Linear Transformation Properties

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})

Fact: Linear over

\mathbb{R}

, not over

\mathbb{C}

With complex scalars,

T(z) = \bar{z}

on

\mathbb{C}

is additive and keeps every real multiple, yet

T(i \cdot 1) = -i

while

i\,T(1) = i

. With only real scalars, the same map is the reflection

(a, b) \mapsto (a, -b)

of the plane, and it is linear.

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.

Which numbers count as scalars is part of the definition. On this page the scalars are real numbers, but vector spaces with complex scalars are just as common, and there the

c

in the homogeneity condition can be

i

or any other complex number. Additivity does not mention scalars at all, so it reads the same in both settings; homogeneity has to be checked for the larger set of scalars.

Linearity Notation

Notation

Linearity Notation

The definition of a linear map is written with one set of symbols standing for two sets of operations. Four marks decide what the equation is actually claiming.

The bracket

[T]

that turns a map into an array, the chaining rule for compositions and the reason the rightmost factor acts first are set out at matrix representation. Coordinate brackets and the script letters used for bases belong to basis, and

\dim

with its field subscript to dimension.

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

T of c u plus d v equals c times T of u plus d times T of v

Four operations appear here and they are not the same four. The addition and the scaling on the left are performed in

V

, before

T

is applied; the ones on the right are performed in

W

, after. One glyph is reused for both, and the equation is precisely the claim that the two structures agree.

CasesSetting

c = d = 1

leaves additivity; setting

\mathbf{v} = \mathbf{0}

leaves homogeneity. The single line is not a shorthand for the pair — it implies both, and both together imply it.

Also writtenSplit into two displayed conditions when each is being checked separately, which is the usual form in a proof and hides that one equation carries both.

Do not confuseReading the line as an identity to be simplified. It is a condition to be tested: a hypothesis about a particular

T

, true for some functions and false for most, and nothing about the symbols guarantees it.

Same glyph elsewhereThe plus sign inside

T(\cdot)

and the plus sign outside it denote different operations in different spaces throughout this page. Nothing marks the difference except position.

T(\mathbf{0}_V) = \mathbf{0}_W, \qquad T^{-1} \circ T = I_V

T of the zero of V is the zero of W; T inverse after T is the identity on V

The subscripts name which space the object belongs to. There are two zero vectors here and two identity maps, and without the labels the first equation reads as though nothing happened and the second as though one identity were being used twice.

CasesWhen

V

and

W

are the same space the subscripts are dropped and nothing is lost. They earn their place exactly when a map crosses between spaces, which is also when the statement stops being obvious.

Also writtenWritten

\mathbf{0}

and

I

unadorned in most sources, leaving the reader to supply the space from the surrounding line.

Do not confuseTaking

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

for part of the definition. It is a consequence, and it works in one direction only: failing it disproves linearity outright, while passing it proves nothing —

T(x) = x|x|

sends

0

to

0

and is not linear.

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

T of the triple x, y, z is the pair two x minus y, x plus three z

The parentheses after

T

are doing two jobs at once. Written strictly the line would begin

T\big((x, y, z)\big)

, with the inner pair building a vector and the outer pair applying the function; one pair is dropped by convention.

T

takes a single argument, not three.

CasesOn the right the parentheses have only the tuple job — there is no function being applied there. The same glyph carries two meanings on the left and one on the right, in a single line.

Also writtenThe argument is written

T(\mathbf{v})

with a bold letter elsewhere on this page. The bold form and the coordinate form name the same thing, and the switch happens whenever a concrete formula is needed.

Do not confuseTreating

T

as a function of three separate variables. It suggests holding one fixed or reordering them, neither of which is defined; the input is one vector and the coordinates are its description, not its arguments.

\mathcal{L}(V, W)

the space of linear maps from V to W

This notation turns functions into elements. Once sums and scalar multiples of transformations are defined pointwise, the collection of all linear maps from

V

to

W

satisfies the vector space conditions itself, and the script letter marks that change of level — its members are maps, and they are being treated as vectors.

CasesIts dimension is

\dim(V) \cdot \dim(W)

, which is the number of entries in a representing matrix. The zero element is the map sending everything to

\mathbf{0}_W

, and the additive inverse of

T

is

(-1)T

.

