(b_n)]_C\bigr].$

Procedure:

1. For each domain basis vector $b_j$, compute $T(b_j)$ in the standard/ambient space.
2. Express $T(b_j)=\sum_i a_{ij}c_i$ (solve for the $C$-coordinates).
3. Column $j$ of the matrix is $(a_{1j},a_{2j},\ldots,a_{mj})^T$.

Change of basis. If $[T]_E$ is the matrix in the standard basis $E$ and $P$ is the matrix whose columns are the new basis vectors $b_j$ (expressed in $E$), then: $[T]_B=P^{-1}[T]_E P.$

Recovering $T$ from a matrix. If $A=[T]_{B,C}$, then the $j$-th column of $A$ holds the $C$-coordinates of $T(b_j)$. Decode: $T(b_j)=\sum_i A_{ij}c_i$. Then use linearity to find $T$ on all inputs.

Question Archetypes

Four patterns cover all questions.

matrix-in-bases (7 question(s); 2013, 2015, 2016, 2019, 2021, 2024)

Recognition Cues

• “Find the matrix of $T$ with respect to the bases $B$ and $C$.”
• Or “Find the matrix of $T$ relative to basis $B$” (domain = codomain, one basis).
• Typical spaces: $\mathbb R^n$, $P_n$ (polynomials), $M_{2\times2}(\mathbb R)$.

Solution Template

1. For each domain basis vector $b_j$, apply $T$ explicitly.
2. Write $T(b_j)$ in terms of codomain basis $C$: solve for coefficients $a_{ij}$.
   • If $C$ is the standard basis: coordinates are simply the coefficients.
   • If $C$ is non-standard: solve a linear system $\sum a_{ij}c_i=T(b_j)$.
3. Assemble: column $j$ = $(a_{1j},\ldots,a_{mj})^T$.
4. Find the null space by solving $[T]x=0$ (if asked).

Worked Example(s)

2021 Paper 1, 2021-P1-Q1b (10 marks)

Matrix of $T(a,b,c)=(a+b,a-b,2c)$ in basis $B=\{(0,1,1),(1,0,1),(1,1,0)\}$.

Step 1 — Compute images in standard coordinates: $T(V_1)=(1,-1,2)$, $T(V_2)=(1,1,2)$, $T(V_3)=(2,0,0)$.

Step 2 — Express each in basis $B$. For $T(V_1)$: solve $c_1(0,1,1)+c_2(1,0,1)+c_3(1,1,0)=(1,-1,2)$.

Sum-trick: adding all three equations $\Rightarrow 2(c_1+c_2+c_3)=2\Rightarrow c_1+c_2+c_3=1$. Then $c_1=0$, $c_2=2$, $c_3=-1$.

Similarly: $T(V_2)\to(1,1,0)$; $T(V_3)\to(-1,1,1)$.

$\boxed{\;[T]_B=\begin{pmatrix}0&1&-1\\2&1&1\\-1&0&1\end{pmatrix}.\;}$

2013 Paper 1, 2013-P1-Q2a-i (10 marks)

Matrix of $T(p)=\int_0^x p(t)\,dt:\,P_2\to P_3$ with domain basis $\{1,x,x^2\}$ and codomain basis $\{1,x,1+x^2,1+x^3\}$.

Images: $T(1)=x$, $T(x)=x^2/2$, $T(x^2)=x^3/3$.

Coordinates in $C=\{1,x,1+x^2,1+x^3\}$: $T(1)=x\to(0,1,0,0)$; $T(x)=x^2/2=(x^2/2-1/2+1/2)=\frac12(1+x^2)-\frac12\cdot1\to(-\frac12,0,\frac12,0)$; $T(x^2)=x^3/3\to(-\frac13,0,0,\frac13)$.

$\boxed{\;[T]=\begin{pmatrix}0&-\frac12&-\frac13\\1&0&0\\0&\frac12&0\\0&0&\frac13\end{pmatrix}.\;}$

Null space: $T(p)=0\Rightarrow\int_0^x p=0\Rightarrow p=0$ (by FTC). $\ker T=\{0\}$.

