Coordinate Vectors and Matrix Action

How do you actually compute with an abstract vector like a polynomial or a function? You must use nothing but arithmetic on lists of numbers. The answer is to fix a basis. Translate every vector into a column of coordinates. Then translate every linear map into a matrix. Applying the map becomes multiplying the matrix by the column. This translation is not a vague analogy. It is a precise, structure-preserving correspondence. This article proves exactly what it preserves and why. The idea matters because it is the hinge on which all of computational linear algebra turns. Without it, "linear map" and "matrix" are related only by example. With it, they become two names for the same object. They are viewed through two different lenses, abstract and concrete. Every time you solve Ax=bAx = b, diagonalize a matrix, or run a numerical algorithm, you are secretly working with an abstract linear map TT on some vector space VV. The matrix arithmetic faithfully simulates the action of TT. This happens because of the theorem proved here. The big picture is a dictionary with three ingredients. First, a basis β\beta of VV produces a coordinate map [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta : V \to \mathbb{F}^n. Second, a basis pair (β,γ)(\beta,\gamma) produces a matrix M(T)M(T) for a linear map T:V→WT : V \to W. Third, a single identity glues them together,

[T(v)]γ=M(T) [v]β.[T(v)]_\gamma = M(T)\,[v]_\beta.

This article states that identity as a theorem and proves it in full. It builds the geometric intuition behind it. It works two nontrivial examples — one with polynomials, one with a genuinely non-diagonal, non-square-friendly setup. This ensures the mechanism is transparent rather than memorized. This article delivers a complete, self-contained proof of the coordinate-matrix identity. It discusses exactly which hypotheses are load-bearing (finite dimensionality, a fixed choice of ordered bases). It offers a worked counterexample showing what fails without an ordered basis. Finally, it bridges forward to how this identity turns L(V,W)\mathcal{L}(V,W) itself into a matrix space. It also turns composition of maps into matrix multiplication.

Key takeaways
  • Fixing an ordered basis β\beta of VV gives a linear bijection [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta: V \to \mathbb{F}^n, the coordinate map.
  • Given The Matrix of a Linear Map M(T)M(T) built from bases β\beta (domain) and γ\gamma (codomain), the identity [T(v)]γ=M(T)[v]β[T(v)]_\gamma = M(T)[v]_\beta holds for every v∈Vv \in V.
  • This identity is exactly what lets you replace an abstract computation (T(v)T(v)) with concrete matrix-vector multiplication.
  • The correspondence T↦M(T)T \mapsto M(T) and v↦[v]βv \mapsto [v]_\beta are both linear bijections (isomorphisms), so no information is lost.
  • Changing the ordered basis changes the coordinate vector and the matrix, but never breaks the identity — it is basis-independent in form even though its numerical output is basis-dependent.
  • Key idea: coordinates and matrices are simultaneously constructed translation devices, and the theorem says the translation commutes with applying the map.

Where this lives today

Every numerical linear algebra library (LAPACK, NumPy, MATLAB) is, under the hood, an implementation of this dictionary: an abstract linear operator on function spaces, signals, or images is discretized by choosing a basis, and all subsequent computation is matrix arithmetic on coordinate vectors, justified exactly by this theorem. In quantum mechanics, states are abstract vectors in a Hilbert space, but physical predictions require expanding a state in a chosen basis (position, momentum, energy eigenbasis) and representing operators as matrices in that same basis — the coordinate-matrix identity is the mathematical content of "changing representation." In computer graphics and robotics, points and orientations are abstract geometric objects, but rendering and control require coordinatizing them relative to a frame and applying transformation matrices, again governed by this identity, including its behavior under a change of frame. In finite element methods for differential equations, an infinite-dimensional function space is approximated by a finite-dimensional subspace with a chosen basis of simple functions, converting a differential operator into a finite matrix acting on coefficient vectors — the entire discretization step is an application of this theorem. In data science and machine learning, feature vectors, principal component coordinates, and learned linear layers all rely on the same fact: a linear transformation of feature space is a matrix acting on coordinate vectors relative to a fixed basis (or fixed feature encoding). This identity is also the precise justification for why composing linear maps corresponds to multiplying matrices, which is the content of Matrix Multiplication as Composition, and it is the starting point for showing that the set of all linear maps L(V,W)\mathcal{L}(V,W) is itself a vector space isomorphic to a matrix space, the subject of The Space L(V,W) as Matrices.

:::quiz[If dim⁡V=3\dim V = 3, dim⁡W=2\dim W = 2, and M(T)M(T) is 2×32\times 3, what is the size of [v]β[v]_\beta for v∈Vv \in V?]

  • 2×12 \times 1
  • 3×13 \times 1 ✓
  • 2×32 \times 3
  • 3×33 \times 3

:::quiz[Which statement correctly describes the coordinate map [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta : V \to \mathbb{F}^n for a finite-dimensional VV with ordered basis β\beta?]

  • It is linear but generally not injective
  • It is a linear bijection ✓
  • It is a bijection but not linear
  • It depends only on the set β\beta, not its ordering

Ask the tutor: How does the coordinate-matrix identity change if I switch from basis β\beta to a different basis β′\beta' — what exactly is the change-of-basis matrix, and how does it interact with M(T)M(T)?