Also writtenWritten

\operatorname{Hom}(V, W)

in more algebraic treatments, and

\mathcal{L}(V)

with one argument when the two spaces coincide.

Do not confuseReading it as a set of matrices. Its elements are maps, and a matrix appears only after a basis has been chosen at each end — see matrix representation.

Same glyph elsewhereA script letter marks a basis at basis, where

\mathcal{B}

is a finite collection of vectors. Here the script letter names a whole space, and the two conventions share only their typeface.

All linear algebra symbols →↑ Linear transformations

The One-Step Test

The two conditions can be combined into a single check.

T

is linear if and only if

Linear Transformation Definition

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

Learn more about this formula: Linear Transformation Definition →

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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

The One-Step Test

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)

:

Proving Linearity

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

Proving Linearity

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

Proving Linearity

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

Proving Linearity

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

Pitfall: Proving linearity with numbers

Testing

T(x, y) = xy

on the pair

(1, 0)

,

(2, 0)

gives

T(3, 0) = 0 = T(1, 0) + T(2, 0)

, which looks like additivity. The pair

(1, 0)

,

(0, 1)

gives

T(1, 1) = 1 \neq 0

, so

T

is not linear.

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.

Every line of the computation holds for all choices of

\mathbf{u}

,

\mathbf{v}

,

c

and

d

at once, because the letters stand for arbitrary values and nothing was assumed about them. That generality is what turns the calculation into a proof: one chain of equalities covers infinitely many cases.

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}

to

\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.

Pitfall: Calling every straight line linear

Treating

T(x) = 2x + 3

as linear because its graph is a line fails additivity:

T(1) + T(1) = 10

, while

T(1 + 1) = T(2) = 7

. Only lines through the origin,

T(x) = mx

, are linear maps.

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

One concrete failure disproves linearity: test zero first, then a single sum or a single multiple.

Consequences of Linearity

Once linearity is established, several useful facts follow automatically. The most basic is that the zero vector maps to the zero vector:

Zero Vector Preservation

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

Learn more about this formula: Zero Vector Preservation →

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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

If

T: U \to V

and

S: V \to W

are both linear, the composition

S \circ T: U \to W

is also linear:

Composition Is Linear

(S \circ T)(\mathbf{u}) = S(T(\mathbf{u})), \qquad [S \circ T] = [S]\,[T]

Learn more about this formula: Composition Is Linear →

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The linearity check:

Composition

(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).

Pitfall: Multiplying in the order of application

Writing

[T][S]

because

T

acts first fails for

T

the projection

\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}

and

S

the quarter-turn

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

:

[T][S]\mathbf{e}_1 = (0, 0)

, but

S(T(\mathbf{e}_1)) = S(1, 0) = (0, 1)

. The matrix is

[S][T] = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}

.

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.

Order irrelevant, then decisive

Two linear maps composed in both orders. Above they agree; below they do not, and the grid ends up somewhere visibly different. Composing linear maps always gives a linear map, but it almost never gives the same one when the order is reversed. Try your own pair on the 2D composition explorer.

Linearity survives composition, which is what allows complicated maps to be assembled from simple ones without leaving the category.

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)

.

THEOREM

T

is invertible if and only if it is bijective: injective (trivial kernel) and surjective (image equals

W

).

Pitfall: Stopping at a trivial kernel

Calling

T(x, y) = (x, y, 0)

from

\mathbb{R}^2

to

\mathbb{R}^3

invertible because its kernel is

\{\mathbf{0}\}

skips surjectivity: no

\mathbf{v}

has

T(\mathbf{v}) = (0, 0, 1)

. Since

2 \neq 3

, no linear map

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

is invertible.

When

\dim(V) = \dim(W)

is finite, 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.

The equivalence for equal dimensions is a counting argument. A trivial kernel gives

\dim \operatorname{im}(T) = \dim(V)

by rank–nullity, and a subspace of

W

with

\dim(W)

dimensions must be all of

W

.

