Linear Maps: Definition and Basic Properties

Suppose you know how a function treats two vectors. Suppose you also know how it scales any single vector. Can you predict what it does to every linear combination? This applies to combinations built from your entire vector space. For a special and enormously important class of functions called linear maps, the answer is yes. Knowing the output on a handful of vectors pins down the output everywhere else. That "everywhere else" means anywhere those vectors can reach by addition and scaling. This rigidity is not a limitation. It is the entire reason linear algebra is computable. It is also why matrices, determinants, eigenvalues, and nearly every tool in the subject exist in the first place. Linear maps matter because they are the structure-preserving functions between vector spaces. This holds in the same sense that continuous functions preserve topological structure. It also holds in the same sense that group homomorphisms preserve algebraic structure. A function between vector spaces that is not linear can do almost anything. A linear map, by contrast, is entirely determined by its action on a basis. This fact turns infinite-dimensional questions into finite computations. It also turns geometric transformations into objects we can manipulate directly. These transformations include rotations, projections, reflections, differentiation, and integration. We can add, scale, and compose them just like the vectors they act on. The big picture is this: fix two vector spaces $V$ and $W$ over the same field. Once we do, the collection of all linear maps from $V$ to $W$ is itself a vector space. It is denoted $\mathcal{L}(V,W)$ . This is a small but structurally deep fact. It means linear maps are not just functions we study one at a time. Instead, they are points in a new space. That space has its own addition, scalar multiplication, zero vector, and additive inverses. Studying $\mathcal{L}(V,W)$ as a vector space is the first step toward representing linear maps by matrices. This correspondence makes the entire subject computational. This article develops the formal definition of a linear map. It proves two basic consequences. Every linear map must send the zero vector to the zero vector. Every linear map must also respect linear combinations. It further establishes that $\mathcal{L}(V,W)$ is a vector space in its own right. These facts are foundational. Everything from Null Space and Injectivity to Range and Surjectivity rests on them. The matrix representation of linear maps also rests on the definitions and lemmas proved here.

Key takeaways

A linear map $T: V \to W$ is a function satisfying additivity $T(u+v) = Tu + Tv$ and homogeneity $T(\lambda v) = \lambda(Tv)$ .
Every linear map sends the zero vector of $V$ to the zero vector of $W$ : this is a consequence of linearity, not part of the definition.
Linearity is equivalent to preserving arbitrary finite linear combinations: $T\left(\sum_i \lambda_i v_i\right) = \sum_i \lambda_i (Tv_i)$ .
The set $\mathcal{L}(V,W)$ of all linear maps from $V$ to $W$ forms a vector space under pointwise addition and scalar multiplication.
A linear map is completely determined by its values on a basis of $V$ , and those values may be prescribed arbitrarily.
Key idea: linearity is exactly the condition that a map commutes with all linear combinations, which is why one linear map can encode infinitely much information in finitely many numbers.

Where this lives today

Linear maps are the working vocabulary of essentially every quantitative discipline. In physics, the transformations relating reference frames (rotations, Lorentz boosts) are linear maps on vector spaces of coordinates, and quantum mechanics represents observables as linear operators on Hilbert spaces — the additivity and homogeneity from this article become the defining axioms of a quantum-mechanical operator. In computer science and machine learning, a single "layer" of a neural network before the nonlinearity is applied is exactly a linear map, and understanding which parts of a network are linear (and hence composable via matrix multiplication) versus nonlinear is central to both theory and efficient implementation. In cryptography, linear maps over finite fields underlie linear feedback shift registers and error-correcting codes, where the vector space structure over $\mathbb{F}_2$ makes the additivity axiom directly load-bearing. In statistics, linear regression and the general linear model are, at their core, the search for a linear map (or affine map, a linear map plus a translation) best approximating observed data, and the vector-space structure of $\mathcal{L}(V,W)$ underlies the fact that sums and scalar multiples of fitted linear models are again linear models. Engineering disciplines from signal processing (linear time-invariant systems) to structural analysis (linear elasticity, small-deformation approximations) rest on the same additivity-and-homogeneity backbone proved rigorously here.

