Linear Maps Determined by Action on a Basis

Suppose you want to build a linear map from the plane to three-dimensional space. Must you specify where every vector goes? That would be an infinite amount of data. The remarkable answer is no. You only ever need to specify where the vectors of a single basis go. The rest of the map is forced. This economy is one of the most useful facts in linear algebra. It converts questions about linear maps — infinite objects defined on infinite sets — into questions about finite lists of vectors. This principle explains why linear algebra is computable at all. Fix a basis of the domain first. Then a linear map between finite-dimensional spaces is completely captured by a finite table of numbers. These numbers are the coordinates of the images of the basis vectors. This is the seed from which the entire theory of matrices grows. Before writing down a single matrix, however, one needs the underlying theorem. That theorem states: assigning arbitrary target vectors to a basis always produces exactly one linear map extending that assignment. This holds no matter what those target vectors are. The result has two halves, and both matter. Existence: no matter how wildly or arbitrarily you choose images for the basis vectors, a linear map always realizes that assignment. They need not be independent, need not span, and can even repeat. Uniqueness: there is never more than one such map. Together these say that a linear map is the same data as a function from a basis to the target space. The map and the assignment determine each other completely. This article proves that theorem carefully, starting from the definitions of linear map, basis, and coordinates. It builds the intuition behind it. It also shows how the theorem is used in two genuinely different flavored examples. It sets up the next natural steps in the subject. One step is encoding the assignment as a literal array of numbers — The Matrix of a Linear Map. Another is using this freedom of assignment to characterize when two spaces are structurally identical, via Isomorphisms and Isomorphic Vector Spaces.

Key takeaways
  • A linear map out of a finite-dimensional space is pinned down entirely by its values on one basis.
  • You may assign the images of basis vectors arbitrarily — any vectors at all, in any target space — and a linear map always exists extending that assignment.
  • That extension is unique: two linear maps agreeing on a basis are identical everywhere.
  • The proof rests on writing every domain vector as a unique linear combination of the basis (coordinates).
  • This is the theoretical root of representing linear maps by matrices.
  • Key idea: specifying a linear map is exactly the same thing as specifying a function on a basis — no more, no less.

Where this lives today

This theorem is the quiet engine behind representing linear maps as matrices: once a basis of the domain is fixed, recording the coordinates of T(v1),…,T(vn)T(v_1),\dots,T(v_n) in a chosen basis of the codomain produces the matrix of TT, and the theorem guarantees this recording loses no information and can always be inverted to reconstruct TT — the subject of The Matrix of a Linear Map. In numerical linear algebra and computer graphics, transformations (rotations, projections, shears) are specified by giving the images of a handful of basis vectors — three columns for a 3×33\times 3 transform — precisely because the theorem guarantees this finite data is equivalent to specifying the transformation on every point in space. In quantum mechanics, linear operators on state spaces are routinely defined by declaring their action on basis states (energy eigenstates, spin states) and then extended linearly to superpositions; this is a direct, everyday invocation of the theorem. In machine learning, a fully connected linear layer is defined by its weight matrix, whose columns are exactly the images of the standard basis vectors under the layer's linear map. In abstract algebra and the study of Isomorphisms and Isomorphic Vector Spaces, the theorem is the tool used to construct explicit isomorphisms between spaces of the same dimension by matching up basis vectors.

Ask the tutor: If I only know TT on a linearly independent set that doesn't span VV, exactly how many different linear extensions of TT to all of VV are possible, and how would I count or parametrize them?

DefinitionLinear map, basis, and the extension problem

Let VV and WW be vector spaces over the same field F\mathbb{F}. A function T:V→WT: V \to W is a linear map if for all u,v∈Vu, v \in V and all scalars λ∈F\lambda \in \mathbb{F}:

T(u+v)=T(u)+T(v),T(λv)=λT(v).T(u+v) = T(u) + T(v), \qquad T(\lambda v) = \lambda T(v).