DefinitionCoordinate vector and the coordinate map

Let VV be a vector space of dimension nn over a field F\mathbb{F}, and let

β=(v1,v2,…,vn)\beta = (v_1, v_2, \dots, v_n)

be an ordered basis of VV (a basis together with a fixed listing order). For v∈Vv \in V, the unique scalars a1,…,an∈Fa_1,\dots,a_n \in \mathbb{F} satisfying

v=a1v1+a2v2+⋯+anvnv = a_1 v_1 + a_2 v_2 + \cdots + a_n v_n

are called the coordinates of vv relative to β\beta. The coordinate vector of vv relative to β\beta is the column

[v]β=(a1a2⋮an)∈Fn.[v]_\beta = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in \mathbb{F}^n.

The map [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta : V \to \mathbb{F}^n sending v↦[v]βv \mapsto [v]_\beta is the coordinate map determined by β\beta. Given finite-dimensional vector spaces VV (dimension nn, ordered basis β=(v1,…,vn)\beta = (v_1,\dots,v_n)) and WW (dimension mm, ordered basis γ=(w1,…,wm)\gamma = (w_1,\dots,w_m)), and a linear map T:V→WT : V \to W, recall that the matrix of TT relative to β,γ\beta,\gamma, written M(T)=M(T)γβM(T) = M(T)_\gamma^\beta, is the m×nm \times n matrix whose jj-th column is [T(vj)]γ[T(v_j)]_\gamma, i.e.

T(vj)=∑i=1mAijwi,M(T)=(Aij).T(v_j) = \sum_{i=1}^m A_{ij} w_i, \qquad M(T) = (A_{ij}).

Intuition

Think of an ordered basis as a set of measuring rods laid down in VV: once you fix them, every vector gets an address, a list of nn numbers telling you how far to go along each rod. That address is the coordinate vector. The coordinate map is nothing more than the act of reading off that address. Because a basis representation is unique (existence and uniqueness of basis coordinates is inherited from the definition of a basis), the coordinate map is a genuine function, well-defined and single-valued. Now suppose TT moves vectors around inside VV, or sends them into a different space WW entirely. If you know the addresses of the basis vectors' images — that is, you know where each rod's tip lands, expressed in the WW-address system — then linearity forces you to know where every vector lands, because every vector is a specific combination of the rods. Encoding "where the rods' tips land" as the columns of a matrix is precisely the construction of M(T)M(T). The theorem below is the formal statement that reading off an address commutes with moving the point: address of image = matrix times address of original point. Multiplying a matrix by a column vector is literally simulating linear combination, so this is really just linearity written out in coordinates.

Figure
Coordinates as addresses, and the matrix as a rule for moving them
V (basis β = (v₁, v₂)), coordinates (a₁,a₂) v₁ v₂ v = 2v₁ + 1v₂ → [v]_β = (2,1)ᵀ W (basis γ = (w₁, w₂, w₃)), coordinates (b₁,b₂,b₃) w₁ w₂ w₃ T(v) → [T(v)]_γ = M(T)·(2,1)ᵀ T

The central theorem

TheoremCoordinate–matrix action identity

Let F\mathbb{F} be a field, let VV be a vector space over F\mathbb{F} with dim⁡V=n<∞\dim V = n < \infty and ordered basis β=(v1,…,vn)\beta = (v_1,\dots,v_n), and let WW be a vector space over F\mathbb{F} with dim⁡W=m<∞\dim W = m < \infty and ordered basis γ=(w1,…,wm)\gamma = (w_1,\dots,w_m). Let T:V→WT : V \to W be a linear map, and let A=M(T)=M(T)γβ∈Fm×nA = M(T) = M(T)_\gamma^\beta \in \mathbb{F}^{m \times n} be the matrix of TT relative to β\beta and γ\gamma, as in The Matrix of a Linear Map. Then for every v∈Vv \in V,

[T(v)]γ  =  A [v]β,[T(v)]_\gamma \;=\; A\,[v]_\beta,

where the right-hand side is ordinary matrix-vector multiplication in Fm\mathbb{F}^{m}. Equivalently, the following diagram commutes: the composite [ ⋅ ]γ∘T[\,\cdot\,]_\gamma \circ T equals the composite (x↦Ax)∘[ ⋅ ]β(x \mapsto Ax) \circ [\,\cdot\,]_\beta, as functions V→FmV \to \mathbb{F}^m. Moreover, both coordinate maps [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta : V \to \mathbb{F}^n and [ ⋅ ]γ:W→Fm[\,\cdot\,]_\gamma : W \to \mathbb{F}^m are linear bijections.

LemmaLinearity and bijectivity of the coordinate map

With V,n,βV, n, \beta as above, the coordinate map [ ⋅ ]β:V→Fn[\,\cdot\,]_\beta : V \to \mathbb{F}^n is a linear bijection.

Proof

Well-defined. Since β=(v1,…,vn)\beta = (v_1,\dots,v_n) is a basis of VV, every v∈Vv \in V can be written as v=∑i=1naiviv = \sum_{i=1}^n a_i v_i for some scalars aia_i, because β\beta spans VV. These scalars are unique: if also v=∑i=1nai′viv = \sum_{i=1}^n a_i' v_i, then

∑i=1n(ai−ai′)vi=0,\sum_{i=1}^n (a_i - a_i') v_i = 0,

and linear independence of β\beta forces ai−ai′=0a_i - a_i' = 0 for all ii. Hence [v]β[v]_\beta is a well-defined single element of Fn\mathbb{F}^n for each vv, and [ ⋅ ]β[\,\cdot\,]_\beta is a genuine function.