2016 Paper 1, 2016-P1-Q2a-i (10 marks)

$T:\,M_2(\mathbb R)\to P_2(x)$, $T\begin{pmatrix}a&b\\c&d\end{pmatrix}=(a+c)+(a-d)x+(b+c)x^2$. Standard bases. Find matrix and null space.

Images of standard basis: $T(E_{11})=1+x\to(1,1,0)^T$; $T(E_{12})=x^2\to(0,0,1)^T$; $T(E_{21})=1+x^2\to(1,0,1)^T$; $T(E_{22})=-x\to(0,-1,0)^T$.

$[T]=\begin{pmatrix}1&0&1&0\\1&0&0&-1\\0&1&1&0\end{pmatrix}\;(3\times4).$

Null space: $a+c=0$, $a-d=0$, $b+c=0\Rightarrow c=-a$, $d=a$, $b=a$. $\ker T=\text{span}\!\left\{\begin{pmatrix}1&1\\-1&1\end{pmatrix}\right\}$, $\dim=1$.

2016 Paper 1, 2016-P1-Q2a-ii (6 marks)

$T(f)=f+5\int_0^x f:\,P_2\to P_3$, domain basis $\{1,1+x,1-x^2\}$, codomain basis $\{1,x,x^2,x^3\}$.

Images (apply to the non-standard domain basis): $T(1)=1+5x$; $T(1+x)=1+6x+\frac52x^2$; $T(1-x^2)=1+5x-x^2-\frac53x^3$.

Codomain basis is standard, so coordinates are the coefficients: $[T]=\begin{pmatrix}1&1&1\\5&6&5\\0&\tfrac52&-1\\0&0&-\tfrac53\end{pmatrix}\;(4\times3).$

2019 Paper 1, 2019-P1-Q1c (10 marks)

$T:\mathbb R^2\to\mathbb R^2$, $T(2,1)=(5,7)$, $T(1,2)=(3,3)$. Find rank of the standard matrix $A$.

Recover $A$. $A[P]=[C]$ where $P=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $C=\begin{pmatrix}5&3\\7&3\end{pmatrix}$. Since $\det P=3\ne0$: $A=CP^{-1}=\frac13\begin{pmatrix}5&3\\7&3\end{pmatrix}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}=\frac13\begin{pmatrix}7&1\\11&-1\end{pmatrix}.$ $\det A=-18/9=-2\ne0$, so $\boxed{\text{rank}(A)=2}$.

Shortcut: images $(5,7),(3,3)$ are independent ($\det=-6\ne0$) $\Rightarrow T$ maps a basis to an independent set $\Rightarrow$ $A$ is invertible $\Rightarrow$ rank $2$.

2024 Paper 1, 2024-P1-Q3a (15 marks)

$\phi:M_{2\times2}\to M_{2\times2}$, $\phi(v)=Av$ with $A=\begin{pmatrix}1&2\\3&-1\end{pmatrix}$. Find matrix, rank, invertibility.

Images of standard basis: $AE_{11}=\begin{pmatrix}1&0\\3&0\end{pmatrix}=E_{11}+3E_{21}$; $AE_{12}=E_{12}+3E_{22}$; $AE_{21}=2E_{11}-E_{21}$; $AE_{22}=2E_{12}-E_{22}$.

$[\phi]=\begin{pmatrix}1&0&2&0\\0&1&0&2\\3&0&-1&0\\0&3&0&-1\end{pmatrix}.$

This is block-diagonal $\begin{pmatrix}A&0\\0&A\end{pmatrix}$ (after reordering basis as $E_{11},E_{21},E_{12},E_{22}$). $\det[\phi]=(\det A)^2=49\ne0$.

$\boxed{\text{rank}=4,\;\phi\text{ is invertible.}}$

2015 Paper 1, 2015-P1-Q3a (12 marks)

Matrix of $T(a_1,a_2,a_3)=(2a_1+5a_2+a_3,-3a_1+a_2-a_3,-a_1+2a_2+3a_3)$ in basis $B=\{(1,0,1),(-1,2,1),(3,-1,1)\}$.

$T(V_1)=(3,-4,2)$, $T(V_2)=(9,4,8)$, $T(V_3)=(2,-11,-2)$.