A list of vectors (v1,…,vn)(v_1, \dots, v_n) in VV is a basis of VV if it is linearly independent and spans VV; equivalently (and this is the fact we use below), every v∈Vv \in V can be written as

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

for a unique choice of scalars a1,…,an∈Fa_1, \dots, a_n \in \mathbb{F}. Given a basis (v1,…,vn)(v_1, \dots, v_n) of VV and an arbitrary list of vectors (w1,…,wn)(w_1, \dots, w_n) in WW (no conditions on the wiw_i at all), the extension problem asks: does there exist a linear map T:V→WT: V \to W with

T(vi)=wifor all i=1,…,n,T(v_i) = w_i \quad \text{for all } i = 1, \dots, n,

and if so, how many such maps are there?

Intuition

Think of a basis as a minimal set of "control knobs" for the vector space. Every vector in VV is some specific combination of the basis vectors — a unique recipe of coordinates — so if you know how a linear map treats each basis vector, linearity forces the map to treat every combination the same way it treats the ingredients: additively and homogeneously. There is no freedom left over once the basis vectors' images are fixed, because every other vector's image is computed from them by a fixed formula. That is the uniqueness half. The existence half is the surprising part: you are allowed to send the basis vectors anywhere, with total disregard for whether the images are independent, whether they span, or whether they are even distinct. The formula T(a1v1+⋯+anvn)=a1w1+⋯+anwnT(a_1v_1+\cdots+a_nv_n) = a_1w_1+\cdots+a_nw_n still defines a bona fide linear map, because the coordinates aia_i of each vv are unique — there's no ambiguity in the recipe, so no ambiguity in the output. Independence of the basis is precisely what guarantees this recipe is well-defined; if the spanning list were not independent, the same vector could be written two different ways, and the formula could try to assign it two different images at once. Picture V=R2V = \mathbb{R}^2 with basis e1,e2e_1, e_2. Choose any two vectors w1,w2∈R3w_1, w_2 \in \mathbb{R}^3, even w1=w2=0w_1 = w_2 = 0, or w1,w2w_1, w_2 linearly dependent. There is still exactly one linear map T:R2→R3T:\mathbb{R}^2 \to \mathbb{R}^3 sending e1↦w1e_1 \mapsto w_1, e2↦w2e_2 \mapsto w_2: namely T(a,b)=aw1+bw2T(a,b) = aw_1 + bw_2.

Figure
The recipe: coordinates in the domain force the output in the codomain
Domain V (basis v₁, v₂) v₁ v₂ v = 3v₁ + 2v₂ Codomain W (arbitrary targets) w₁ = T(v₁) w₂ = T(v₂) T(v) = 3w₁ + 2w₂ Same coordinates (3, 2) carry over — this is the whole proof.
TheoremLinear maps are determined by action on a basis

Let VV and WW be vector spaces over a field F\mathbb{F}, and suppose VV is finite-dimensional with basis (v1,…,vn)(v_1, \dots, v_n). Let (w1,…,wn)(w_1, \dots, w_n) be any list of nn vectors in WW (not assumed independent, spanning, or distinct). Then: 1. (Existence) There exists a linear map T:V→WT: V \to W such that T(vi)=wiT(v_i) = w_i for every i=1,…,ni = 1,\dots,n. 2. (Uniqueness) This map is unique: if S:V→WS: V \to W is linear and S(vi)=wiS(v_i) = w_i for every ii, then S=TS = T. Equivalently: for a fixed basis (v1,…,vn)(v_1,\dots,v_n) of VV, the map

Φ:L(V,W)→Wn,Φ(T)=(T(v1),…,T(vn))\Phi : \mathcal{L}(V,W) \to W^n, \qquad \Phi(T) = (T(v_1), \dots, T(v_n))

is a bijection, where L(V,W)\mathcal{L}(V,W) denotes the set of all linear maps from VV to WW.

Proof

We prove existence and uniqueness separately. The technique for existence is direct construction via coordinates; the technique for uniqueness is a linearity unwinding argument.

Setup (coordinates are well-defined). Since (v1,…,vn)(v_1,\dots,v_n) is a basis of VV, it spans VV, so every v∈Vv \in V can be written as

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

for some scalars a1,…,an∈Fa_1,\dots,a_n \in \mathbb{F}. These scalars are unique: if also v=b1v1+⋯+bnvnv = b_1v_1+\cdots+b_nv_n, then subtracting gives

(a1−b1)v1+⋯+(an−bn)vn=0,(a_1-b_1)v_1 + \cdots + (a_n-b_n)v_n = 0,

and linear independence of (v1,…,vn)(v_1,\dots,v_n) forces ai−bi=0a_i - b_i = 0 for every ii, i.e. ai=bia_i = b_i. So each v∈Vv\in V has a well-defined, unique coordinate tuple (a1,…,an)(a_1,\dots,a_n) relative to this basis. Call this tuple coord(v)\mathrm{coord}(v).