Linear. Let v,v′∈Vv, v' \in V with v=∑iaiviv = \sum_i a_i v_i, v′=∑iai′viv' = \sum_i a_i' v_i, and let c∈Fc \in \mathbb{F}. Then

v+v′=∑i=1n(ai+ai′)vi,cv=∑i=1n(cai)vi,v + v' = \sum_{i=1}^n (a_i + a_i') v_i, \qquad c v = \sum_{i=1}^n (c a_i) v_i,

using distributivity and associativity of scalar multiplication in VV. By the uniqueness just established, these are the coordinate representations of v+v′v+v' and cvcv, so

[v+v′]β=[v]β+[v′]β,[cv]β=c[v]β.[v + v']_\beta = [v]_\beta + [v']_\beta, \qquad [cv]_\beta = c[v]_\beta.

Thus [ ⋅ ]β[\,\cdot\,]_\beta is linear.

Injective. Suppose [v]β=0[v]_\beta = 0, i.e., all coordinates of vv are 00. Then v=∑i0⋅vi=0v = \sum_i 0 \cdot v_i = 0. Since [ ⋅ ]β[\,\cdot\,]_\beta is linear, its kernel is {0}\{0\}, which is equivalent to injectivity for a linear map between vector spaces (if [u]β=[v]β[u]_\beta = [v]_\beta then [u−v]β=0[u - v]_\beta = 0, so u−v=0u - v = 0).

Surjective. Given any (a1,…,an)∈Fn(a_1,\dots,a_n) \in \mathbb{F}^n, set v=∑i=1naivi∈Vv = \sum_{i=1}^n a_i v_i \in V; this is a legitimate element of VV since VV is closed under linear combinations. By construction [v]β=(a1,…,an)T[v]_\beta = (a_1,\dots,a_n)^T. Hence every element of Fn\mathbb{F}^n is hit. Injective and surjective together give bijective. ■\blacksquare

∎
Proof

Proof of the main theorem. Step 1: Reduce to basis vectors using linearity of TT and of the coordinate map. Fix v∈Vv \in V and write it in coordinates relative to β\beta:

v=∑j=1ncjvj,so [v]β=(c1,…,cn)T.v = \sum_{j=1}^n c_j v_j, \qquad \text{so } [v]_\beta = (c_1, \dots, c_n)^T.

This representation exists and is unique because β\beta is a basis, exactly as used in the Lemma. Apply TT and use linearity of TT:

T(v)=T(∑j=1ncjvj)=∑j=1ncj T(vj).T(v) = T\left(\sum_{j=1}^n c_j v_j\right) = \sum_{j=1}^n c_j\, T(v_j).

This step uses only that TT is a linear map, i.e. T(x+y)=T(x)+T(y)T(x+y) = T(x)+T(y) and T(cx)=cT(x)T(cx) = cT(x), extended to finite sums by induction on nn. Step 2: Take γ\gamma-coordinates of both sides and use linearity of the coordinate map on WW. By the Lemma applied to WW with basis γ\gamma, the coordinate map [ ⋅ ]γ:W→Fm[\,\cdot\,]_\gamma : W \to \mathbb{F}^m is linear. Apply it to the identity from Step 1:

[T(v)]γ=[∑j=1ncj T(vj)]γ=∑j=1ncj [T(vj)]γ,[T(v)]_\gamma = \left[\sum_{j=1}^n c_j\, T(v_j)\right]_\gamma = \sum_{j=1}^n c_j\, [T(v_j)]_\gamma,

where the last equality is linearity of [ ⋅ ]γ[\,\cdot\,]_\gamma applied repeatedly (additivity across the sum, then scalar homogeneity on each term). Step 3: Identify [T(vj)]γ[T(v_j)]_\gamma with the jj-th column of AA. By the definition of M(T)=A=(Aij)M(T) = A = (A_{ij}), the jj-th column of AA is exactly [T(vj)]γ[T(v_j)]_\gamma, i.e.

[T(vj)]γ=(A1jA2j⋮Amj).[T(v_j)]_\gamma = \begin{pmatrix} A_{1j} \\ A_{2j} \\ \vdots \\ A_{mj} \end{pmatrix}.

This is not an extra assumption; it is the literal content of how M(T)M(T) was constructed in The Matrix of a Linear Map — the columns of M(T)M(T) are defined to be the γ\gamma-coordinate vectors of the images of the β\beta-basis vectors. Step 4: Assemble and recognize matrix-vector multiplication. Combining Steps 2 and 3,

[T(v)]γ=∑j=1ncj(A1j⋮Amj).[T(v)]_\gamma = \sum_{j=1}^n c_j \begin{pmatrix} A_{1j} \\ \vdots \\ A_{mj}\end{pmatrix}.