Express each in basis $B$ via $P^{-1}$ (computed via cofactor expansion, $\det P=-2$): Column 1: $(17/2,-7/2,-3)^T$; Column 2: $(-3/2,9/2,5)^T$; Column 3: $(14,-9,-7)^T$.

$\boxed{\;[T]_B=\begin{pmatrix}17/2&-3/2&14\\-7/2&9/2&-9\\-3&5&-7\end{pmatrix}.\;}$

Common Traps

• Columns = images, not rows. The $j$-th column of $[T]$ is the coordinate vector of $T(b_j)$.
• Matrix dimensions: $[T]$ is $m\times n$ where $m=\dim W$ (codomain) and $n=\dim V$ (domain). For $T:M_{2\times2}\to P_2$: $3\times4$. For $T:P_2\to P_3$: $4\times3$.
• Non-standard codomain basis: if $C$ is not the standard basis, you must express $T(b_j)$ as a linear combination of the $c_i$‘s. Many students silently use the standard basis for the codomain and get the wrong matrix.
• $T(E_{22})=-x$ (not $+x$) in the 2016-Q2a-i problem: $a-d=0-1=-1$, so the coefficient of $x$ is $-1$.
• For the 2019 rank question: don’t compute $A$ in full if you can check independence of the images directly.

change-of-basis (1 question; 2023)

Recognition Cues

• “The matrix of $T$ in the standard basis is $[T]_E$. Find the matrix in the new basis $B$.”
• Given: the new basis vectors expressed in the old (standard) basis.

Solution Template

1. Form $P$ whose columns are the new basis vectors in the standard basis.
2. Compute $P^{-1}$ (by row reduction or adjugate).
3. $[T]_B=P^{-1}[T]_E P$.

Worked Example(s)

2023 Paper 1, 2023-P1-Q2a (15 marks)

$[T]_E=\begin{pmatrix}1&1&2\\-1&2&1\\0&1&3\end{pmatrix}$. New basis $B=\{(1,1,1),(0,1,1),(0,0,1)\}$. Find $[T]_B$.

$P=\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}$ (lower triangular). Inverse: $P^{-1}=\begin{pmatrix}1&0&0\\-1&1&0\\0&-1&1\end{pmatrix}$ (subtract each row from the next).

Compute $[T]_E P$ column by column: $[T]_E(1,1,1)^T=(4,2,4)^T$; $[T]_E(0,1,1)^T=(3,3,4)^T$; $[T]_E(0,0,1)^T=(2,1,3)^T$.

$[T]_E P=\begin{pmatrix}4&3&2\\2&3&1\\4&4&3\end{pmatrix}.$

Apply $P^{-1}$ (row operations — row 2 $\leftarrow$ row 2 $-$ row 1; row 3 $\leftarrow$ row 3 $-$ row 2):

$\boxed{\;[T]_B=P^{-1}[T]_E P=\begin{pmatrix}4&3&2\\-2&0&-1\\2&1&2\end{pmatrix}.\;}$

Common Traps

• $P$ columns are the new basis vectors in the old (standard) basis — not the change-of-basis matrix in the opposite direction.
• The formula is $[T]_B=P^{-1}[T]_E P$, not $P[T]_E P^{-1}$.
• For a triangular $P$ (as here), $P^{-1}$ is also triangular and is computed by inspection — no need for full Gaussian elimination.

recover-map-from-action (1 question; 2025)

Recognition Cues

• ”$T$ is a linear map from $\mathbb R^n\to\mathbb R^m$ with $T(v_1)=w_1$, $T(v_2)=w_2$, $T(v_3)=w_3$.”
• The $v_i$ are not the standard basis vectors. Find $T(x,y,z)$.

Solution Template

1. Verify $\{v_1,v_2,v_3\}$ is linearly independent (check $\det\ne0$).
2. Set up $AB=C$ where $A=[T]$ (standard matrix), $B=[v_1\,v_2\,v_3]$, $C=[w_1\,w_2\,w_3]$.
3. Solve: $A=CB^{-1}$.
4. Write $T(x)=A(x,y,z)^T$.

