Difficulty: Intermediate Prerequisites: Basis and dimension (L16–17), linear independence, span, vector spaces and subspaces, matrix inverses.
This lecture formalises something you have been doing informally since you first wrote a vector as a column of numbers: expressing a vector in terms of a basis. The key insight is that once you fix a basis, every vector in the space has exactly one set of "coordinates" relative to that basis, and those coordinates live in Rⁿ. That connection turns abstract vector spaces (polynomials, matrices, function spaces) into something you can compute with using the same row-reduction toolkit you already know. If you are comfortable with basis, span, and linear independence, you are ready for this material.
Every vector in a vector space can be written as a unique linear combination of basis vectors; the coefficients are called coordinates. The coordinate mapping is itself a linear transformation (and an isomorphism), which means you can translate problems in exotic vector spaces into equivalent problems in Rⁿ, solve them there, and translate back.
Basis
A set B = {b₁, ..., bₙ} of vectors in a vector space V that (1) spans V and (2) is linearly independent. Think of it as the minimal "ingredient list" that can build every vector in the space.
Coordinates (of x relative to B)
The unique scalars c₁, ..., cₙ such that x = c₁b₁ + ... + cₙbₙ. In simple terms, these are the weights you need on each basis vector to reconstruct x.
Coordinate vector ([x]_B, the B-coordinate vector)
The column vector [c₁, ..., cₙ]ᵀ formed by stacking those coordinates. It lives in Rⁿ even when the original vector space is something else entirely (polynomials, matrices, etc.).
Change-of-coordinate matrix (P_B)
The matrix whose columns are the basis vectors b₁, ..., bₙ. It converts B-coordinates back to standard coordinates via x = P_B [x]_B, and its inverse converts the other way: [x]_B = P_B⁻¹ x.
Coordinate mapping
The function T : V → Rⁿ defined by T(x) = [x]_B. It is a linear transformation that is one-to-one and onto.
Isomorphism
A transformation T : V → W that is both one-to-one and onto. When such a map exists, V and W are called isomorphic (V ≅ W), meaning they are structurally identical as vector spaces. Think of it as a perfect "dictionary" between two spaces: every element in one has exactly one counterpart in the other, and the algebra is preserved.
Isomorphic (V ≅ W)
Two vector spaces are isomorphic when there exists an isomorphism between them. In practical terms, any linear algebra question you can ask in one space has an equivalent answer in the other.
If B = {b₁, ..., bₙ} is a basis for V, then every x in V can be written as x = c₁b₁ + ... + cₙbₙ for a unique set of scalars c₁, ..., cₙ.
Uniqueness is the critical word here. Two different sets of coefficients cannot produce the same vector relative to the same basis.
The proof relies on both properties of a basis: spanning guarantees at least one representation exists; linear independence guarantees at most one exists.
In Rⁿ with the standard basis E = {e₁, ..., eₙ}, the coordinate vector [x]_E is just x itself. The standard basis coordinates are the ordinary components you are used to.
In P₃ with the standard basis E = {1, t, t², t³}, the polynomial x(t) = a₀ + a₁t + a₂t² + a₃t³ has coordinate vector [x]_E = [a₀, a₁, a₂, a₃]ᵀ.
In M₂ₓ₂ with the standard basis of four elementary matrices, a 2×2 matrix maps to a vector in R⁴ by reading off its entries.
From B-coordinates to x (reconstruction): Given [x]_B, compute x = P_B [x]_B, i.e. multiply the change-of-coordinate matrix by the coordinate vector.
Example: B = {[1,1]ᵀ, [−1,2]ᵀ} and [x]_B = [−3,2]ᵀ. Then x = (−3)[1,1]ᵀ + (2)[−1,2]ᵀ = [−5,1]ᵀ.
From x to B-coordinates: Solve [x]_B = P_B⁻¹ x, or equivalently row-reduce the augmented matrix [P_B | x].
Example: Same basis, x = [2,8]ᵀ. Solve the system to find the weights c₁, c₂ such that c₁[1,1]ᵀ + c₂[−1,2]ᵀ = [2,8]ᵀ.
The map x ↦ [x]_B is linear: it preserves addition and scalar multiplication.
It is one-to-one (different vectors produce different coordinate vectors) and onto (every vector in Rⁿ is somebody's coordinate vector).
Because it satisfies all three properties, it is an isomorphism.
If V has a basis of n vectors, then V ≅ Rⁿ. This is the payoff: any finite-dimensional vector space can be "translated" into Rⁿ.
Pₙ ≅ Rⁿ⁺¹ (polynomials of degree at most n map to vectors in Rⁿ⁺¹).
M₂ₓ₂ ≅ R⁴ (2×2 matrices map to vectors in R⁴).
Linear independence, span, and basis questions in V become the same questions in Rⁿ, where you can row-reduce.
To check whether {p₁, p₂, p₃} is linearly independent in P₂, convert each polynomial to its coordinate vector in R³ using the standard basis {1, t, t²}.
Then check whether those coordinate vectors are linearly independent (row-reduce the matrix whose columns are the coordinate vectors and look for pivot columns).
Example: p₁(t) = 1 − t, p₂(t) = 2 − t + t², p₃(t) = 2t + 3t². Their coordinate vectors are [1,−1,0]ᵀ, [2,−1,1]ᵀ, [0,2,3]ᵀ. Row-reduce to determine independence.