Fact: Hilbert's hotel as a linear map

Moving every guest up one room is the shift

(x_1, x_2, \ldots) \mapsto (0, x_1, x_2, \ldots)

. Its kernel is

\{\mathbf{0}\}

, yet room 1 stays empty:

(1, 0, 0, \ldots)

is never reached. The map is injective but not surjective, from a space to itself.

Counting of this kind needs the dimensions to be finite numbers. The space of all infinite sequences

(x_1, x_2, x_3, \ldots)

of real numbers has no finite basis, and for maps of that space into itself the argument has nothing to count. Polynomials of every degree and continuous functions on an interval form further spaces with no finite basis, and the same caution applies to each of them.

Sums and Scalar Multiples of Transformations

If

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

to

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

is

(-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.

Linear maps between two spaces are themselves vectors, so the whole theory of vector spaces applies to them too.

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.

Calling every straight line linear

A function whose graph is a straight line, such as

T(x) = 2x + 3

, is called linear, because in school algebra a "linear function" means exactly that. For a linear map the word is narrower: the line must pass through the origin.

Wrong working:

T(x) = 2x + 3

has a constant rate of change, so it is taken to be linear, and

T(1 + 1)

is computed as

T(1) + T(1) = 5 + 5 = 10

.

Right working: directly,

T(1 + 1) = T(2) = 2 \cdot 2 + 3 = 7 \neq 10

, so additivity fails and

T

is not linear. The map

x \mapsto 2x

without the constant passes:

2(1 + 1) = 4 = 2 + 2

.

Why it happens: the same word has two meanings. In algebra and calculus, "linear" means first degree, and any constant term is allowed. In linear algebra a constant term is a translation, and translations are not linear.

How to check: evaluate at zero. A linear map must give

T(0) = 0

, while here

T(0) = 3

.

Multiplying in the order of application

In

S \circ T

the map

T

acts first, so its matrix is written first, and the matrix of the composition is taken to be

[T][S]

instead of

[S][T]

.

Wrong working: let

T

be the projection onto the first axis and

S

the quarter-turn,

[T] = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad [S] = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \qquad [T][S] = \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix}

which sends

\mathbf{e}_1

to

(0, 0)

.

Right working: the matrix that acts first stands on the right.

[S][T] = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}

Directly,

S(T(\mathbf{e}_1)) = S(1, 0) = (0, 1)

and

S(T(\mathbf{e}_2)) = S(0, 0) = (0, 0)

, which are exactly the columns of

[S][T]

.

Why it happens: the maps are described in the order they act, first

T

and then

S

, and the matrices are written down in that same reading order. But a matrix product acts on a vector from the right, so the factor next to

\mathbf{x}

acts first.

How to check: apply the product to

\mathbf{e}_1

and compare it with

S(T(\mathbf{e}_1))

computed one map at a time. Here

[T][S]\mathbf{e}_1 = (0, 0)

does not match

(0, 1)

.

Stopping at a trivial kernel

Injectivity is checked, the kernel turns out to be

\{\mathbf{0}\}

, and

T

is declared invertible. The shortcut is used even though

V

and

W

have different dimensions, so surjectivity is never checked.

Wrong working: for

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

,

T(x, y) = (x, y, 0)

, the equation

T(x, y) = \mathbf{0}

forces

x = y = 0

. The kernel is trivial, so

T

is called invertible.

Right working: the image of

T

is the plane of vectors with third entry

0

, so

(0, 0, 1)

is not

T(\mathbf{v})

for any

\mathbf{v}

.

T

is injective but not surjective, hence not invertible. By rank–nullity,

\dim \operatorname{im}(T) = 2 - 0 = 2 < 3

, and no linear map from

\mathbb{R}^2

to

\mathbb{R}^3

can reach all of

\mathbb{R}^3

.

Why it happens: "either condition is enough" is true only when

\dim(V) = \dim(W)

, and the condition on the dimensions is easy to forget once the shortcut is familiar.

How to check: compare the dimensions first. If

\dim(V) \neq \dim(W)

,

T

is not invertible; if they are equal, either the kernel test or the image test settles it.

Summary: How Linear Maps Combine

The four sections above on consequences of linearity 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.