Worked Example(s)

2025 Paper 1, 2025-P1-Q2a (15 marks)

$T:\mathbb R^3\to\mathbb R^2$, $T(1,1,-1)=(1,0)$, $T(4,1,1)=(0,1)$, $T(1,-1,2)=(1,1)$. Find $T$.

$B=[(1,1,-1)^T,(4,1,1)^T,(1,-1,2)^T]$, $\det B=1\ne0$ (inputs form a basis).

$A=CB^{-1}=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}\begin{pmatrix}3&-7&-5\\-1&3&2\\2&-5&-3\end{pmatrix}=\begin{pmatrix}5&-12&-8\\1&-2&-1\end{pmatrix}$.

$\boxed{\;T(x,y,z)=(5x-12y-8z,\;x-2y-z).\;}$

Common Traps

• Check $\det B\ne0$ first — if the given vectors are dependent, $T$ is either underdetermined or inconsistent.
• Equation is $AB=C$, so $A=CB^{-1}$ — not $A=B^{-1}C$ or $A=C B^T$.

recover-map-from-matrix (1 question; 2016)

Recognition Cues

• Given the matrix $A=[T]_{B,C}$ with both $B$ (domain basis) and $C$ (codomain basis) non-standard.
• Find the explicit formula $T(p(x))=\ldots$

Solution Template

1. Decode columns: column $j$ gives $T(b_j)$ in basis $C$: $T(b_j)=\sum_i A_{ij}c_i$.
2. Express the standard monomials $1,x,x^2$ as linear combinations of $b_j$ (solve a linear system).
3. Use linearity: $T(1)=\ldots$, $T(x)=\ldots$, $T(x^2)=\ldots$.
4. Write $T(p+qx+rx^2)=p\,T(1)+q\,T(x)+r\,T(x^2)$.

Worked Example(s)

2016 Paper 1, 2016-P1-Q2c (18 marks)

$A=\begin{pmatrix}1&-1&2\\-2&1&-1\\1&2&3\end{pmatrix}=[T]_{B,C}$, domain $B=\{1-x,x-x^2,x+x^2\}$, codomain $C=\{1,1+x,1+x^2\}$.

Decode columns: $T(1-x)=1\cdot1+(-2)(1+x)+1(1+x^2)=x^2-2x$; $T(x-x^2)=2x^2+x+2$; $T(x+x^2)=3x^2-x+4$.

Express standard monomials in $B$: $1=b_1+\frac12b_2+\frac12b_3$; $x=0\cdot b_1+\frac12b_2+\frac12b_3$; $x^2=-\frac12b_2+\frac12b_3$.

Apply linearity: $T(1)=\frac72x^2-2x+3$; $T(x)=\frac52x^2+3$; $T(x^2)=\frac12x^2-x+1$.

$\boxed{\;T(p+qx+rx^2)=(3p+3q+r)+(-2p-r)x+\tfrac12(7p+5q+r)x^2.\;}$

Common Traps

• Both bases are non-standard — column $j$ encodes $T(b_j)$ in the codomain basis $C$, not in standard coordinates.
• Expressing the standard monomials in terms of $B$ requires solving a $3\times3$ system; fractional entries ($\pm\frac12$) propagate throughout.

Marks-Aware Writing

6-mark questions (2016-Q2a-ii): Compute images of 3 domain-basis vectors; write columns; no null space required.

10-mark questions (2013, 2016-Q2a-i, 2019, 2021): Compute all images, express in codomain basis (show the system for non-standard bases), assemble matrix, state null space. Verify one column by reconstruction.

12-15-mark questions (2015, 2023, 2024, 2025): For $P^{-1}[T]P$ (2023): compute $P$, invert it, perform both multiplications. For matrix-in-basis (2015): compute $P^{-1}$ explicitly via adjugate or row reduction. For recovering $T$ (2025): show $A=CB^{-1}$, compute $B^{-1}$, multiply.

18-mark questions (2016-Q2c): Full derivation — decode all three columns, solve for the monomials in the domain basis, apply linearity to all three standard monomials, state the general formula. Verify by plugging in the domain basis vectors.

Practice Set