Step 1: Existence. Define T:V→WT: V \to W by the formula

T(v):=a1w1+a2w2+⋯+anwn,where (a1,…,an)=coord(v).T(v) := a_1 w_1 + a_2 w_2 + \cdots + a_n w_n, \quad \text{where } (a_1,\dots,a_n) = \mathrm{coord}(v).

This is a legitimate function V→WV \to W precisely because coord(v)\mathrm{coord}(v) is unique for each vv (shown above) — there is no ambiguity in which output T(v)T(v) names. TT hits the basis correctly. Fix j∈{1,…,n}j \in \{1,\dots,n\}. Writing vjv_j itself in the basis, its coordinate tuple is the standard tuple with a 11 in position jj and 00 elsewhere, i.e. vj=0v1+⋯+1vj+⋯+0vnv_j = 0v_1+\cdots+1v_j+\cdots+0v_n; this is a valid representation, and by uniqueness of coordinates it is the representation. Hence

T(vj)=0⋅w1+⋯+1⋅wj+⋯+0⋅wn=wj,T(v_j) = 0\cdot w_1 + \cdots + 1\cdot w_j + \cdots + 0\cdot w_n = w_j,

exactly as required. TT is additive. Let u,v∈Vu, v \in V with coordinate tuples (a1,…,an)(a_1,\dots,a_n) and (b1,…,bn)(b_1,\dots,b_n) respectively, so u=∑iaiviu = \sum_i a_i v_i and v=∑ibiviv = \sum_i b_i v_i. Then

u+v=∑i=1naivi+∑i=1nbivi=∑i=1n(ai+bi)vi,u + v = \sum_{i=1}^n a_i v_i + \sum_{i=1}^n b_i v_i = \sum_{i=1}^n (a_i+b_i) v_i,

using the vector space axioms (commutativity/associativity of addition and distributivity of scalar multiplication over vector addition) to regroup. Since (ai+bi)(a_i+b_i) are scalars and the viv_i form a basis, this is the coordinate representation of u+vu+v (unique, by the setup above), so coord(u+v)=(a1+b1,…,an+bn)\mathrm{coord}(u+v) = (a_1+b_1,\dots,a_n+b_n). Therefore, by definition of TT,

T(u+v)=∑i=1n(ai+bi)wi=∑i=1naiwi+∑i=1nbiwi=T(u)+T(v),T(u+v) = \sum_{i=1}^n (a_i+b_i) w_i = \sum_{i=1}^n a_i w_i + \sum_{i=1}^n b_i w_i = T(u) + T(v),

where the middle equality again uses distributivity and commutativity of addition in WW. TT is homogeneous. Let v∈Vv \in V have coordinates (a1,…,an)(a_1,\dots,a_n), so v=∑iaiviv = \sum_i a_i v_i, and let λ∈F\lambda \in \mathbb{F}. Then

λv=λ∑i=1naivi=∑i=1n(λai)vi,\lambda v = \lambda \sum_{i=1}^n a_i v_i = \sum_{i=1}^n (\lambda a_i) v_i,

using distributivity of scalar multiplication over the sum and associativity of scalar multiplication. By uniqueness of coordinates this is the representation of λv\lambda v, so coord(λv)=(λa1,…,λan)\mathrm{coord}(\lambda v) = (\lambda a_1,\dots,\lambda a_n), and hence

T(λv)=∑i=1n(λai)wi=λ∑i=1naiwi=λT(v).T(\lambda v) = \sum_{i=1}^n (\lambda a_i) w_i = \lambda \sum_{i=1}^n a_i w_i = \lambda T(v).

All three required properties (well-defined function, additivity, homogeneity) hold, so TT is a linear map with T(vi)=wiT(v_i) = w_i for all ii. This proves existence.