Ask the tutor: Why must a linear map's behavior on a basis determine it everywhere, and what breaks if the spanning set isn't linearly independent?

V

Intuition

Think of a linear map as a function that "does not know" how to distinguish between performing an operation on the input and performing it on the output. If you add two vectors first and then apply $T$ , you get the same answer as applying $T$ to each vector separately and then adding the results. If you scale a vector first and then apply $T$ , you get the same answer as applying $T$ first and then scaling by the same amount. This is a very strong compatibility requirement, and most functions you can write down do not satisfy it: $f(x) = x^2$ , $f(x) = \sin x$ , $f(x) = x + 1$ are all non-linear on $\mathbb{R}$ , while $f(x) = 3x$ is linear. The picture to hold in mind is that a linear map is completely controlled by what it does to a basis. If $v_1, \dots, v_n$ is a basis of $V$ , then every $v \in V$ can be written uniquely as $v = a_1 v_1 + \cdots + a_n v_n$ , and additivity plus homogeneity force

Tv = a_1(Tv_1) + \cdots + a_n(Tv_n).

So the entire function $T$ , defined a priori on the (possibly infinite) set $V$ , collapses into a finite list of data: the vectors $Tv_1, \dots, Tv_n$ in $W$ . This is the seed from which matrix representations grow. The figure below shows a linear map $T: \mathbb{R}^2 \to \mathbb{R}^2$ sending basis vectors $e_1 = (1,0)$ and $e_2 = (0,1)$ to $Te_1 = (2,1)$ and $Te_2 = (-1,1)$ . The vector $v = 2e_1 + 1e_2 = (2,1)$ is sent to $Tv = 2(2,1) + 1(-1,1) = (3,3)$ , illustrating that the map's effect on an arbitrary vector is dictated entirely by its effect on the basis.

Figure

A linear map is determined by where it sends a basis

The zero-vector consequence: a picture worth keeping

Before stating the main theorem, notice something conceptually important: nowhere in the definition above does it say $T(0) = 0$ . That fact is a theorem, proved from additivity and homogeneity, not an axiom bolted onto the definition. This distinction matters because it shows that linearity is a genuinely restrictive pair of conditions from which other properties can be derived, rather than a checklist of unrelated requirements.

V

T(0_V) = 0_W

\mathcal{L}(V,W)

\mathcal{L}(V,W)

Worked examples

Example 1: Verifying linearity directly, and computing with basis data. Let $V = \mathbb{R}^3$ and $W = \mathbb{R}^2$ , both over $\mathbb{R}$ . Define $T: \mathbb{R}^3 \to \mathbb{R}^2$ by

T(x,y,z) = (2x - y + 3z, \; x + 4z).

Checking linearity. Let $u = (x_1,y_1,z_1)$ , $v = (x_2,y_2,z_2)$ , and $\lambda \in \mathbb{R}$ . Additivity:

T(u+v) = T(x_1+x_2,\, y_1+y_2,\, z_1+z_2) = \big(2(x_1+x_2) - (y_1+y_2) + 3(z_1+z_2),\; (x_1+x_2)+4(z_1+z_2)\big).

Regrouping each coordinate (using distributivity and commutativity in $\mathbb{R}$ ):

= \big((2x_1-y_1+3z_1) + (2x_2-y_2+3z_2),\; (x_1+4z_1)+(x_2+4z_2)\big) = Tu + Tv.

Homogeneity:

T(\lambda u) = T(\lambda x_1, \lambda y_1, \lambda z_1) = (2\lambda x_1 - \lambda y_1 + 3\lambda z_1,\; \lambda x_1 + 4\lambda z_1) = \lambda(2x_1-y_1+3z_1,\, x_1+4z_1) = \lambda(Tu).