Now recall the definition of the matrix-vector product AxAx for A∈Fm×nA \in \mathbb{F}^{m\times n} and x=(c1,…,cn)T∈Fnx = (c_1,\dots,c_n)^T \in \mathbb{F}^n: the ii-th entry of AxAx is ∑j=1nAijcj\sum_{j=1}^n A_{ij} c_j, which is precisely the statement that AxAx is the linear combination of the columns of AA with coefficients cjc_j:

Ax=∑j=1ncj(A1j⋮Amj).Ax = \sum_{j=1}^n c_j \begin{pmatrix} A_{1j}\\ \vdots \\ A_{mj}\end{pmatrix}.

Since x=[v]βx = [v]_\beta by definition of cjc_j, we conclude

[T(v)]γ=A[v]β=M(T) [v]β,[T(v)]_\gamma = A[v]_\beta = M(T)\,[v]_\beta,

which is the desired identity. It holds for the arbitrary v∈Vv \in V fixed at the start of Step 1, so it holds for all v∈Vv \in V. Diagram commutativity is a restatement: the two functions V→FmV \to \mathbb{F}^m given by v↦[T(v)]γv \mapsto [T(v)]_\gamma and v↦A[v]βv \mapsto A[v]_\beta agree at every input, hence are equal as functions. The final clause (both coordinate maps are linear bijections) is exactly the Lemma applied once with (V,n,β)(V,n,\beta) and once with (W,m,γ)(W,m,\gamma). ■\blacksquare Proof technique used: direct computation via the defining properties of a basis (existence/uniqueness of coordinates) combined with linearity of TT and of the coordinate maps; no result depending on this theorem was invoked, so the argument is non-circular.

∎
CorollaryComposition and identity checks

With hypotheses as in the theorem, if IV:V→VI_V : V \to V is the identity map and β=β′\beta = \beta' is used on both sides, then M(IV)=InM(I_V) = I_n (the n×nn\times n identity matrix), and for every vv, [v]β=In[v]β[v]_\beta = I_n [v]_\beta, consistent with the theorem. More generally, if U→SV→TWU \xrightarrow{S} V \xrightarrow{T} W are linear maps between finite-dimensional spaces with ordered bases α,β,γ\alpha,\beta,\gamma respectively, applying the theorem twice gives

[T(S(u))]γ=M(T)[S(u)]β=M(T)(M(S)[u]α)=(M(T)M(S))[u]α[T(S(u))]_\gamma = M(T)[S(u)]_\beta = M(T)\big(M(S)[u]_\alpha\big) = \big(M(T)M(S)\big)[u]_\alpha

for all u∈Uu \in U, foreshadowing the identification of matrix multiplication with composition of linear maps.

Worked examples

Example 1 (Differentiation on polynomials, non-square matrix). Let V=P3(R)V = P_3(\mathbb{R}), real polynomials of degree at most 33, with ordered basis

β=(1,x,x2,x3),dim⁡V=4.\beta = (1, x, x^2, x^3), \qquad \dim V = 4.

Let W=P2(R)W = P_2(\mathbb{R}) with ordered basis

γ=(1,x,x2),dim⁡W=3.\gamma = (1, x, x^2), \qquad \dim W = 3.

Let T:V→WT : V \to W be differentiation, T(p)=p′T(p) = p'. This is linear because (p+q)′=p′+q′(p+q)' = p'+q' and (cp)′=c p′(cp)' = c\,p' for polynomials, standard facts from calculus. Compute the matrix. We need TT applied to each basis vector of β\beta, expressed in γ\gamma-coordinates:

T(1)=0=0⋅1+0⋅x+0⋅x2,[T(1)]γ=(0,0,0)T,T(1) = 0 = 0\cdot 1 + 0\cdot x + 0\cdot x^2, \qquad [T(1)]_\gamma = (0,0,0)^T,
T(x)=1=1⋅1+0⋅x+0⋅x2,[T(x)]γ=(1,0,0)T,T(x) = 1 = 1\cdot 1 + 0\cdot x + 0\cdot x^2, \qquad [T(x)]_\gamma = (1,0,0)^T,
T(x2)=2x=0⋅1+2⋅x+0⋅x2,[T(x2)]γ=(0,2,0)T,T(x^2) = 2x = 0\cdot 1 + 2\cdot x + 0\cdot x^2, \qquad [T(x^2)]_\gamma = (0,2,0)^T,
T(x3)=3x2=0⋅1+0⋅x+3⋅x2,[T(x3)]γ=(0,0,3)T.T(x^3) = 3x^2 = 0\cdot 1 + 0\cdot x + 3\cdot x^2, \qquad [T(x^3)]_\gamma = (0,0,3)^T.

Stacking these as columns,

A=M(T)=(010000200003)∈R3×4.A = M(T) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \in \mathbb{R}^{3 \times 4}.

Now test the theorem on a specific, asymmetric vector. Let

p(x)=5−2x+4x2+7x3.p(x) = 5 - 2x + 4x^2 + 7x^3.

Its coordinate vector is

[p]β=(5−247).[p]_\beta = \begin{pmatrix} 5 \\ -2 \\ 4 \\ 7 \end{pmatrix}.

Directly, by calculus,

p′(x)=−2+8x+21x2,[p′]γ=(−2821).p'(x) = -2 + 8x + 21x^2, \qquad [p']_\gamma = \begin{pmatrix} -2 \\ 8 \\ 21 \end{pmatrix}.

