What are the two conditions for a linear transformation?

A function T is linear if it satisfies additivity (T(u + v) = T(u) + T(v)) and homogeneity (T(cv) = cT(v)) for all vectors u, v and all scalars c. Equivalently, T(cu + dv) = cT(u) + dT(v) for all vectors and scalars — the one-step test.

How do you prove a transformation is linear?

Take arbitrary vectors u, v and scalars c, d. Compute T(cu + dv) using T's formula, then simplify and show the result equals cT(u) + dT(v). The key is that every component must be a linear expression in the input coordinates — no constants, products, powers, or nonlinear functions.

How do you disprove linearity?

Find a single counterexample. The fastest first check is T(0) = 0 — every linear map sends zero to zero, so T(0) ≠ 0 immediately disqualifies. If that passes, test additivity or homogeneity with specific inputs. One failure is enough.

Is the composition of two linear transformations linear?

Yes. If T and S are both linear, the composition S ∘ T is linear: (S ∘ T)(cu + dv) = S(cT(u) + dT(v)) = cS(T(u)) + dS(T(v)). When represented by matrices, composition corresponds to matrix multiplication.

When is a linear transformation invertible?

A linear transformation is invertible if and only if it is bijective — injective (trivial kernel) and surjective (image equals the codomain). When domain and codomain have equal dimension, checking either condition establishes both. The inverse is itself linear.

Properties of Linear Transformation

The Two Linearity Conditions

The One-Step Test

Proving Linearity

Disproving Linearity

Consequences of Linearity

Composition

Invertibility

Sums and Scalar Multiples of Transformations

Common Pitfalls

Summary: How Linear Maps Combine

What Linearity Gives You and How to Verify It

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 \to W

is linear if it satisfies two conditions for all

\mathbf{u}, \mathbf{v} \in V

and all scalars

c \in \mathbb{R}

:

Additivity:

T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})

Homogeneity:

T(c\mathbf{v}) = cT(\mathbf{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(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v})

for all vectors

\mathbf{u}, \mathbf{v} \in V

and all scalars

c, d \in \mathbb{R}

. Setting

c = d = 1

recovers additivity. Setting

\mathbf{v} = \mathbf{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

T(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k) = c_1T(\mathbf{v}_1) + c_2T(\mathbf{v}_2) + \cdots + c_kT(\mathbf{v}_k)

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

\mathbf{u}, \mathbf{v}

and arbitrary scalars

c, d

. Compute

T(c\mathbf{u} + d\mathbf{v})

using the formula that defines

T

, then simplify and show the result equals

cT(\mathbf{u}) + dT(\mathbf{v})

Worked Example

Let

T: \mathbb{R}^3 \to \mathbb{R}^2

be defined by

T(x, y, z) = (2x - y, x + 3z)

. Take

\mathbf{u} = (u_1, u_2, u_3)

and

\mathbf{v} = (v_1, v_2, v_3)

T(c\mathbf{u} + d\mathbf{v}) = T(cu_1 + dv_1, \; cu_2 + dv_2, \; cu_3 + dv_3)

= (2(cu_1 + dv_1) - (cu_2 + dv_2), \; (cu_1 + dv_1) + 3(cu_3 + dv_3))

= (c(2u_1 - u_2) + d(2v_1 - v_2), \; c(u_1 + 3u_3) + d(v_1 + 3v_3))

= c(2u_1 - u_2, \; u_1 + 3u_3) + d(2v_1 - v_2, \; v_1 + 3v_3) = cT(\mathbf{u}) + dT(\mathbf{v})

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(\mathbf{0}) = \mathbf{0}

. Every linear transformation sends

\mathbf{0}

\mathbf{0}

, so if

T(\mathbf{0}) \neq \mathbf{0}

, the function is immediately disqualified. This eliminates translations (

T(\mathbf{v}) = \mathbf{v} + \mathbf{b}

), constant functions (

T(\mathbf{v}) = \mathbf{c}

), and any formula with a constant term.