So $T$ is linear. Sanity check via Theorem (a). We must have $T(0,0,0) = (0,0)$ . Indeed $T(0,0,0) = (2(0)-0+3(0),\ 0+4(0)) = (0,0)$ . ✓. Using the basis-determination principle. The standard basis of $\mathbb{R}^3$ is $e_1=(1,0,0)$ , $e_2=(0,1,0)$ , $e_3=(0,0,1)$ . Compute:

Te_1 = (2,1), \qquad Te_2 = (-1,0), \qquad Te_3 = (3,4).

By Theorem (c), for any $v = (x,y,z) = xe_1+ye_2+ze_3$ ,

Tv = x(2,1) + y(-1,0) + z(3,4) = (2x-y+3z,\ x+4z),

which matches the original formula exactly — confirming that $T$ is fully reconstructed from just three vectors $Te_1, Te_2, Te_3 \in \mathbb{R}^2$ , in agreement with the existence/uniqueness clause of the theorem.

Example 2: Building a linear map from scratch by prescribing basis images, and testing a non-example. Let $V = \mathcal{P}_2(\mathbb{R})$ , the space of real polynomials of degree at most $2$ , with basis $1, x, x^2$ . Let $W = \mathbb{R}^2$ . Define $T \in \mathcal{L}(V,W)$ by prescribing

T(1) = (1,0), \qquad T(x) = (0,-2), \qquad T(x^2) = (5,1).

By the existence/uniqueness clause of the Theorem, this data determines a unique linear map on all of $V$ , given explicitly by extending via Theorem (c): for $p(x) = a_0 + a_1 x + a_2 x^2$ ,

T(p) = a_0(1,0) + a_1(0,-2) + a_2(5,1) = (a_0+5a_2,\ -2a_1+a_2).

Numerical check. Take $p(x) = 3 - 4x + 2x^2$ , so $(a_0,a_1,a_2) = (3,-4,2)$ . Then

T(p) = (3 + 5(2),\ -2(-4)+2) = (3+10,\ 8+2) = (13, 10).

Let's verify additivity numerically with $q(x) = 1 + x - x^2$ , so $(b_0,b_1,b_2)=(1,1,-1)$ :

T(q) = (1 + 5(-1),\ -2(1)+(-1)) = (1-5,\ -2-1) = (-4,-3).

Now $p+q = 4 - 3x + x^2$ , i.e. coefficients $(4,-3,1)$ , so

T(p+q) = (4+5(1),\ -2(-3)+1) = (9, 7).

Compare to $T(p)+T(q) = (13,10)+(-4,-3) = (9,7)$ . They agree, confirming additivity in this instance (as guaranteed by the theorem in general). A deliberate non-example, to see where linearity fails. Define $S: \mathcal{P}_2(\mathbb{R}) \to \mathbb{R}^2$ by $S(p) = (p(0)^2,\ p(1))$ , i.e. $S(a_0+a_1x+a_2x^2) = (a_0^2,\ a_0+a_1+a_2)$ . Test additivity with the same $p,q$ above: $p(0)=3$ , so first coordinate of $S(p)$ is $9$ ; $q(0)=1$ , giving $S(q)$ first coordinate $1$ . Sum of first coordinates: $9+1=10$ . But $(p+q)(0) = 4$ , so $S(p+q)$ has first coordinate $4^2=16 \neq 10$ . Since $S(p+q) \neq S(p)+S(q)$ , $S$ fails additivity and is not linear — the squaring operation breaks the compatibility with vector addition that Theorem (a)–(c) rely on.

