Linear transformations

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2013, 2022)
• Priority tier: T3
• Marks (count): 10 (1), 8 (1)
• Average solve time: ~4 min
• Difficulty mix: easy 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Both questions on this atom are rated “easy” and average under 5 minutes — the fastest high-confidence marks in the linear algebra section. The 2022 question is a linearity-reconstruction computation (express a target vector in terms of the given basis, apply the given values, add up) and takes 3–4 minutes. The 2013 question is a two-sentence structural proof using injectivity and surjectivity. Mastering this atom costs almost no time and guarantees at least 18 marks whenever either question appears.

Minimum Theory

Linear transformation. $T:V\to W$ is a linear transformation (also called a linear map or linear operator when $V=W$) if for all $u,v\in V$ and all $c\in\mathbb F$: $T(u+v)=T(u)+T(v)\quad\text{and}\quad T(cu)=cT(u).$ Equivalently: $T(au+bv)=aT(u)+bT(v)$ for all scalars $a,b$.

Kernel and image. $\ker T=\{v\in V:T(v)=0\}$ (a subspace of $V$); $\operatorname{Im}T=\{T(v):v\in V\}$ (a subspace of $W$). The Rank–Nullity Theorem: $\dim V=\dim\ker T+\dim\operatorname{Im}T$.

Matrix of $T$. If $V=\mathbb R^n$ with standard basis $\{e_1,\ldots,e_n\}$, then $[T]$ is the matrix whose $j$-th column is $T(e_j)$. Knowing $T$ on any basis determines $T$ completely by linearity.

Invertible $T$ maps bases to bases. If $T:V\to V$ is invertible (i.e. bijective), then: (i) $T$ is injective ($\ker T=\{0\}$) so it preserves linear independence; (ii) $T$ is surjective so its image spans $V$. Combining: if $\{v_1,\ldots,v_n\}$ is a basis of $V$, then $\{Tv_1,\ldots,Tv_n\}$ is also a basis.

Question Archetypes

determine-map-from-action (1 question(s); 2022)

Recognition Cues

• ”$T(v_1)=w_1$ and $T(v_2)=w_2$ are given; find $T(w)$ for a new vector $w$.”
• The two given vectors $v_1,v_2$ form a basis (or span) of the domain.
• Method: express $w=av_1+bv_2$, then $T(w)=aT(v_1)+bT(v_2)$.

Solution Template

1. Extract $T$ on the standard basis. Use the given data to compute $T(e_j)$ for each standard basis vector (by expressing each $e_j$ as a linear combination of the given vectors).
2. Express $w$ as a linear combination of the known vectors: $w=av_1+bv_2$ (solve for $a,b$).
3. Apply linearity: $T(w)=aT(v_1)+bT(v_2)$.
4. Compute the numerical result.

Worked Example

2022 Paper 1, 2022-P1-Q1b (10 marks)

Let $T:\mathbb R^2\to\mathbb R^3$ be a linear transformation with $T(1,0)^T=(1,2,3)^T$ and $T(1,1)^T=(-3,2,8)^T$. Find $T(2,4)^T$.

Step 1 — Find $T(0,1)$. Since $(1,1)-(1,0)=(0,1)$: $T(0,1)^T=T(1,1)^T-T(1,0)^T=(-3,2,8)^T-(1,2,3)^T=(-4,0,5)^T.$

Step 2 — Decompose. $(2,4)=2(1,0)+4(0,1)$.

Step 3 — Apply linearity. $T(2,4)^T=2T(1,0)^T+4T(0,1)^T=2(1,2,3)^T+4(-4,0,5)^T=(2,4,6)^T+(-16,0,20)^T.$

$\boxed{\;T(2,4)^T=(-14,\;4,\;26)^T.\;}$

Verification. Matrix form: columns $T(e_1)=(1,2,3)^T$ and $T(e_2)=(-4,0,5)^T$, so $[T]=\begin{pmatrix}1&-4\\2&0\\3&5\end{pmatrix}$. Then $[T](2,4)^T=(-14,4,26)^T$ ✓.

Alternative route. Directly: $(2,4)=-2(1,0)+4(1,1)$ (solve $a(1,0)+b(1,1)=(2,4)$: $b=4$, $a=-2$). So $T(2,4)=-2(1,2,3)+4(-3,2,8)=(-2-12,-4+8,-6+32)=(-14,4,26)$ ✓.