Now via the matrix:

A[p]β=(010000200003)(5−247)=((0)(5)+(1)(−2)+(0)(4)+(0)(7)(0)(5)+(0)(−2)+(2)(4)+(0)(7)(0)(5)+(0)(−2)+(0)(4)+(3)(7))=(−2821).A [p]_\beta = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 5\\-2\\4\\7 \end{pmatrix} = \begin{pmatrix} (0)(5)+(1)(-2)+(0)(4)+(0)(7)\\ (0)(5)+(0)(-2)+(2)(4)+(0)(7)\\ (0)(5)+(0)(-2)+(0)(4)+(3)(7) \end{pmatrix} = \begin{pmatrix} -2 \\ 8 \\ 21 \end{pmatrix}.

Both computations agree exactly, confirming [T(p)]γ=A[p]β[T(p)]_\gamma = A[p]_\beta, as guaranteed by the theorem.

Example 2 (A linear map on R3\mathbb{R}^3 with a non-standard basis, showing the identity is not tied to standard coordinates). Let V=W=R3V = W = \mathbb{R}^3, and let T:R3→R3T : \mathbb{R}^3 \to \mathbb{R}^3 be given in standard coordinates by

T(x,y,z)=(2x+y,  y−z,  x+3z).T(x,y,z) = (2x + y,\; y - z,\; x + 3z).

This is linear: writing T(x,y,z)=(2,1,0)⋅(x,y,z)TT(x,y,z) = (2,1,0)\cdot(x,y,z)^T etc., each output coordinate is a linear combination of x,y,zx,y,z, and sums/scalar multiples check out immediately from the formula (linear in each coordinate separately, hence linear on R3\mathbb{R}^3). Instead of the standard basis, use the non-standard ordered basis

β=γ=(u1,u2,u3),u1=(1,1,0),u2=(0,1,1),u3=(1,0,1).\beta = \gamma = (u_1, u_2, u_3), \quad u_1 = (1,1,0), \quad u_2 = (0,1,1), \quad u_3 = (1,0,1).

First check β\beta is a basis: the matrix with these as columns,

(101110011),\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix},

has determinant 1(1⋅1−0⋅1)−0(1⋅1−0⋅0)+1(1⋅1−1⋅0)=1−0+1=2≠01(1\cdot1 - 0\cdot1) - 0(1\cdot1-0\cdot0) + 1(1\cdot1 - 1\cdot0) = 1 - 0 + 1 = 2 \ne 0, so the three vectors are linearly independent, and three independent vectors in a 33-dimensional space form a basis. Compute T(u1),T(u2),T(u3)T(u_1), T(u_2), T(u_3):

T(u1)=T(1,1,0)=(2⋅1+1, 1−0, 1+0)=(3,1,1),T(u_1) = T(1,1,0) = (2\cdot1+1,\ 1-0,\ 1+0) = (3,1,1),
T(u2)=T(0,1,1)=(0+1, 1−1, 0+3)=(1,0,3),T(u_2) = T(0,1,1) = (0+1,\ 1-1,\ 0+3) = (1,0,3),
T(u3)=T(1,0,1)=(2+0, 0−1, 1+3)=(2,−1,4).T(u_3) = T(1,0,1) = (2+0,\ 0-1,\ 1+3) = (2,-1,4).

Now express each of these as combinations of u1,u2,u3u_1,u_2,u_3 to get γ\gamma-coordinates. Solve au1+bu2+cu3=(3,1,1)a u_1 + b u_2 + c u_3 = (3,1,1):

a+c=3,a+b=1,b+c=1.a + c = 3,\quad a+b = 1,\quad b+c = 1.

From the first, c=3−ac = 3-a. Substitute into the third: b+3−a=1⇒b=a−2b + 3 - a = 1 \Rightarrow b = a - 2. Substitute into the second: a+(a−2)=1⇒2a=3⇒a=3/2a + (a-2) = 1 \Rightarrow 2a = 3 \Rightarrow a = 3/2. Then b=−1/2b = -1/2, c=3/2c = 3/2. Check: a+c=3/2+3/2=3a+c = 3/2+3/2=3 ✓, a+b=3/2−1/2=1a+b=3/2-1/2=1 ✓, b+c=−1/2+3/2=1b+c=-1/2+3/2=1 ✓. So [T(u1)]γ=(3/2, −1/2, 3/2)T[T(u_1)]_\gamma = (3/2,\,-1/2,\,3/2)^T. Solve au1+bu2+cu3=(1,0,3)a u_1+bu_2+cu_3 = (1,0,3): a+c=1a+c=1, a+b=0a+b=0, b+c=3b+c=3. From second, b=−ab=-a. Third: −a+c=3⇒c=a+3-a+c=3 \Rightarrow c = a+3. First: a+a+3=1⇒2a=−2⇒a=−1a + a+3 = 1 \Rightarrow 2a = -2 \Rightarrow a=-1. Then b=1b=1, c=2c=2. Check: a+c=−1+2=1a+c=-1+2=1✓, a+b=−1+1=0a+b=-1+1=0✓, b+c=1+2=3b+c=1+2=3✓. So [T(u2)]γ=(−1, 1, 2)T[T(u_2)]_\gamma = (-1,\,1,\,2)^T. Solve au1+bu2+cu3=(2,−1,4)a u_1+bu_2+cu_3=(2,-1,4): a+c=2a+c=2, a+b=−1a+b=-1, b+c=4b+c=4. From second b=−1−ab=-1-a. Third: −1−a+c=4⇒c=a+5-1-a+c=4 \Rightarrow c=a+5. First: a+a+5=2⇒2a=−3⇒a=−3/2a+a+5=2 \Rightarrow 2a=-3 \Rightarrow a=-3/2. Then b=−1+3/2=1/2b=-1+3/2=1/2, c=−3/2+5=7/2c=-3/2+5=7/2. Check: a+c=−3/2+7/2=2a+c=-3/2+7/2=2✓, a+b=−3/2+1/2=−1a+b=-3/2+1/2=-1✓, b+c=1/2+7/2=4b+c=1/2+7/2=4✓. So [T(u3)]γ=(−3/2, 1/2, 7/2)T[T(u_3)]_\gamma = (-3/2,\,1/2,\,7/2)^T. Assembling columns,