\mathbb{R}

Scope, edge cases & common traps

Situation	What holds	What can go wrong
$V, W$ over the same field $\mathbb{F}$	Definition applies directly	If fields differ (e.g. $V$ real, $W$ complex without restriction of scalars) "linear map" is undefined until scalars are matched
$V$ infinite-dimensional	Theorem (c) still holds for finite sums; basis-extension needs care with infinite Hamel bases	Cannot naively extend to infinite linear combinations (convergence is a topological notion, absent in pure algebra)
Only additivity assumed, scalars = $\mathbb{Q}$	Additivity implies $\mathbb{Q}$ -homogeneity automatically	Over $\mathbb{R}$ or $\mathbb{C}$ , additivity alone does not imply homogeneity (counterexample above)
$T(0)=0$	Always true for linear $T$	Common trap: treating $T(0)=0$ as part of the definition; it is a consequence, and checking it is a quick necessary (not sufficient) test for linearity
Prescribing images of a spanning set that is not independent	Extension may not exist or may be non-unique	Basis-extension theorem requires linear independence and spanning; a mere spanning set overdetermines $T$

Common trapThe classic error

Many students verify $T(0) = 0$ and conclude $T$ is linear. This is backwards: $T(0)=0$ is a necessary consequence of linearity (proved in part (a) above), not a sufficient condition for it. The function $T(x,y) = (x + y^2, \, 0)$ on $\mathbb{R}^2$ satisfies $T(0,0)=(0,0)$ but is not linear, since $T(0,1) = (1,0)$ while $2\cdot T(0, \tfrac12) = 2(\tfrac14, 0) = (\tfrac12, 0) \neq T(0,1)$ under scaling $v=(0,\frac12)$ by $\lambda=2$ , violating homogeneity. Always verify both additivity and homogeneity (or the equivalent combined identity in Theorem (d)) — never infer linearity from $T(0)=0$ alone.

The mathematician behind it

Origins · 1 minHermann Grassmann and how this was discovered

Hermann Grassmann

1809–1877 · Prussian mathematician, linguist, and independent scholar working largely outside the mainstream mathematical establishment

Grassmann developed, in his 1844 treatise Die lineale Ausdehnungslehre ("The Theory of Linear Extension"), an abstract algebra of vectors, linear combinations, and multi-dimensional extension that anticipated the modern axiomatic notion of vector space and linear map by more than half a century. His work was so far ahead of the notational and conceptual conventions of his time that it was largely ignored by contemporaries, and Grassmann spent much of his career as a secondary-school teacher, later turning to Sanskrit philology, where he achieved more immediate recognition. It was only after his death, as mathematicians like Giuseppe Peano formalized vector spaces axiomatically, that the depth of Grassmann's ideas — including a precursor to the very additivity-and-homogeneity structure central to this article — was fully appreciated.

History

- 1844: Hermann Grassmann publishes Die lineale Ausdehnungslehre, introducing an abstract calculus of linear combinations far ahead of contemporary algebra.

- 1888: Giuseppe Peano gives the first fully modern axiomatic definition of a vector space, in Calcolo Geometrico.

- 1918: Hermann Weyl, in Raum, Zeit, Materie, uses the axiomatic vector space and linear map framework to formalize the mathematics of general relativity.

- 1920s–1930s: Stefan Banach and others extend linear map theory to infinite-dimensional normed spaces, founding functional analysis.

- 1958: Paul Halmos's Finite-Dimensional Vector Spaces popularizes the clean, basis-free treatment of linear maps now standard in undergraduate courses.

Why isn't

T(0)=0

included directly in the definition of a linear map?

Mathematical definitions aim to state the minimal independent conditions from which everything else follows by proof, avoiding redundancy. Additivity and homogeneity alone already force $T(0_V)=0_W$ , as shown in Theorem (a): setting $u=v=0_V$ in additivity, or setting $\lambda=0$ in homogeneity, both derive it. Including $T(0)=0$ as a third axiom would be logically redundant — it adds no new information beyond what additivity and homogeneity already guarantee. Keeping the definition to exactly two conditions also clarifies why the property holds (it is forced by compatibility with vector addition and scaling), rather than treating it as a separate assumption to check by hand every time.

Is every additive function (

T(u+v)=Tu+Tv

) automatically linear?

