Source: MATH 416 Exam 1, Abstract Linear Algebra, UIUC
Tags: linear map, linear transformation, coordinate vector, change of basis, matrix representation, nullspace, kernel, range, image, one-to-one, injective, onto, surjective, subspace proof, rank-nullity
Difficulty: Intermediate to Advanced | Prerequisites: Study Notes 01 (row reduction) and 02 (linear independence, bases, dimension).
This is where MATH 416 moves from computing with specific matrices to reasoning about abstract structure. A linear map is a function between vector spaces that respects addition and scalar multiplication. Once you choose bases for the domain and codomain, every linear map can be represented as a matrix, and the matrix does the same job as the map. The nullspace and range capture what the map "kills" and what it "hits," and their dimensions are linked by the rank-nullity theorem. Proofs in this section (subspace proofs, injectivity arguments) are a major source of exam marks.
A coordinate vector records a vector's coefficients relative to a chosen basis. The matrix representation [T]^β_γ lets you compute T(v) via matrix multiplication: [T(v)]_β = [T]^β_γ [v]_γ. The nullspace is everything T sends to zero; the range is everything T can output. If the nullspace is trivial ({0}), T is one-to-one.
Linear map (linear transformation)
A function T: V → W between vector spaces satisfying T(u + v) = T(u) + T(v) and T(cv) = cT(v) for all vectors u, v and scalars c.
In simple terms, T "plays nicely" with the two operations that define a vector space.
Coordinate vector [v]_γ
Given a basis γ = {v₁, ..., v_n} for V and a vector v ∈ V, the coordinate vector [v]_γ is the column of scalars (a₁, ..., a_n)^t such that v = a₁v₁ + ... + a_nv_n.
Think of it as the "address" of v in the coordinate system defined by γ.
Matrix representation [T]^β_γ
The matrix whose j-th column is [T(γ_j)]_β, where γ is the basis for the domain and β is the basis for the codomain.
In simple terms, apply T to each basis vector of γ, then write the result in β-coordinates. Those coordinate columns, side by side, form the matrix.
Nullspace (kernel) N(T)
N(T) = {v ∈ V : T(v) = 0_W}. The set of all vectors in the domain that T maps to the zero vector in the codomain.
Think of it as everything the map "destroys."
Range (image) R(T)
R(T) = {T(v) : v ∈ V}. The set of all vectors in W that are actually hit by T.
Think of it as the "output" of the map, the portion of W that T can reach.
One-to-one (injective)
T is one-to-one if T(v₁) = T(v₂) implies v₁ = v₂. Equivalently, N(T) = {0_V}.
Onto (surjective)
T is onto if R(T) = W, meaning every vector in the codomain is the image of something in the domain.
Isomorphism
A linear map that is both one-to-one and onto. If such a map exists between V and W, the two spaces are isomorphic.
To find [v]_γ for v = (1, 2, 3) with γ = {(1,0,1), (1,0,0), (0,1,0)} in R³:
Write v = a₁(1,0,1) + a₂(1,0,0) + a₃(0,1,0).
This gives the system:
a₁ + a₂ = 1
a₃ = 2
a₁ = 3
Solving: a₁ = 3, a₂ = -2, a₃ = 2.
So [v]_γ = (3, -2, 2)^t.
The key step is expressing v as a linear combination of the basis vectors and reading off the coefficients.
Given T: R³ → R³ defined by T(x₁, x₂, x₃) = (x₂ + x₃, x₁, x₁), with β = {e₁, e₂, e₃} (standard) and γ = {(1,0,1), (1,0,0), (0,1,0)}:
Apply T to each vector in γ:
T(1,0,1) = (0+1, 1, 1) = (1, 1, 1)
T(1,0,0) = (0+0, 1, 1) = (0, 1, 1)
T(0,1,0) = (1+0, 0, 0) = (1, 0, 0)
Since β is the standard basis, [w]_β = w for any w in R³.