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Transformations · summary

How linear maps combine

Every operation here produces another linear map — that closure is what lets the set of all linear maps be a vector space in its own right, rather than merely a collection.

06operations

Unconditional2

iv

Scalar multiple§ 8

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

Scaling a linear map keeps it linear, including at c = 0, which gives the zero map — the…

Unconditional

Requires: any scalar $c$

Scaling a linear map keeps it linear, including at $c = 0$ , which gives the zero map — the additive identity of $\mathcal{L}(V, W)$ . Together with the sum above, this is the second of the two operations a vector space needs.

vi

The zero vector is fixed§ 5

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

Follows from T(0 · v) = 0 · T(v).

Unconditional

Requires: $T$ linear
Common error: treating this as a definition — it is a consequence, and its failure is a fast disproof

Follows from $T(0 \cdot \mathbf{v}) = 0 \cdot T(\mathbf{v})$ . Useful in the contrapositive: any map sending $\mathbf{0}$ somewhere else is not linear, no further work needed. This is why an affine map $\mathbf{x} \mapsto A\mathbf{x} + \mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$ fails immediately.

Fast disproofT(x, y) = (x + 1, y)
T(0, 0) = (1, 0) ≠ (0, 0) → not linear

Conditional4

i

Composition§ 6

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

Apply T’s linearity inside S’s and the result falls out.

Stays linear

Holds when: the codomain of $T$ matches the domain of $S$

Apply $T$ ’s linearity inside $S$ ’s and the result falls out. In matrix terms this is exactly matrix multiplication — which is why that product is defined the way it is, and why it fails to commute: $S \circ T$ and $T \circ S$ are different maps.

ii

Inverse§ 7

T^{-1} \circ T = I, \quad T \circ T^{-1} = I

The inverse of a linear bijection is itself linear — not obvious, but true.

Stays linear

Holds when: $T$ is a bijection
Fails when: $T$ has nontrivial kernel or misses part of the codomain

The inverse of a linear bijection is itself linear — not obvious, but true. An invertible map between spaces of equal dimension is an isomorphism, meaning the two spaces are structurally the same object written twice. See the invertibility equivalence for the matrix form of the condition.

iii

Sum§ 8

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

Defined pointwise, and linear because both summands are.

Stays linear

Holds when: $S$ and $T$ share domain and codomain

Defined pointwise, and linear because both summands are. This is what supplies the addition on $\mathcal{L}(V, W)$ — the operation exists because linear maps are closed under it, not by stipulation.

v

The space of linear maps§ 8

\dim \mathcal{L}(V, W) = \dim(V) \cdot \dim(W)

Closure under sum and scalar multiple makes L(V, W) a vector space whose elements happen to be…

Conditional

Holds when: both $V$ and $W$ finite-dimensional

Closure under sum and scalar multiple makes $\mathcal{L}(V, W)$ a vector space whose elements happen to be functions. Its dimension is the product of the two, which is the same count as the entries of an $m \times n$ matrix — the matrix representation identifies it with the space of $m \times n$ matrices.

Unconditional

Conditional

Composition§ 6Stays linear

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

Holds when: the codomain of $T$ matches the domain of $S$

Apply

T

’s linearity inside

S

’s and the result falls out. In matrix terms this is exactly matrix multiplication — which is why that product is defined the way it is, and why it fails to commute:

S \circ T

and

T \circ S

are different maps.

Read the full section

Inverse§ 7Stays linear

T^{-1} \circ T = I, \quad T \circ T^{-1} = I

Holds when: $T$ is a bijection
Fails when: $T$ has nontrivial kernel or misses part of the codomain

The inverse of a linear bijection is itself linear — not obvious, but true. An invertible map between spaces of equal dimension is an isomorphism, meaning the two spaces are structurally the same object written twice. See the invertibility equivalence for the matrix form of the condition.

Read the full section

Sum§ 8Stays linear

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

Holds when: $S$ and $T$ share domain and codomain

Defined pointwise, and linear because both summands are. This is what supplies the addition on

\mathcal{L}(V, W)

— the operation exists because linear maps are closed under it, not by stipulation.