A=M(T)=(3/2−1−3/2−1/211/23/227/2).A = M(T) = \begin{pmatrix} 3/2 & -1 & -3/2 \\ -1/2 & 1 & 1/2 \\ 3/2 & 2 & 7/2 \end{pmatrix}.

Now test the theorem on a vector that is not one of the basis vectors: let v=(4,−1,2)∈R3v = (4,-1,2) \in \mathbb{R}^3, in standard coordinates. First find [v]β[v]_\beta: solve au1+bu2+cu3=(4,−1,2)a u_1+bu_2+cu_3=(4,-1,2): a+c=4a+c=4, a+b=−1a+b=-1, b+c=2b+c=2. From second b=−1−ab=-1-a. Third: −1−a+c=2⇒c=a+3-1-a+c=2 \Rightarrow c=a+3. First: a+a+3=4⇒2a=1⇒a=1/2a+a+3=4 \Rightarrow 2a=1 \Rightarrow a=1/2. Then b=−3/2b=-3/2, c=7/2c=7/2. Check: a+c=1/2+7/2=4a+c=1/2+7/2=4✓, a+b=1/2−3/2=−1a+b=1/2-3/2=-1✓, b+c=−3/2+7/2=2b+c=-3/2+7/2=2✓.

[v]β=(1/2, −3/2, 7/2)T.[v]_\beta = (1/2,\,-3/2,\,7/2)^T.

Direct computation: T(4,−1,2)=(2⋅4+(−1), −1−2, 4+3⋅2)=(7,−3,10)T(4,-1,2) = (2\cdot4+(-1),\ -1-2,\ 4+3\cdot2) = (7,-3,10). Express in γ\gamma-coordinates: solve a+c=7a+c=7, a+b=−3a+b=-3, b+c=10b+c=10. From second b=−3−ab=-3-a. Third: −3−a+c=10⇒c=a+13-3-a+c=10 \Rightarrow c=a+13. First: a+a+13=7⇒2a=−6⇒a=−3a+a+13=7 \Rightarrow 2a=-6 \Rightarrow a=-3. Then b=0b=0, c=10c=10. Check: a+c=−3+10=7a+c=-3+10=7✓, a+b=−3+0=−3a+b=-3+0=-3✓, b+c=0+10=10b+c=0+10=10✓.

[T(v)]γ=(−3, 0, 10)T.[T(v)]_\gamma = (-3,\,0,\,10)^T.

Now via the matrix:

A[v]β=(3/2−1−3/2−1/211/23/227/2)(1/2−3/27/2).A[v]_\beta = \begin{pmatrix} 3/2 & -1 & -3/2 \\ -1/2 & 1 & 1/2 \\ 3/2 & 2 & 7/2 \end{pmatrix}\begin{pmatrix} 1/2\\-3/2\\7/2 \end{pmatrix}.

First entry: 32⋅12+(−1)⋅(−32)+(−32)⋅72=34+32−214=34+64−214=−124=−3.\frac32\cdot\frac12 + (-1)\cdot(-\frac32) + (-\frac32)\cdot\frac72 = \frac34 + \frac32 - \frac{21}{4} = \frac{3}{4}+\frac{6}{4}-\frac{21}{4} = -\frac{12}{4} = -3. Second entry: −12⋅12+1⋅(−32)+12⋅72=−14−32+74=−14+74−32=64−64=0.-\frac12\cdot\frac12 + 1\cdot(-\frac32) + \frac12\cdot\frac72 = -\frac14 - \frac32 + \frac74 = -\frac14+\frac74-\frac32 = \frac{6}{4} - \frac{6}{4} = 0. Third entry: 32⋅12+2⋅(−32)+72⋅72=34−3+494=34−124+494=404=10.\frac32\cdot\frac12 + 2\cdot(-\frac32) + \frac72\cdot\frac72 = \frac34 - 3 + \frac{49}{4} = \frac{3}{4}-\frac{12}{4}+\frac{49}{4} = \frac{40}{4}=10. So A[v]β=(−3,0,10)TA[v]_\beta = (-3,0,10)^T, matching [T(v)]γ[T(v)]_\gamma computed directly. The identity holds even with an entirely non-standard, non-orthogonal, fraction-producing basis, confirming that the theorem is genuinely basis-independent in form.