Step 2: Uniqueness. Suppose S:V→WS: V \to W is linear and satisfies S(vi)=wiS(v_i) = w_i for all i=1,…,ni = 1,\dots,n. Let v∈Vv \in V be arbitrary, with (unique) coordinates (a1,…,an)(a_1,\dots,a_n) so that v=a1v1+⋯+anvnv = a_1v_1 + \cdots + a_nv_n. Using linearity of SS — first additivity applied repeatedly to split the nn-term sum, then homogeneity on each term — we get

S(v)=S ⁣(∑i=1naivi)=∑i=1nS(aivi)=∑i=1naiS(vi)=∑i=1naiwi.S(v) = S\!\left(\sum_{i=1}^n a_i v_i\right) = \sum_{i=1}^n S(a_i v_i) = \sum_{i=1}^n a_i S(v_i) = \sum_{i=1}^n a_i w_i.

(The first equality of sums is proved by induction on nn from the two-term additivity axiom S(x+y)=S(x)+S(y)S(x+y)=S(x)+S(y), applied n−1n-1 times to peel off one summand at a time.) But the right-hand side, ∑iaiwi\sum_i a_iw_i, is exactly T(v)T(v) by the defining formula for TT constructed in Step 1. Hence S(v)=T(v)S(v) = T(v) for every v∈Vv \in V, i.e. S=TS = T as functions. This shows any linear map agreeing with the assignment vi↦wiv_i \mapsto w_i must equal the specific TT built above; combined with Step 1's existence of TT, this proves the assignment determines a unique linear map.

Step 3: The bijection statement. Steps 1–2 show that for every (w1,…,wn)∈Wn(w_1,\dots,w_n) \in W^n there is exactly one T∈L(V,W)T \in \mathcal{L}(V,W) with Φ(T)=(T(v1),…,T(vn))=(w1,…,wn)\Phi(T) = (T(v_1),\dots,T(v_n)) = (w_1,\dots,w_n). This says precisely that Φ:L(V,W)→Wn\Phi: \mathcal{L}(V,W) \to W^n is both surjective (existence, Step 1) and injective (uniqueness, Step 2 shows Φ(T)=Φ(S)⇒T=S\Phi(T)=\Phi(S) \Rightarrow T=S), hence bijective. ■\blacksquare

∎
CorollaryTwo linear maps agreeing on a basis are equal

If S,T:V→WS, T : V \to W are linear, VV finite-dimensional with basis (v1,…,vn)(v_1,\dots,v_n), and S(vi)=T(vi)S(v_i) = T(v_i) for all ii, then S=TS = T as functions on all of VV. (This is exactly uniqueness in the theorem, applied to the common assignment wi:=S(vi)=T(vi)w_i := S(v_i) = T(v_i).)

Worked examples

Example 1 — a genuinely "wild" assignment. Let V=R3V = \mathbb{R}^3 with the standard basis (e1,e2,e3)(e_1,e_2,e_3), and let W=R2W = \mathbb{R}^2. Choose the assignment

e1↦w1=(1,4),e2↦w2=(2,8),e3↦w3=(−3,0).e_1 \mapsto w_1 = (1, 4), \qquad e_2 \mapsto w_2 = (2, 8), \qquad e_3 \mapsto w_3 = (-3, 0).

Notice w2=2w1w_2 = 2w_1, so (w1,w2,w3)(w_1,w_2,w_3) is linearly dependent and does not span R2\mathbb{R}^2 on its own in an independent way — the theorem places no restriction on the wiw_i at all, so this is perfectly admissible. By the theorem there is exactly one linear map T:R3→R2T:\mathbb{R}^3\to\mathbb{R}^2 realizing this, given explicitly by the formula from the proof:

T(a,b,c)=a w1+b w2+c w3=a(1,4)+b(2,8)+c(−3,0)=(a+2b−3c, 4a+8b).T(a,b,c) = a\,w_1 + b\,w_2 + c\,w_3 = a(1,4) + b(2,8) + c(-3,0) = (a+2b-3c,\ 4a+8b).

Check it reproduces the assignment: T(1,0,0)=(1,4)=w1T(1,0,0) = (1,4) = w_1 ✓, T(0,1,0)=(2,8)=w2T(0,1,0) = (2,8) = w_2 ✓, T(0,0,1)=(−3,0)=w3T(0,0,1) = (-3,0)=w_3 ✓. Check linearity directly on a specific pair to see the mechanism at work: take u=(1,−1,2)u = (1,-1,2) and v=(3,0,−1)v=(3,0,-1), λ=5\lambda = 5.