Common Traps

• Two routes work: (A) find $T(e_1),T(e_2)$ from $T(v_1)$ and $T(v_1)-T(v_2)$, then use the matrix; (B) directly express the target as a linear combination of the given vectors. Route A is more systematic; Route B is faster if you spot the combination immediately.
• Always verify with the matrix form — it is a quick check and earns the verification mark.
• The given vectors $(1,0)$ and $(1,1)$ are a basis of $\mathbb R^2$, so the map is fully determined by their images. If the given vectors did not span the domain, the map might not be uniquely determined.

transformation-property-proof (1 question(s); 2013)

Recognition Cues

• “Prove that an invertible linear operator maps a basis to a basis.”
• “Show that $T$ is injective iff $\ker T=\{0\}$.”
• “If $T$ is invertible and $v_1,\ldots,v_n$ is a basis, show $Tv_1,\ldots,Tv_n$ is a basis.”

Solution Template

1. Prove linear independence of the image set using injectivity ($\ker T=\{0\}$).
2. Prove spanning of the image set using surjectivity ($\operatorname{Im}T=V$).
3. Conclude the image set is a basis (independent + spans $V$, and has $n=\dim V$ elements).

Worked Example

2013 Paper 1, 2013-P1-Q2a-ii (8 marks)

Let $V$ be an $n$-dimensional vector space and $T:V\to V$ an invertible linear operator. If $\beta=\{X_1,\ldots,X_n\}$ is a basis of $V$, show that $\beta'=\{TX_1,\ldots,TX_n\}$ is also a basis of $V$.

Step 1 — $\beta'$ is linearly independent. Suppose $\sum_{i=1}^{n}c_i\,TX_i=0$ for some $c_i\in\mathbb F$. By linearity: $T\!\left(\sum_{i=1}^{n}c_i X_i\right)=0.$ Since $T$ is invertible, it is injective, so $\ker T=\{0\}$: $\sum_{i=1}^{n}c_i X_i=0.$ Since $\beta$ is a basis (hence independent), $c_1=\cdots=c_n=0$. So $\beta'$ is independent. $\checkmark$

Step 2 — $\beta'$ spans $V$. Let $v\in V$. Since $T$ is surjective, there exists $u\in V$ with $T(u)=v$. Since $\beta$ spans $V$, write $u=\sum c_i X_i$. Then: $v=T(u)=T\!\left(\sum c_i X_i\right)=\sum c_i\,TX_i\in\operatorname{span}(\beta').$

Conclusion. $\beta'$ is a linearly independent set of $n=\dim V$ vectors that spans $V$; hence it is a basis of $V$. $\boxed{\;\beta'=\{TX_1,\ldots,TX_n\}\text{ is a basis of }V.\;}\qquad\blacksquare$

Common Traps

• Both injectivity and surjectivity are needed. Injectivity alone preserves independence but doesn’t guarantee spanning; surjectivity alone guarantees spanning but not independence. The proof explicitly uses both.
• In a finite-dimensional space, “linearly independent set of size $\dim V$” automatically implies “spans $V$” — so Step 2 is logically redundant given Step 1, but it is expected for full marks and makes the proof complete.
• The result fails without invertibility. If $T$ is the zero map, every $TX_i=0$, which is not a basis.
• State the theorem structure clearly: $T$ invertible $\Rightarrow$ $T$ injective (for Step 1) + $T$ surjective (for Step 2). Don’t just say “by invertibility.”

Marks-Aware Writing

8-mark proof (transformation-property): The proof has exactly two steps — independence and spanning. Write each as a numbered paragraph with a $\checkmark$ at the end. The crucial micro-proofs are: ”$T(\sum c_i X_i)=0\Rightarrow\sum c_i X_i=0$ because $\ker T=\{0\}$” and ”$v=T(u)=\sum c_i TX_i$ because $T$ is surjective.” Four to five lines per step is the expected length.

10-mark computation (determine map): Write the decomposition of the target vector, apply linearity, compute. Include both a direct calculation and a matrix-form verification. The verification is worth 2 marks on its own.

Practice Set

• 2024-P1-Q1b (10 m) — — Hint: determine $T$ from its values on a basis; use linearity to find $T$ at the target point.
• 2023-P1-Q1b (10 m) — — Hint: linear map; express target in terms of given vectors.
• 2013-P1-Q2a-i (10 m) — — Hint: kernel and image; Rank–Nullity Theorem.
• 2020-P1-Q1b (10 m) — — Hint: linear transformation computation.
• 2017-P1-Q3a (15 m) — — Hint: larger transformation proof; use injectivity/surjectivity.
• 2016-P1-Q2a-i (10 m) — — Hint: linear map definition check.
• 2016-P1-Q2a-ii (6 m) — — Hint: quick property of linear maps.
• 2016-P1-Q2c (18 m) — — Hint: extended transformation theorem at 18 marks.
• 2025-P1-Q1b (10 m) — — Hint: determine $T$ or prove a property.
• 2025-P1-Q2a (15 m) — — Hint: longer transformation proof or matrix representation.