MTH2021 · Linear Algebra with Applications
Linear Transformations & the Rank-Nullity Theorem
Weeks 6–7 study linear transformations T : V → W through their kernel and range, the Rank–Nullity Theorem rank(T) + nullity(T) = dim(V), and the notions of injective, surjective and isomorphism. Every linear map is captured by a matrix [T]_{B',B} relative to chosen bases. These computations appear in Quiz 7 and the tail of Test 2 (Weeks 3–7), and constructing an isomorphism or applying rank–nullity is prime handwritten-section material in the final.
What this chapter covers
- 01Linear map: T(u + kv) = T(u) + kT(v); consequence T(0) = 0; matrix transformations T_A(x) = Ax are linear
- 02Kernel ker(T) = {v : T(v) = 0} and range ran(T); both are subspaces; for T_A, ker(T_A) = ker(A), ran(T_A) = col(A)
- 03rank(T) = dim ran(T); nullity(T) = dim ker(T)
- 04Rank–Nullity Theorem: rank(T) + nullity(T) = n = dim(V)
- 05T injective ⇔ ker(T) = {0}; for equal finite dimensions, injective ⇔ surjective
- 06Isomorphism = bijective linear map; V ≅ W; dim(U) = dim(V) ⇒ U ≅ V
- 07Matrix of a map relative to bases: [T]_{B',B} with columns [T(vⱼ)]_{B'}; [T]_{B',B}[x]_B = [T(x)]_{B'}
- 08Composition: [T₂ ∘ T₁] = [T₂][T₁]; a linear map is determined by its values on a basis
Matrix of a map, kernel, and the Rank–Nullity Theorem
- +1Compute T on each basis vector of B. T(1) = 1 (a₀=1): coordinates (1, 0). T(x) = 1 − 2x (a₁=1): coordinates (1, −2). T(x²) = −3x (a₂=1): coordinates (0, −3).
- +1Assemble the columns [T(vⱼ)]_{B'}: [T]_{B',B} = [[1, 1, 0], [0, −2, −3]] (a 2×3 matrix, as expected for P₂ → P₁).
- +1Find ker(T): solve [T](a₀,a₁,a₂)ᵀ = 0, i.e. a₀ + a₁ = 0 and 2a₁ + 3a₂ = 0. Let a₂ = 2 (to clear fractions): a₁ = −3, a₀ = 3. So the kernel is spanned by 3 − 3x + 2x².
- +1Read off dimensions. nullity(T) = dim ker(T) = 1. The matrix [T] has 2 pivots (rows independent), so rank(T) = 2, i.e. ran(T) = P₁.
- +1Verify Rank–Nullity: rank(T) + nullity(T) = 2 + 1 = 3 = dim(P₂) ✓. Because rank = 2 = dim(P₁), T is surjective; because nullity = 1 ≠ 0, T is not injective.
Key terms
- Linear transformation
- A map T : V → W with T(u + kv) = T(u) + kT(v) for all u, v and scalars k; it necessarily sends 0 to 0 and preserves linear combinations.
- Kernel and nullity
- ker(T) = {v : T(v) = 0}, a subspace of the domain; nullity(T) = dim ker(T). T is injective iff ker(T) = {0}.
- Range and rank
- ran(T) = {T(v)}, a subspace of the codomain; rank(T) = dim ran(T). For a matrix map, ran(T_A) = col(A).
- Rank–Nullity Theorem
- rank(T) + nullity(T) = dim(V): the dimensions of the range and kernel add to the dimension of the domain.
- Isomorphism
- A bijective linear map T : V → W; then V ≅ W. Any two finite-dimensional spaces of equal dimension are isomorphic.
- Matrix of a linear map
- [T]_{B',B}, whose j-th column is [T(vⱼ)]_{B'}; it satisfies [T]_{B',B}[x]_B = [T(x)]_{B'}, turning the abstract map into matrix multiplication.
Linear Transformations & the Rank-Nullity Theorem FAQ
How do I build the matrix of a linear map?
Apply T to each vector of the domain basis B, express each image in the codomain basis B', and use those coordinate vectors as the columns of [T]_{B',B}. The resulting matrix satisfies [T][x]_B = [T(x)]_{B'}, so once you have it, evaluating T is just matrix multiplication.
What does the Rank–Nullity Theorem actually tell me?
It says the domain's dimension splits exactly between what the map collapses (the kernel) and what it reaches (the range): rank(T) + nullity(T) = dim(V). So if you know any two of the three numbers you get the third for free, and it is the fastest way to check whether a map can be injective or surjective.
How are injective, surjective and isomorphism connected here?
T is injective iff ker(T) = {0} (nullity 0), and surjective iff rank(T) = dim(W). When the domain and codomain have the same finite dimension, injective and surjective are equivalent, so either one gives an isomorphism. That is why proving ker(T) = {0} plus equal dimensions is enough to conclude V ≅ W.
Why is the kernel of the map the same as the kernel of its matrix?
Because [T][x]_B = [T(x)]_{B'}, and coordinate vectors are zero exactly when the vector is zero. So T(v) = 0 iff [T][v]_B = 0, letting you compute the kernel, rank and nullity of an abstract map (on polynomials or matrices) by row-reducing an ordinary matrix.
Where is this assessed?
Kernel/range, rank–nullity and the matrix of a map are examined in Quiz 7 and the closing weeks of Test 2 (Weeks 3–7, since Week 7 is included). The final's handwritten section commonly asks you to construct an isomorphism or apply rank–nullity in a proof, so rehearse those arguments.
Exam move
Drill the three-step matrix construction (apply T to each domain basis vector, coordinatise in the codomain basis, stack as columns), then get ker, rank and nullity by row-reducing that matrix. Always close with the Rank–Nullity check rank + nullity = dim(domain). Rehearse the standard proof that ker(T) = {0} plus equal dimensions gives an isomorphism, because that is a favourite handwritten-exam item. This is examined in Quiz 7 and Test 2 (Weeks 3–7). When a rank–nullity count doesn't balance, ask Sia to recheck your kernel.
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