The two linearity conditions — preservation of addition and scalar multiplication — are the gateway to the entire theory. Proving them unlocks every tool of linear algebra; failing them shuts the door. This page covers how to verify linearity, what consequences follow from it, and how linear transformations interact through composition and inversion.
The Two Linearity Conditions
A function T:V→W is linear if it satisfies two conditions for all u,v∈V and all scalars c∈R:
Additivity: T(u+v)=T(u)+T(v)
Homogeneity: T(cv)=cT(v)
Both conditions must hold for every choice of vectors and scalars — not just for specific examples. Verifying linearity on a few particular vectors does not suffice. Disproving linearity requires only a single counterexample: one pair of vectors where additivity fails, or one scalar where homogeneity fails.
The One-Step Test
The two conditions can be combined into a single check. T is linear if and only if
T(cu+dv)=cT(u)+dT(v)
for all vectors u,v∈V and all scalars c,d∈R. Setting c=d=1 recovers additivity. Setting v=0 recovers homogeneity. So the combined condition implies both individual conditions, and the converse is straightforward.
The one-step test extends to arbitrary linear combinations: if T is linear, then
Linear transformations commute with linear combinations of any length. This is the operational content of linearity — it says that T can be "pulled through" any finite sum of scaled vectors.
Proving Linearity
To prove a given function T is linear, take arbitrary vectors u,v and arbitrary scalars c,d. Compute T(cu+dv) using the formula that defines T, then simplify and show the result equals cT(u)+dT(v).
Worked Example
Let T:R3→R2 be defined by T(x,y,z)=(2x−y,x+3z). Take u=(u1,u2,u3) and v=(v1,v2,v3):
Linearity holds. The key observation is that every component of T is a linear expression in the input coordinates — no products, powers, or constant terms appear.
Disproving Linearity
To show T is not linear, find a single concrete counterexample.
The fastest first check is T(0)=0. Every linear transformation sends 0 to 0, so if T(0)=0, the function is immediately disqualified. This eliminates translations (T(v)=v+b), constant functions (T(v)=c), and any formula with a constant term.
If T(0)=0 passes, test additivity or homogeneity with specific inputs. For T(x,y)=(x2,y): T(1,0)+T(1,0)=(1,0)+(1,0)=(2,0), but T((1,0)+(1,0))=T(2,0)=(4,0)=(2,0). One failure is enough.
For T(x)=x+1: T(0)=1=0. Done — no need to check anything else.
Products of unknowns, powers higher than 1, absolute values, square roots, and constant terms all break linearity. If the formula for T involves any of these, the function is almost certainly not linear (and a counterexample is usually easy to construct).
Consequences of Linearity
Once linearity is established, several useful facts follow automatically.
T(0)=0: the zero vector always maps to the zero vector. Proof: T(0)=T(0⋅v)=0⋅T(v)=0.
T(−v)=−T(v): negation is preserved. Proof: T(−v)=T((−1)v)=(−1)T(v)=−T(v).
T(u−v)=T(u)−T(v): subtraction is preserved. Follows from combining additivity with the negation result.
If {v1,…,vk} is linearly dependent, then {T(v1),…,T(vk)} is also dependent. Linear transformations can destroy independence (by collapsing vectors to zero) but cannot create it. A nonzero dependence relation ∑civi=0 maps to ∑ciT(vi)=0 with the same coefficients.
These are not additional axioms — they are free consequences of the two linearity conditions.
Composition
If T:U→V and S:V→W are both linear, the composition S∘T:U→W defined by (S∘T)(u)=S(T(u)) is also linear:
The first equality applies T's linearity inside S; the second applies S's linearity.
Composition is associative: (R∘S)∘T=R∘(S∘T). It is not commutative: S∘T and T∘S are generally different (and may not even have compatible domains).
When both transformations are represented by matrices, composition corresponds to matrix multiplication: the matrix of S∘T is the product of the matrix of S with the matrix of T. This is the fundamental reason matrix multiplication is defined as it is — it encodes the composition of the maps the matrices represent.
Invertibility
A linear transformation T:V→W is invertible if there exists a linear map T−1:W→V satisfying T−1∘T=IV and T∘T−1=IW.
The inverse, when it exists, is itself linear. If T−1(w1)=v1 and T−1(w2)=v2, then T(cv1+dv2)=cT(v1)+dT(v2)=cw1+dw2, so T−1(cw1+dw2)=cv1+dv2=cT−1(w1)+dT−1(w2).
T is invertible if and only if it is bijective: injective (trivial kernel) and surjective (image equals W). When dim(V)=dim(W), these two conditions are equivalent — checking either one establishes both. For matrix transformations, invertibility of T corresponds to invertibility of the matrix A.
An invertible linear transformation between spaces of the same dimension is called an isomorphism. It establishes that the two spaces are structurally identical as vector spaces.
Sums and Scalar Multiples of Transformations
If S,T:V→W are both linear, new transformations can be built from them.
The sum (S+T)(v)=S(v)+T(v) is linear: (S+T)(cu+dv)=S(cu+dv)+T(cu+dv)=cS(u)+dS(v)+cT(u)+dT(v)=c(S+T)(u)+d(S+T)(v).
The scalar multiple (cT)(v)=cT(v) is linear by a similar argument.
These operations give the set of all linear transformations from V to W the structure of a vector space, denoted L(V,W). The zero element is the zero transformation, and the additive inverse of T is (−1)T. When V and W are finite-dimensional, this space has dimension dim(V)⋅dim(W), matching the number of entries in the representing matrix.
Common Pitfalls
Verifying linearity on a handful of specific vectors does not prove linearity. The conditions must hold for all vectors and all scalars simultaneously. A function can satisfy T(u+v)=T(u)+T(v) for certain pairs while failing for others.
The condition T(0)=0 is necessary but not sufficient. Passing it eliminates many non-examples quickly, but a function with T(0)=0 can still fail additivity or homogeneity. The function T(x)=x∣x∣ satisfies T(0)=0 but is not linear.
A component-by-component check works for maps defined by explicit formulas. If T(x1,…,xn)=(f1(x),…,fm(x)), then T is linear if and only if each component function fi is a linear expression in x1,…,xn — meaning fi=ai1x1+ai2x2+⋯+ainxn with no constant term. Any constant, product, power, or nonlinear function appearing in any component disqualifies the entire map.