The columns of [T]^β_γ are these images:
[T]^β_γ = [ 1 0 1 ]
[ 1 1 0 ]
[ 1 1 0 ]
[T(v)]_β = [T]^β_γ [v]_γ
This says: to compute T(v) in β-coordinates, multiply the matrix representation by the coordinate vector of v in γ-coordinates.
Verification from the exam. With v = (1,2,3):
Left side: T(1,2,3) = (2+3, 1, 1) = (5, 1, 1). In standard coordinates, [T(v)]_β = (5, 1, 1)^t.
Right side: [T]^β_γ [v]_γ = [[1,0,1],[1,1,0],[1,1,0]] × (3,-2,2)^t = (3+0+2, 3-2+0, 3-2+0)^t = (5, 1, 1)^t.
Both sides match.
Proving R(T) is a subspace of W. You need three things:
Zero vector. T(0_V) = 0_W (because T is linear), so 0_W ∈ R(T).
Closed under addition. If w₁, w₂ ∈ R(T), then w₁ = T(v₁) and w₂ = T(v₂) for some v₁, v₂ ∈ V. Then w₁ + w₂ = T(v₁) + T(v₂) = T(v₁ + v₂), so w₁ + w₂ ∈ R(T).
Closed under scalar multiplication. If w₁ ∈ R(T), then w₁ = T(v₁). So cw₁ = cT(v₁) = T(cv₁), hence cw₁ ∈ R(T).
The proof for N(T) being a subspace of V follows the same template.
Claim. If N(T) = {0_V}, then T is one-to-one.
Proof. Suppose T(v₁) = T(v₂). Then T(v₁) - T(v₂) = 0_W, so T(v₁ - v₂) = 0_W by linearity. This means v₁ - v₂ ∈ N(T) = {0_V}, so v₁ - v₂ = 0_V, giving v₁ = v₂.
This is one of the most frequently tested proof fragments in the course.
The matrix-coordinate formula
[T(v)]_β = [T]^β_γ [v]_γ
Rank-nullity theorem
dim(V) = dim(N(T)) + dim(R(T)), equivalently dim(V) = nullity(T) + rank(T)
Column j of [T]^β_γ
[T(γ_j)]_β
Matrix representations of linear maps are how computer graphics systems encode transformations (rotations, reflections, projections). Choosing a good basis can make a complicated transformation diagonal or nearly so, which is the idea behind eigenvalue decomposition, used in everything from Google's PageRank to quantum mechanics.
Students often confuse the two bases in [T]^β_γ. The subscript γ is the domain basis (what you express the input in), and the superscript β is the codomain basis (what you express the output in). Swapping them gives a different, incorrect matrix.
"N(T) = {0}" does not mean "T is onto." Injectivity and surjectivity are independent properties unless you know the dimensions of V and W are equal.
When proving R(T) is a subspace, students sometimes forget to verify the zero vector condition. It is often the easiest step, but omitting it costs marks.
Students sometimes think "one-to-one" means "bijective." One-to-one (injective) only means distinct inputs give distinct outputs. It says nothing about whether every element in the codomain is hit.
⚠️ "Compute [v]_γ" and "Compute [T]^β_γ" are standard 10-point questions. Practise until the process is automatic.
⚠️ The verification [T(v)]_β = [T]^β_γ [v]_γ is a favourite "compute both sides" problem. Do each side independently, then confirm they agree.
⚠️ Subspace proofs (show R(T) or N(T) is a subspace) require all three conditions: contains 0, closed under addition, closed under scalar multiplication. State each one explicitly.
⚠️ The proof that N(T) = {0} implies T is 1-1 is a classic short proof. Memorise the logical chain: T(v₁) = T(v₂) → T(v₁ - v₂) = 0 → v₁ - v₂ ∈ N(T) → v₁ - v₂ = 0 → v₁ = v₂.