To show B = {p₁, p₂, p₃} is a basis for P₂, convert to coordinate vectors in R³ and verify the resulting 3×3 matrix is invertible (i.e. has three pivots, or a nonzero determinant).
To find [p]_B for a given polynomial p, solve the system P_B · c = [p]_E for c, where P_B is the matrix of coordinate columns and [p]_E is p's standard coordinate vector.
Unique representation: x = c₁b₁ + c₂b₂ + ... + cₙbₙ
Change-of-coordinate matrix: P_B = [b₁ b₂ ... bₙ]
Standard-to-B conversion: x = P_B [x]_B
B-coordinate recovery: [x]_B = P_B⁻¹ x
Coordinate systems underpin computer graphics, where the same object is described in different frames of reference (camera coordinates, world coordinates, screen coordinates) and the change-of-coordinate matrix is how the GPU transforms between them. Robotics uses exactly the same idea: each joint of a robot arm defines its own basis, and forward kinematics is a sequence of change-of-coordinate transformations.
Students sometimes think the coordinate vector [x]_B changes the vector itself. It does not. The vector x is the same object in V regardless of which basis you use to describe it; only the list of numbers (the coordinates) changes.
Confusing P_B with P_B⁻¹ is common. P_B converts from B-coordinates to standard coordinates. To go the other direction (standard to B), you need the inverse.
Students sometimes forget that [x]_E = x only when E is the standard basis of Rⁿ. For a non-standard basis, the coordinate vector and the vector itself are different columns.
Isomorphism does not mean the two spaces are "the same space." It means they behave identically under linear algebra operations. P₂ and R³ contain different kinds of objects (polynomials vs. columns of numbers), but their linear algebra structure is interchangeable.
⚠️ The Unique Representation Theorem is frequently tested by asking you to find coordinates in a non-standard basis, either in Rⁿ or in a polynomial/matrix space.
⚠️ Expect at least one problem that asks you to show a set is a basis for P₂ (or similar) by converting to coordinate vectors and row-reducing.
⚠️ Know both directions: reconstructing x from [x]_B, and finding [x]_B from x. Be comfortable setting up and solving the augmented matrix [P_B | x].
⚠️ Be prepared to state why P_B is always invertible (its columns are linearly independent because they form a basis).
True or false: If B is a basis for V, a vector x in V can be expressed as a linear combination of the basis vectors in more than one way.
Fill in the blank: The change-of-coordinate matrix P_B has the ______ as its columns.
True or false: The coordinate mapping x ↦ [x]_B is a linear transformation.
Fill in the blank: If V has a basis of n vectors, then V is isomorphic to ______.
True or false: To convert from standard coordinates to B-coordinates, you multiply by P_B.
Answers: 1. False (uniqueness is guaranteed). 2. Basis vectors b₁, ..., bₙ. 3. True. 4. Rⁿ. 5. False (you multiply by P_B⁻¹).
Q: Let B = {b₁, b₂} be a basis for R², where b₁ = [1, 1]ᵀ and b₂ = [−1, 2]ᵀ. If [x]_B = [−3, 2]ᵀ, find x.
A: x = (−3)[1,1]ᵀ + (2)[−1,2]ᵀ = [−3−2, −3+4]ᵀ = [−5, 1]ᵀ.
Q: Using the same basis B, find [x]_B when x = [2, 8]ᵀ.
A: Solve c₁[1,1]ᵀ + c₂[−1,2]ᵀ = [2,8]ᵀ. The system is c₁ − c₂ = 2 and c₁ + 2c₂ = 8. From the first equation c₁ = 2 + c₂; substituting gives 2 + c₂ + 2c₂ = 8, so c₂ = 2 and c₁ = 4. Thus [x]_B = [4, 2]ᵀ.
Q: Let p₁(t) = 1 − t, p₂(t) = 2 − t + t², p₃(t) = 2t + 3t². Write each polynomial as a coordinate vector in R³ using the standard basis {1, t, t²} for P₂.
A: [p₁]_E = [1, −1, 0]ᵀ, [p₂]_E = [2, −1, 1]ᵀ, [p₃]_E = [0, 2, 3]ᵀ.
Q: How would you determine whether {p₁, p₂, p₃} above is linearly independent in P₂?
A: Form the matrix A = [[1, 2, 0], [−1, −1, 2], [0, 1, 3]] (columns are the coordinate vectors) and row-reduce. If every column has a pivot, the set is linearly independent.
Q: Why is the change-of-coordinate matrix P_B always invertible?
A: The columns of P_B are the vectors of a basis, so they are linearly independent. A square matrix with linearly independent columns is invertible.
Q: Explain why P₂ ≅ R³.
A: The standard basis for P₂ is {1, t, t²}, which has 3 elements. The coordinate mapping T : P₂ → R³ defined by T(p) = [p]_E is a linear transformation that is one-to-one and onto, so it is an isomorphism.
This material connects directly to change of basis (the next step is converting between two non-standard bases, not just between a non-standard basis and the standard one). It also sets up the matrix representation of linear transformations: once you fix bases for the domain and codomain, any linear transformation can be represented as a matrix acting on coordinate vectors. If you have studied eigenvalues, the idea of choosing a "good" basis (e.g. eigenvectors) to simplify a matrix is a direct application of change of coordinates.
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