T(u)=(1−2−6, 4−8)=(−7,−4),T(v)=(3+0+3, 12+0)=(6,12).T(u) = (1 - 2 - 6,\ 4-8) = (-7,-4), \qquad T(v) = (3+0+3,\ 12+0) = (6,12).
u+v=(4,−1,1):T(u+v)=(4−2−3, 16−8)=(−1,8)=(−7,−4)+(6,12)=T(u)+T(v).✓u+v = (4,-1,1): \quad T(u+v) = (4 - 2 -3,\ 16-8) = (-1, 8) = (-7,-4)+(6,12) = T(u)+T(v). \checkmark
5v=(15,0,−5):T(5v)=(15+0+15, 60+0)=(30,60)=5(6,12)=5T(v).✓5v = (15,0,-5): \quad T(5v) = (15+0+15,\ 60+0) = (30,60) = 5(6,12) = 5T(v). \checkmark

The theorem guarantees this is not a coincidence for these particular vectors: it must hold for all u,v,λu,v,\lambda, because TT was built by the coordinate formula, and the proof shows that formula is always additive and homogeneous, regardless of how dependent or degenerate the chosen images wiw_i are.

Example 2 — a non-standard domain basis, testing uniqueness. Let V=R2V = \mathbb{R}^2, but use the non-standard basis (v1,v2)=((1,1), (1,−1))(v_1,v_2) = \big((1,1),\,(1,-1)\big) — one should check this is indeed a basis: it is independent since (1,1)(1,1) and (1,−1)(1,-1) are not scalar multiples of each other, and any two independent vectors in a 22-dimensional space automatically span it. Let W=R3W = \mathbb{R}^3 and assign

T(v1)=(2,0,1),T(v2)=(0,3,−1).T(v_1) = (2,0,1), \qquad T(v_2) = (0,3,-1).

To find T(x,y)T(x,y) for a general vector, first find the coordinates of (x,y)(x,y) relative to (v1,v2)(v_1,v_2): solve (x,y)=a(1,1)+b(1,−1)=(a+b, a−b)(x,y) = a(1,1)+b(1,-1) = (a+b,\ a-b), giving a=x+y2a = \frac{x+y}{2}, b=x−y2b = \frac{x-y}{2}. Then by the formula in the proof,

T(x,y)=a T(v1)+b T(v2)=x+y2(2,0,1)+x−y2(0,3,−1).T(x,y) = a\,T(v_1) + b\,T(v_2) = \frac{x+y}{2}(2,0,1) + \frac{x-y}{2}(0,3,-1).

Expanding,

T(x,y)=(x+y, 3(x−y)2, x+y2−x−y2)=(x+y, 32(x−y), y).T(x,y) = \left(x+y,\ \frac{3(x-y)}{2},\ \frac{x+y}{2} - \frac{x-y}{2}\right) = \left(x+y,\ \tfrac{3}{2}(x-y),\ y\right).

Sanity check on the basis vectors: T(1,1)=(2,0,1)T(1,1) = (2, 0, 1) ✓ matches T(v1)T(v_1); T(1,−1)=(0,3,−1)T(1,-1) = (0, 3, -1) ✓ matches T(v2)T(v_2). Now test uniqueness concretely: suppose someone proposes a different rule S(x,y)=(x+y,32(x−y),y+(x−y)(x+y−2))S(x,y) = (x+y, \frac32(x-y), y + (x-y)(x+y-2)) hoping it also sends v1↦(2,0,1)v_1\mapsto(2,0,1) and v2↦(0,3,−1)v_2 \mapsto (0,3,-1). Check: at (1,1)(1,1), the extra term is (1−1)(1+1−2)=0(1-1)(1+1-2) = 0, so S(1,1)=T(1,1)S(1,1)=T(1,1); at (1,−1)(1,-1), the extra term is (1−(−1))(1+(−1)−2)=(2)(−2)=−4≠0(1-(-1))(1+(-1)-2) = (2)(-2) = -4 \ne 0, so S(1,−1)=(0,3,−1−4)=(0,3,−5)≠(0,3,−1)S(1,-1) = (0,3,-1-4) = (0,3,-5) \ne (0,3,-1). So SS actually fails to extend the assignment — and indeed SS is not linear at all (the extra term is quadratic), consistent with the theorem: no function other than the specific linear TT above can extend this assignment linearly, and this attempted "different" extension simply isn't linear.