⚠️ True/false traps: "If T: V → W is 1-1, then T is onto" is false in general (consider T: R² → R³ defined by T(x,y) = (x,y,0), which is injective but not surjective).
True or false: The nullspace of a linear map is always a subspace of the domain.
Fill in the blank: The j-th column of [T]^β_γ is obtained by applying T to the j-th vector of ___ and writing the result in ___-coordinates.
True or false: If T: R² → R³ is one-to-one, then T must be onto.
True or false: The set {A ∈ M_{m×n} : a₁₁ = 0} is a subspace of M_{m×n}.
True or false: C⁰(R), the space of continuous functions on R, has a finite basis.
Answers: 1. True. 2. γ; β. 3. False (dimensions differ, so an injective map from a smaller space cannot be surjective onto a larger one). 4. True (contains the zero matrix, closed under addition and scalar multiplication). 5. False (it contains all polynomials as a subspace, and the polynomial space is infinite-dimensional).
Q: Let β = {e₁, e₂, e₃} and γ = {(1,0,1), (1,0,0), (0,1,0)}. Compute [v]_γ for v = (2, 5, 1).
A: Solve a₁(1,0,1) + a₂(1,0,0) + a₃(0,1,0) = (2, 5, 1). From the third component: a₁ = 1. From the first: 1 + a₂ = 2, so a₂ = 1. From the second: a₃ = 5. Thus [v]_γ = (1, 1, 5)^t.
Q: Give the definitions of N(T) and R(T) for a linear map T: V → W.
A: N(T) = {v ∈ V : T(v) = 0_W}. R(T) = {T(v) : v ∈ V}.
Q: Prove that if N(T) = {0_V}, then T is one-to-one.
A: Suppose T(v₁) = T(v₂). Then T(v₁ - v₂) = T(v₁) - T(v₂) = 0_W, so v₁ - v₂ ∈ N(T). Since N(T) = {0_V}, we have v₁ - v₂ = 0_V, hence v₁ = v₂. Therefore T is one-to-one.
Q: True or false: "If a linear map T: V → W is 1-1, then T is onto." Justify.
A: False. Counterexample: T: R² → R³ defined by T(x, y) = (x, y, 0) is injective (distinct inputs give distinct outputs) but not surjective (for instance, (0, 0, 1) is not in the range).
Q: True or false: "The space of continuous functions C⁰(R) has a finite basis." Justify.
A: False. C⁰(R) contains the space P of all polynomials as a subspace, and P has the infinite linearly independent set {1, x, x², x³, ...}. A space with an infinite linearly independent subset cannot have a finite basis.
Q: Prove that {A ∈ M_{m×n} : a₁₁ = 0} is a subspace of M_{m×n}.
A: (i) The zero matrix has a₁₁ = 0, so it is in the set. (ii) If A, B are in the set, then a₁₁ = 0 and b₁₁ = 0, so (A + B)₁₁ = 0, meaning A + B is in the set. (iii) If A is in the set and c is a scalar, then (cA)₁₁ = c · 0 = 0, so cA is in the set. All three subspace conditions hold.
Coordinate vectors rely on the basis and dimension theory from Study Notes 02. The matrix representation [T]^β_γ converts abstract linear maps into concrete matrices, linking this topic back to the matrix arithmetic of Study Notes 01. The rank-nullity theorem ties the nullspace and range dimensions together and connects to the theory of systems of equations (the nullspace of a matrix A is the solution set of Ax = 0). Eigenvalues and diagonalisation, coming later in the course, build directly on change-of-basis and matrix representations.
linear transformation, linear map, coordinate vector, change of basis matrix, matrix representation, domain, codomain, nullspace, kernel, null space, range, image, column space, injective, surjective, bijective, one-to-one, onto, isomorphism, rank, nullity, rank-nullity theorem, subspace proof, subspace test, M_{m×n}, continuous functions, C^0, finite basis, infinite-dimensional