If

T(\mathbf{0}) = \mathbf{0}

passes, test additivity or homogeneity with specific inputs. For

T(x, y) = (x^2, y)

T(1, 0) + T(1, 0) = (1, 0) + (1, 0) = (2, 0)

, but

T((1, 0) + (1, 0)) = T(2, 0) = (4, 0) \neq (2, 0)

. One failure is enough.

For

T(x) = x + 1

T(0) = 1 \neq 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).

Strategy	When to reach for it	What it catches	Example
Check T(0) = 0	first, always — the fastest single test	constant terms, translations, affine offsets	T(x) = x + 1: T(0) = 1 ≠ 0
Scan components for non-linear terms	explicit formula T(x₁,…,xₙ) = (f₁,…,fₘ) given	products, powers > 1, \|·\|, √, constants	T(x, y) = (xy, y): product xy disqualifies
Test additivity with specific u, v	formula passes T(0)=0 but has a power or product	nonlinear interaction between inputs	T(x,y) = (x², y): T(1,0)+T(1,0) = (2,0) ≠ T(2,0) = (4,0)
Test homogeneity with specific c	formula involves \|·\| or odd-degree behavior	nonlinear scaling (often via sign)	T(x) = \|x\|: T(−1·2) = 2 ≠ −1·T(2) = −2

Consequences of Linearity

Once linearity is established, several useful facts follow automatically.

T(\mathbf{0}) = \mathbf{0}

: the zero vector always maps to the zero vector. Proof:

T(\mathbf{0}) = T(0 \cdot \mathbf{v}) = 0 \cdot T(\mathbf{v}) = \mathbf{0}

T(-\mathbf{v}) = -T(\mathbf{v})

: negation is preserved. Proof:

T(-\mathbf{v}) = T((-1)\mathbf{v}) = (-1)T(\mathbf{v}) = -T(\mathbf{v})

T(\mathbf{u} - \mathbf{v}) = T(\mathbf{u}) - T(\mathbf{v})

: subtraction is preserved. Follows from combining additivity with the negation result.

If

\{\mathbf{v}_1, \dots, \mathbf{v}_k\}

is linearly dependent, then

\{T(\mathbf{v}_1), \dots, T(\mathbf{v}_k)\}

is also dependent. Linear transformations can destroy independence (by collapsing vectors to zero) but cannot create it. A nonzero dependence relation

\sum c_i \mathbf{v}_i = \mathbf{0}

maps to

\sum c_i T(\mathbf{v}_i) = \mathbf{0}

with the same coefficients.

These are not additional axioms — they are free consequences of the two linearity conditions.

Identity	What it says	Source
T(0) = 0	the zero vector always maps to the zero vector	homogeneity with c = 0
T(−v) = −T(v)	negation is preserved	homogeneity with c = −1
T(u − v) = T(u) − T(v)	subtraction is preserved	additivity combined with negation identity
T(Σ cᵢ vᵢ) = Σ cᵢ T(vᵢ)	linear combinations of any length are preserved	iterating additivity and homogeneity
Linear dependence is preserved	if {vᵢ} is dependent, then {T(vᵢ)} is dependent (independence may be destroyed but never created)	applying T to a dependence relation Σ cᵢ vᵢ = 0

Composition

T: U \to V

and

S: V \to W

are both linear, the composition

S \circ T: U \to W

defined by

(S \circ T)(\mathbf{u}) = S(T(\mathbf{u}))

is also linear:

(S \circ T)(c\mathbf{u} + d\mathbf{v}) = S(T(c\mathbf{u} + d\mathbf{v})) = S(cT(\mathbf{u}) + dT(\mathbf{v})) = cS(T(\mathbf{u})) + dS(T(\mathbf{v}))

The first equality applies

T

's linearity inside

S

; the second applies

S

's linearity.