CounterexampleWhy the ordering (and fixing) of the basis is essential

Drop the requirement that β\beta is an ordered basis and instead treat it as an unordered set, or worse, let the "basis used to build [v]β[v]_\beta" silently differ from the "basis used to build M(T)M(T)." Take V=W=R2V = W = \mathbb{R}^2, T(x,y)=(x+2y,  3x−y)T(x,y) = (x+2y,\; 3x - y), and the standard basis e1=(1,0),e2=(0,1)e_1=(1,0), e_2=(0,1). Build M(T)M(T) using the ordering γ=(e1,e2)\gamma = (e_1,e_2) for the codomain and β=(e1,e2)\beta = (e_1,e_2) for the domain:

T(e1)=(1,3),T(e2)=(2,−1),A=(123−1).T(e_1) = (1,3), \quad T(e_2) = (2,-1), \quad A = \begin{pmatrix}1 & 2\\ 3 & -1\end{pmatrix}.

Now suppose, when computing a specific coordinate vector, someone instead lists the domain basis in the swapped order β′=(e2,e1)\beta' = (e_2, e_1) — the same set, different order — and still writes the coordinate vector of v=(5,1)=5e1+1e2v = (5,1) = 5e_1 + 1e_2 using this swapped listing convention, getting [v]β′=(1,5)T[v]_{\beta'} = (1,5)^T (coefficient of e2e_2 first, then e1e_1), while continuing to use the matrix AA that was built assuming the order (e1,e2)(e_1,e_2). Direct computation gives the true answer:

T(5,1)=(5+2, 15−1)=(7,14).T(5,1) = (5+2,\ 15-1) = (7,14).

But multiplying the mismatched pairing,

A(15)=(1⋅1+2⋅53⋅1+(−1)⋅5)=(11−2)≠(714).A \begin{pmatrix}1\\5\end{pmatrix} = \begin{pmatrix}1\cdot1+2\cdot5\\ 3\cdot1+(-1)\cdot5\end{pmatrix} = \begin{pmatrix}11\\-2\end{pmatrix} \ne \begin{pmatrix}7\\14\end{pmatrix}.

The identity [T(v)]γ=A[v]β[T(v)]_\gamma = A[v]_\beta fails, not because linearity broke, but because [v]β′[v]_{\beta'} was computed relative to a different ordered basis than the one used to construct AA's columns. The theorem's hypothesis that a single, fixed, ordered basis β\beta is used consistently on both sides of the identity is not decorative: violate it and the arithmetic silently computes the wrong vector while looking exactly like a valid matrix-vector product.

Scope, edge cases & common traps

SituationWhat happens
VV or WW infinite-dimensionalCoordinate vectors and matrices as finite columns/arrays are not defined this way; the theorem as stated requires dim⁡V,dim⁡W<∞\dim V, \dim W < \infty
Using the same basis for domain and codomain when T:V→VT:V\to VAllowed and common (β=γ\beta = \gamma); the identity still reads [T(v)]β=M(T)[v]β[T(v)]_\beta = M(T)[v]_\beta
Reordering a basis without rebuilding M(T)M(T)Breaks the identity, as shown in the counterexample; order is part of the data of an ordered basis
Zero vector v=0v = 0[0]β=0∈Fn[0]_\beta = 0 \in \mathbb{F}^n; identity gives [T(0)]γ=A⋅0=0[T(0)]_\gamma = A\cdot 0 = 0, consistent with T(0)=0T(0)=0 for linear TT
TT not linearTheorem's proof breaks at Step 1 (cannot pull the sum through TT); no matrix representation of this kind exists
F=C\mathbb{F} = \mathbb{C} or a general fieldProof goes through verbatim; nothing used R\mathbb{R}-specific structure (no inner product, no ordering of F\mathbb{F})

The mathematician behind it

Origins · 1 minHermann Günther Grassmann and how this was discovered
HG
Hermann Günther Grassmann
1809–1877 · 19th-century Germany, linear algebra and exterior algebra

Grassmann was a schoolteacher in Stettin (then Prussia, now Szczecin, Poland) who worked largely outside the professional mathematical establishment of his time. His 1844 treatise Die lineale Ausdehnungslehre ("The Theory of Linear Extension") introduced, in essence, abstract vector spaces, linear independence, dimension, and bases decades before these notions were standard — laying the conceptual groundwork that makes a statement like "fix an ordered basis, then every vector has unique coordinates" a theorem rather than an assumption. His work was written in dense, idiosyncratic notation and largely ignored during his lifetime; he was better known then for contributions to comparative linguistics and Sanskrit studies. Only after his death did mathematicians such as Giuseppe Peano recognize the full generality of his ideas and recast them in the language that eventually became modern linear algebra.