Read the full section

Scalar multiple§ 8Unconditional

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

Requires: any scalar $c$

Scaling a linear map keeps it linear, including at

c = 0

, which gives the zero map — the additive identity of

\mathcal{L}(V, W)

. Together with the sum above, this is the second of the two operations a vector space needs.

Read the full section

The space of linear maps§ 8Conditional

\dim \mathcal{L}(V, W) = \dim(V) \cdot \dim(W)

Holds when: both $V$ and $W$ finite-dimensional

Closure under sum and scalar multiple makes

\mathcal{L}(V, W)

a vector space whose elements happen to be functions. Its dimension is the product of the two, which is the same count as the entries of an

m \times n

matrix — the matrix representation identifies it with the space of

m \times n

matrices.

Read the full section

The zero vector is fixed§ 5Unconditional

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

Requires: $T$ linear
Common error: treating this as a definition — it is a consequence, and its failure is a fast disproof

Follows from

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

. Useful in the contrapositive: any map sending

\mathbf{0}

somewhere else is not linear, no further work needed. This is why an affine map

\mathbf{x} \mapsto A\mathbf{x} + \mathbf{b}

with

\mathbf{b} \neq \mathbf{0}

fails immediately.

Fast disproofT(x, y) = (x + 1, y)
T(0, 0) = (1, 0) ≠ (0, 0) → not linear

Read the full section

Read downward and the entries build on each other: composition and inversion give

\mathcal{L}(V,V)

the structure of an algebra, while sum and scalar multiple give

\mathcal{L}(V,W)

the structure of a vector space. The last entry is the two facts combined.

How linear maps combine·/linear-algebra/transformations/propertiesLearn Math Class

Linearity Properties FAQ

Does $T(\mathbf{0}) = \mathbf{0}$ prove that a map is linear?

+

No, the implication runs one way only. Every linear map sends zero to zero, so failing the test disproves linearity outright, but passing it proves nothing. The map

T(x) = x|x|

sends

0

to

0

and is not linear. Use it to rule maps out quickly, never to rule them in.Read more →

Is the single-line linearity condition shorthand for the two separate ones?

+

It is genuinely equivalent rather than an abbreviation. Setting

c = d = 1

recovers additivity and setting

\mathbf{v} = \mathbf{0}

recovers homogeneity, so the one line implies both and both imply it. Read it as a hypothesis to be tested about a particular

T

, not an identity to be simplified.Read more →

Is $\mathcal{L}(V, W)$ a set of matrices?

+

No, its elements are maps, and a matrix appears only after a basis has been chosen at each end. The space is itself a vector space, with dimension

\dim(V) \cdot \dim(W)

, which is exactly the number of entries a representing matrix would have. Its zero element is the map sending everything to

\mathbf{0}_W

.Read more →

Properties of Linear Transformation

What Linearity Gives You and How to Verify It

The Two Linearity Conditions

Linearity Notation

Linearity Notation

The One-Step Test

Proving Linearity

Worked Example

Disproving Linearity

Consequences of Linearity

Composition

Invertibility

Sums and Scalar Multiples of Transformations

Common Pitfalls

Summary: How Linear Maps Combine

How linear maps combine

Linearity Properties FAQ

Does $T(\mathbf{0}) = \mathbf{0}$ prove that a map is linear?

Is the single-line linearity condition shorthand for the two separate ones?

Is $\mathcal{L}(V, W)$ a set of matrices?

Properties of Linear Transformation

What Linearity Gives You and How to Verify It

The Two Linearity Conditions

Linearity Notation

Linearity Notation

The One-Step Test

Proving Linearity

Worked Example

Disproving Linearity

Consequences of Linearity

Composition

Invertibility

Sums and Scalar Multiples of Transformations

Common Pitfalls

Summary: How Linear Maps Combine

How linear maps combine

Linearity Properties FAQ

Does T(0)=0T(\mathbf{0}) = \mathbf{0}T(0)=0 prove that a map is linear?

Is the single-line linearity condition shorthand for the two separate ones?

Is L(V,W)\mathcal{L}(V, W)L(V,W) a set of matrices?

Does $T(\mathbf{0}) = \mathbf{0}$ prove that a map is linear?

Is $\mathcal{L}(V, W)$ a set of matrices?