Not in general, though it depends on the field. Over $\mathbb{Q}$ , additivity alone implies homogeneity for all rational scalars, by an induction and division argument, so additive maps between $\mathbb{Q}$ -vector spaces are automatically linear. Over $\mathbb{R}$ or $\mathbb{C}$ , however, additivity does not imply homogeneity for irrational or non-rational scalars in general; pathological additive-but-non-linear functions exist, constructed using a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ (see the counterexample above). Such functions are extremely irregular — non-measurable and discontinuous everywhere — and never arise in ordinary calculations, but their existence shows homogeneity is a logically independent axiom, not a free consequence of additivity over general fields.

Why does the vector space structure on

\mathcal{L}(V,W)

matter, beyond being a curiosity?

Once $\mathcal{L}(V,W)$ is known to be a vector space, all the machinery of linear algebra — bases, dimension, subspaces, linear maps between spaces of linear maps — applies to collections of linear maps, not just to individual ones. This is the first step toward showing that $\mathcal{L}(V,W)$ is isomorphic to a space of matrices once bases are chosen, with $\dim \mathcal{L}(V,W) = (\dim V)(\dim W)$ in the finite-dimensional case. It also justifies operations engineers and physicists use constantly, such as superposing two linear systems ( $S+T$ ) or rescaling a linear transformation ( $\lambda T$ ), confirming these operations stay within the same well-behaved class of maps.

Does the basis-extension theorem work for infinite-dimensional vector spaces?

Yes, with the same proof structure, provided $V$ has a basis in the (possibly infinite) Hamel-basis sense: every vector space has a basis, a fact that for infinite-dimensional spaces requires the axiom of choice (via Zorn's Lemma). Given a basis $\{v_\alpha\}_{\alpha \in A}$ of $V$ (any index set $A$ ) and arbitrary vectors $\{w_\alpha\}_{\alpha \in A}$ in $W$ , there is a unique linear map $T$ with $Tv_\alpha = w_\alpha$ , because every $v \in V$ is still a finite linear combination of basis vectors (this is part of the definition of basis even in infinite dimensions), so Theorem (c)'s finite-sum identity applies unchanged.

Can a linear map send a nonzero vector to zero?

Yes — this does not violate any part of the definition or the theorem proved here. A linear map sending some nonzero $v$ to $0_W$ is not injective, but injectivity is a separate property, not implied by linearity itself. For example, the map $T:\mathbb{R}^2\to\mathbb{R}$ , $T(x,y)=x-y$ , is linear and sends the nonzero vector $(1,1)$ to $0$ . The set of vectors a linear map sends to zero is called its null space (or kernel), and its structure is the subject of Null Space and Injectivity.

How does this relate to matrices?

Once $V$ and $W$ are finite-dimensional with chosen bases, the basis-extension theorem above shows a linear map is completely encoded by the finite list of vectors $Tv_1,\dots,Tv_n$ (the images of a basis of $V$ ), each expressed in coordinates relative to a basis of $W$ . Arranging these coordinate vectors as the columns of a rectangular array produces the matrix representing $T$ . This correspondence, which turns abstract linear maps into concrete arrays of numbers amenable to computation (row reduction, determinants, eigenvalues), is developed in Examples of Linear Maps and the matrix representation theory that follows.

Why does the counterexample use

\mathbb{Q}

versus

\mathbb{R}

instead of a simpler example?

Because no simple formula-based function can serve as a counterexample here: any additive function built from elementary formulas on $\mathbb{R}$ (polynomials, exponentials, trigonometric combinations) that is additive turns out to automatically be homogeneous too (in fact, of the form $f(x) = cx$ ), a consequence of additivity plus continuity or measurability. To genuinely separate additivity from full homogeneity over $\mathbb{R}$ , one is forced into a non-constructive object, built from an uncountable Hamel basis via the axiom of choice, since these pathological solutions are provably impossible to describe by any explicit formula, continuous or measurable.