History
- 1844: Grassmann publishes the Ausdehnungslehre, introducing an abstract theory of linear extension, independence, and dimension.
- 1858: Arthur Cayley's A Memoir on the Theory of Matrices systematizes matrix algebra, including matrix multiplication, independently of Grassmann's abstract framework.
- 1888: Giuseppe Peano gives the first modern axiomatic definition of a vector space, explicitly building on and clarifying Grassmann's ideas.
- Early 1900s: The coordinate-vector/matrix dictionary becomes standard in linear algebra pedagogy as courses formalize the link between abstract linear maps and matrix computation.
- Mid-20th century: Textbook treatments (Halmos, Hoffman–Kunze, and later Axler) crystallize the modern statement of [Tv]γ=M(T)[v]β[Tv]_\gamma = M(T)[v]_\beta as the bridge theorem between abstract and computational linear algebra.
Why do we need an ordered basis instead of just a basis?

A basis as a set has no notion of "first coordinate" or "second coordinate" — coordinates only make sense once you fix an order for listing the basis vectors, because the coordinate vector's entries are matched positionally to that list. Two different orderings of the same set of basis vectors give two different, though related (permuted), coordinate vectors for the same vv, and correspondingly two different matrices for the same map TT (with permuted rows or columns). The theorem's identity [Tv]γ=M(T)[v]β[Tv]_\gamma = M(T)[v]_\beta is only guaranteed when the ordering used to build the coordinate vectors matches the ordering used to build the matrix's rows and columns, as the counterexample demonstrates explicitly.

Does the theorem require V=WV = W?

No. The theorem is stated for a general linear map T:V→WT : V \to W between possibly different finite-dimensional spaces, with independently chosen ordered bases β\beta for VV and γ\gamma for WW. The matrix M(T)M(T) is then m×nm \times n where m=dim⁡Wm = \dim W and n=dim⁡Vn = \dim V, generally non-square. Example 1 in this article (differentiation from P3P_3 to P2P_2) is exactly such a case, with a 3×43\times 4 matrix, and the identity holds without any modification.

What breaks if VV is infinite-dimensional?

The construction of a coordinate vector as a finite column, and of M(T)M(T) as a finite array, relies on a finite ordered basis so that every vector is a finite linear combination with a fixed, bounded list of coefficients. In infinite dimensions, one can still define coordinates relative to a Hamel basis, but the "matrix" would need infinitely many rows or columns, and multiplying an infinite matrix by an infinite coordinate vector requires infinite sums, which may not converge without extra structure (like a topology or an inner product, as in Hilbert space theory). The clean finite-dimensional identity proved here does not directly generalize without such additional machinery.

Is the correspondence T↦M(T)T \mapsto M(T) itself linear and bijective, like the coordinate map is?

Yes — this is the natural next fact built on top of this theorem. Fixing β\beta and γ\gamma, one shows M(T1+T2)=M(T1)+M(T2)M(T_1+T_2) = M(T_1)+M(T_2) and M(cT)=cM(T)M(cT) = cM(T) by checking columns, and one shows T↦M(T)T \mapsto M(T) is a bijection from L(V,W)\mathcal{L}(V,W) onto Fm×n\mathbb{F}^{m\times n} by explicitly inverting the construction (any matrix defines a unique linear map via the coordinate identity). This upgrades the present pointwise identity into a full isomorphism of vector spaces, which is exactly the content of The Space L(V,W) as Matrices.

How does this identity relate to matrix multiplication representing composition of maps?

Applying the theorem twice, once for S:U→VS : U \to V and once for T:V→WT: V \to W, shows that [T(S(u))]γ=M(T)(M(S)[u]α)[T(S(u))]_\gamma = M(T)\big(M(S)[u]_\alpha\big) for every uu, as derived in the Corollary above. Separately, one shows M(T)(M(S)x)=(M(T)M(S))xM(T)\big(M(S)x\big) = \big(M(T)M(S)\big)x for every column xx, which is just associativity of matrix multiplication baked into its definition. Comparing, M(T∘S)=M(T)M(S)M(T\circ S) = M(T) M(S) follows because both matrices send every coordinate vector [u]α[u]_\alpha to the same output and the coordinate map is a bijection (so matrices agreeing on all coordinate vectors of a spanning-and-independent set of columns, namely all of Fn\mathbb{F}^n, must be equal). This is exactly the substance of Matrix Multiplication as Composition.

Can I use this theorem with the standard basis only, and ignore ordered/non-standard bases?

You certainly can restrict attention to the standard basis of Fn\mathbb{F}^n for computational purposes, and in that case coordinate vectors equal the vectors themselves. But the power of the theorem is its validity for any ordered basis, including highly non-standard ones (as in Example 2, with fractional coordinates and a skewed basis). Many applications — change of basis, diagonalization, choosing an eigenbasis, finite element bases — crucially rely on using a non-standard basis where the matrix of TT becomes simple (e.g., diagonal), and the theorem guarantees the same coordinate-matrix identity holds in that basis too.

What is the practical shortcut for remembering the identity's direction?

Read [T(v)]γ=M(T)[v]β[T(v)]_\gamma = M(T)[v]_\beta left to right: "coordinates of the image, in the codomain's basis, equal the matrix times coordinates of the input, in the domain's basis." The matrix always eats a domain-basis coordinate vector and produces a codomain-basis coordinate vector, matching its shape m×nm \times n with n=dim⁡(domain)n = \dim(\text{domain}) and m=dim⁡(codomain)m = \dim(\text{codomain}). If your matrix multiplication doesn't type-check dimensionally, you have mismatched a basis to the wrong side.

Before this

Where this leads