CounterexampleWhy the domain list must be a basis — dependence breaks well-definedness

The theorem needs (v1,…,vn)(v_1,\dots,v_n) to be a basis of VV, not merely a spanning list. Drop independence and existence itself can fail — the assignment becomes contradictory. Let V=R2V = \mathbb{R}^2 and consider the spanning-but-dependent list (v1,v2,v3)=((1,0), (0,1), (1,1))(v_1, v_2, v_3) = \big((1,0),\,(0,1),\,(1,1)\big). This list spans R2\mathbb{R}^2 but is not independent, since v3=v1+v2v_3 = v_1+v_2. Attempt the assignment

T(v1)=(1,0),T(v2)=(0,1),T(v3)=(5,5).T(v_1) = (1,0), \qquad T(v_2) = (0,1), \qquad T(v_3) = (5,5).

If a linear TT satisfying all three existed, linearity applied to the dependence relation v3=v1+v2v_3 = v_1+v_2 would force

T(v3)=T(v1+v2)=T(v1)+T(v2)=(1,0)+(0,1)=(1,1).T(v_3) = T(v_1+v_2) = T(v_1)+T(v_2) = (1,0)+(0,1) = (1,1).

But the assignment demands T(v3)=(5,5)≠(1,1)T(v_3) = (5,5) \ne (1,1). Contradiction — no linear map can satisfy this assignment. The existence half of the theorem fails completely once the spanning list is allowed to be dependent, because dependent lists impose hidden consistency conditions on any admissible assignment (here, the assigned images must satisfy the same relation the vectors do), and an arbitrary assignment generically violates them. Contrast this with a basis: because a basis has coordinates that are always unique, there is no relation like v3=v1+v2v_3 = v_1+v_2 lurking among basis vectors to violate, so every assignment is automatically consistent. This is exactly why the theorem restricts to bases.

Scope, edge cases & common traps

SituationWhat happens
VV infinite-dimensional, (vi)(v_i) a (Hamel) basisTheorem still holds verbatim; every vv still has a unique finite coordinate combination
Domain list spans but is dependentExistence can fail (see counterexample); uniqueness is also moot since no valid TT may even exist
Domain list independent but does not spanThe "assignment" only pins down TT on the subspace spanned by the viv_i; TT is not unique on all of VV — many linear extensions to the rest of VV are possible
Images (w1,…,wn)(w_1,\dots,w_n) chosen dependent/repeated/zeroNo problem: existence and uniqueness hold regardless, exactly as the theorem states
W={0}W = \{0\}The unique linear map is the zero map, consistent with the theorem (wi=0w_i = 0 for all ii forced)

The mathematician behind it

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

Grassmann, a schoolteacher and self-taught mathematician in Stettin, developed in his 1844 Die lineale Ausdehnungslehre an abstract theory of vector spaces, linear dependence, dimension, and linear combinations decades before these ideas became standard. His work was so far ahead of the notation and expectations of his time that it was largely ignored by his contemporaries; only later mathematicians, including Peano who gave the first modern axiomatic definition of a vector space, recognized how much of what we now call linear algebra — bases, dimension, and the extension property behind this very theorem — Grassmann had already anticipated.

History
- 1844 — Hermann Grassmann publishes Die Ausdehnungslehre, introducing linear independence, span, and dimension in essentially modern generality.
- 1888 — Giuseppe Peano gives the first fully axiomatic definition of a real vector space, making statements like the extension theorem precisely provable.
- 1900s — David Hilbert and Erhard Schmidt develop coordinate and basis methods for infinite-dimensional (Hilbert) spaces, extending these ideas beyond finite dimension.
- 1920s–1930s — Formal linear algebra with matrices, linear maps, and bases becomes standard university material, with the extension theorem as a foundational lemma.
- Present — The theorem underlies every construction of a matrix representation of a linear map in modern linear algebra courses and computational libraries.
Why does the domain list need to be a basis and not just any spanning set?