Composition is associative:

(R \circ S) \circ T = R \circ (S \circ T)

. It is not commutative:

S \circ T

and

T \circ 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 \circ 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 \to W

is invertible if there exists a linear map

T^{-1}: W \to V

satisfying

T^{-1} \circ T = I_V

and

T \circ T^{-1} = I_W

.

The inverse, when it exists, is itself linear. If

T^{-1}(\mathbf{w}_1) = \mathbf{v}_1

and

T^{-1}(\mathbf{w}_2) = \mathbf{v}_2

, then

T(c\mathbf{v}_1 + d\mathbf{v}_2) = cT(\mathbf{v}_1) + dT(\mathbf{v}_2) = c\mathbf{w}_1 + d\mathbf{w}_2

, so

T^{-1}(c\mathbf{w}_1 + d\mathbf{w}_2) = c\mathbf{v}_1 + d\mathbf{v}_2 = cT^{-1}(\mathbf{w}_1) + dT^{-1}(\mathbf{w}_2)

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

S, T: V \to W

are both linear, new transformations can be built from them.

The sum

(S + T)(\mathbf{v}) = S(\mathbf{v}) + T(\mathbf{v})

is linear:

(S + T)(c\mathbf{u} + d\mathbf{v}) = S(c\mathbf{u} + d\mathbf{v}) + T(c\mathbf{u} + d\mathbf{v}) = cS(\mathbf{u}) + dS(\mathbf{v}) + cT(\mathbf{u}) + dT(\mathbf{v}) = c(S + T)(\mathbf{u}) + d(S + T)(\mathbf{v})

.

The scalar multiple

(cT)(\mathbf{v}) = cT(\mathbf{v})

is linear by a similar argument.

These operations give the set of all linear transformations from

V

W

the structure of a vector space, denoted

\mathcal{L}(V, W)

. The zero element is the zero transformation, and the additive inverse of

T

(-1)T

. When

V

and

W

are finite-dimensional, this space has dimension

\dim(V) \cdot \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(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})

for certain pairs while failing for others.

The condition

T(\mathbf{0}) = \mathbf{0}

is necessary but not sufficient. Passing it eliminates many non-examples quickly, but a function with

T(\mathbf{0}) = \mathbf{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(x_1, \dots, x_n) = (f_1(\mathbf{x}), \dots, f_m(\mathbf{x}))

, then

T

is linear if and only if each component function

f_i

is a linear expression in

x_1, \dots, x_n

— meaning

f_i = a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n

with no constant term. Any constant, product, power, or nonlinear function appearing in any component disqualifies the entire map.

Summary: How Linear Maps Combine

The four consequences-cluster sections above isolate distinct ways that linearity propagates through operations on the transformations themselves. The table below collects them in one place: composition produces another linear map, inversion preserves linearity, and the pointwise sum and scalar multiple of linear maps are themselves linear — giving the full set of all linear transformations from V to W the structure of a vector space in its own right.

Operation on linear maps	Definition	Result is linear?	Structural payoff
Composition S ∘ T	(S ∘ T)(u) = S(T(u))	yes — apply T's linearity inside S's	corresponds to matrix multiplication; associative, not commutative
Inverse T⁻¹	T⁻¹ ∘ T = I, T ∘ T⁻¹ = I (when T is bijective)	yes — the inverse of a linear bijection is linear	an invertible T between equal-dim spaces is an isomorphism
Sum S + T	(S + T)(v) = S(v) + T(v)	yes — pointwise sum of linear maps is linear	gives 𝓛(V, W) its addition
Scalar multiple cT	(cT)(v) = c · T(v)	yes — scaling a linear map keeps it linear	gives 𝓛(V, W) its scalar multiplication
The whole set 𝓛(V, W)	all linear maps V → W with + and scalar mult.	— it is itself a vector space	dim 𝓛(V, W) = dim(V) · dim(W) (finite-dimensional case)