A spanning-but-dependent list has hidden linear relations among its vectors, such as v3=v1+v2v_3 = v_1+v_2. Linearity of any map TT automatically forces the same relation among the images: T(v3)T(v_3) must equal T(v1)+T(v2)T(v_1)+T(v_2). If you assign images that violate this forced relation, no linear map can possibly satisfy your assignment, so existence fails. A basis has no such relations — independence guarantees every vector has a unique coordinate representation — so there are no hidden constraints to violate, and any assignment whatsoever is automatically consistent. That is exactly why the theorem requires a basis, not merely a spanning set.

Do the images w1,…,wnw_1,\dots,w_n need to be linearly independent or span WW?

No — this is one of the most important and surprising parts of the theorem. The images can be zero, repeated, or all multiples of each other; none of this affects existence or uniqueness of the linear map TT with T(vi)=wiT(v_i)=w_i. What the images do determine is the properties of TT itself: if the images are independent, TT turns out to be injective; if they span WW, TT turns out to be surjective. But none of that is required for TT to exist and be unique in the first place.

Does this theorem work for infinite-dimensional vector spaces?

Yes, provided "basis" is understood in the Hamel sense: a linearly independent set such that every vector is a finite linear combination of basis elements. Every vector space has such a basis (a consequence of the Axiom of Choice, via Zorn's Lemma), and the same coordinate argument used in the proof works verbatim, because coordinates are always finite sums even in infinite dimension. The subtlety in infinite dimensions lies elsewhere — such bases are typically impossible to write down explicitly — but the logical structure of the existence/uniqueness proof is unchanged.

How does this relate to writing a linear map as a matrix?

Once you fix a basis (v1,…,vn)(v_1,\dots,v_n) of VV and a basis of WW, the theorem says a linear map TT is completely and losslessly encoded by the list (T(v1),…,T(vn))(T(v_1),\dots,T(v_n)). Writing each T(vj)T(v_j) in coordinates relative to the chosen basis of WW and stacking these coordinate columns side by side produces exactly the matrix of TT. The theorem is precisely the statement that this encoding is a bijection between linear maps and matrices (of the appropriate size), which is why matrix algebra faithfully mirrors the algebra of linear maps.

Can the same reasoning define a map on part of a basis and extend it however I like elsewhere?

Yes, and this is a useful trick. If (v1,…,vk)(v_1,\dots,v_k) is a linearly independent list that is not a full basis, you can extend it to a full basis (v1,…,vk,vk+1,…,vn)(v_1,\dots,v_k,v_{k+1},\dots,v_n) (a standard fact about extending independent lists in finite-dimensional spaces), assign the first kk images however you like to satisfy some condition, and then assign the remaining vk+1,…,vnv_{k+1},\dots,v_n completely freely, even arbitrarily. The theorem guarantees a single linear map realizes all of these choices simultaneously, which is exactly why non-uniqueness appears whenever your original list under-spans VV.

Is the map TT constructed in the proof actually computable, or just theoretically guaranteed to exist?

It is fully computable, and the proof is constructive, not just an abstract existence argument. Given any vector vv, you find its unique coordinates a1,…,ana_1,\dots,a_n relative to the basis (a linear algebra computation, e.g. solving a linear system), and then T(v)T(v) is simply a1w1+⋯+anwna_1w_1+\cdots+a_nw_n. Both worked examples in this article carry out exactly this recipe explicitly, producing closed-form formulas for TT.

What goes wrong if I only check that TT is linear on the basis vectors themselves, instead of on all of VV?

Checking T(vi+vj)=T(vi)+T(vj)T(v_i+v_j) = T(v_i)+T(v_j) for basis vectors alone says nothing about linearity on general vectors like 2v1−3v2+v52v_1 - 3v_2 + v_5; linearity is a universally quantified statement over all vectors and scalars in VV, not just over the finitely many basis vectors. The content of the theorem's proof is precisely that verifying compatibility of an assignment on a basis, combined with the coordinate uniqueness of the basis, is enough to force full linearity everywhere — but this requires the construction and proof given above, not merely eyeballing the basis vectors.

Before this

Where this leads