Skip to content
VidiMaster it, module by module
Module 1/Linear Maps

Linear Maps: Definition

A linear map is a function between vector spaces that preserves addition and scalar multiplication. This lesson defines it precisely, shows how to test a formula for linearity, and separates genuine linear maps from affine and nonlinear impostors.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Which of the following maps R2→R2\mathbb{R}^2\to\mathbb{R}^2 is linear?

A linear map T:R2→RT:\mathbb{R}^2\to\mathbb{R} satisfies T(1,0)=4T(1,0)=4 and T(0,1)=−2T(0,1)=-2. Compute T(3,5)T(3,5).

What you’ll be able to do

  • State the definition of a linear map in terms of additivity and homogeneity, and recognize the equivalent "preserves linear combinations" form.
  • Check whether a map given by a formula is linear, and identify precisely which axiom a non-example violates.
  • Derive the consequence T(0)=0T(\mathbf{0})=\mathbf{0} and use it as a quick rejection test.
  • Evaluate a linear map at a vector, including by exploiting linearity from the known images of basis vectors.
  • Distinguish linear maps (projection, rotation) from affine or nonlinear maps (translation, squaring, absolute value).

In your course

· MATH2015 · Linear Algebra & Probability
§3.1 Linear maps
  • Definition 3.1Linear map
    f:Rn→Rmf:\mathbb{R}^n\to\mathbb{R}^m is linear if f(v+w)=f(v)+f(w)f(v+w)=f(v)+f(w) and f(c v)=c f(v)f(c\,v)=c\,f(v) for all v,wv,w and scalars cc.
  • Remark 3.1A linear map sends 0↦0\mathbf 0\mapsto\mathbf 0
  • Definition 3.2Injective, surjective, bijective
Watch the affine trap (Example 3.2): y=ax+by=ax+b is linear only when b=0b=0.
1

What a Linear Map Is (Intuition)

A linear map (also called a linear transformation) is a function T:V→WT:V\to W between two vector spaces that respects the two operations that make a vector space what it is: adding vectors and scaling them by a number.

Throughout, let:

  • VV and WW be vector spaces over the real numbers R\mathbb{R} (the field of scalars),
  • u,vu, v denote vectors in VV,
  • cc denote a scalar (a real number).

Geometrically, a linear map is the kind of transformation that keeps grid lines straight and evenly spaced and keeps the origin fixed. Scalings, rotations about the origin, reflections through the origin, projections, and shears are all linear.

What is not linear? The clearest example is a translation — sliding every point by a fixed vector, like T(x)=x+1T(x)=x+1. It moves the origin, so it breaks the structure even though its graph is still a straight line. "Linear" in this course means something stricter than "its graph is a line."

2

The Definition: Two Axioms

A map T:V→WT:V\to W is linear if and only if it satisfies both of the following for all vectors u,v∈Vu,v\in V and all scalars c∈Rc\in\mathbb{R}:

  1. Additivity:   T(u+v)=T(u)+T(v)\;T(u+v)=T(u)+T(v)
  2. Homogeneity:   T(cu)=c T(u)\;T(cu)=c\,T(u)

Read the symbols carefully. On the left of additivity, u+vu+v is addition in VV; on the right, T(u)+T(v)T(u)+T(v) is addition in WW. The map carries one structure to the other.

These two axioms can be fused into a single condition that is often the fastest to use. A map TT is linear if and only if it preserves linear combinations: T(cu+dv)=c T(u)+d T(v)for all u,v∈V, c,d∈R.T(cu+dv)=c\,T(u)+d\,T(v)\quad\text{for all }u,v\in V,\ c,d\in\mathbb{R}.

Why equivalent? Setting c=d=1c=d=1 recovers additivity, and setting d=0d=0 recovers homogeneity, so the single condition implies both axioms. Conversely, if both axioms hold then T(cu+dv)=T(cu)+T(dv)=cT(u)+dT(v)T(cu+dv)=T(cu)+T(dv)=cT(u)+dT(v), using additivity on the sum and homogeneity on each piece.

3

Immediate Consequences

The axioms force some facts for free.

Every linear map sends zero to zero:   T(0V)=0W.\;T(\mathbf{0}_V)=\mathbf{0}_W.

Proof. Using homogeneity with scalar c=0c=0: for any vv,   T(0)=T(0⋅v)=0⋅T(v)=0.\;T(\mathbf{0})=T(0\cdot v)=0\cdot T(v)=\mathbf{0}. (Alternatively, T(0)=T(0+0)=T(0)+T(0)T(\mathbf{0})=T(\mathbf{0}+\mathbf{0})=T(\mathbf{0})+T(\mathbf{0}), and subtracting T(0)T(\mathbf{0}) from both sides gives T(0)=0T(\mathbf{0})=\mathbf{0}.)

This gives a powerful rejection test (the contrapositive): if T(0)≠0, then T is NOT linear.\text{if } T(\mathbf{0})\neq\mathbf{0},\ \text{then } T \text{ is NOT linear.} It instantly disqualifies translations and anything with a constant term.

Two more consequences:

  • Negatives: T(−v)=−T(v)T(-v)=-T(v) (take c=−1c=-1 in homogeneity).
  • General combinations: by induction, T ⁣(∑i=1ncivi)=∑i=1nci T(vi).T\!\left(\sum_{i=1}^n c_i v_i\right)=\sum_{i=1}^n c_i\,T(v_i). This is why knowing TT on a basis determines TT everywhere.
4

Checking a Formula; Examples vs Non-Examples

When a map is given by formulas on coordinates, there is a clean rule of thumb:

TT is linear iff every output coordinate is a homogeneous degree-one expression in the input variables — a sum of constant multiples of the inputs, with no constant term, no powers or products of variables, and no absolute values or other nonlinear functions.

Linear (pass the axioms):

  • Projection onto the xx-axis: P(x,y)=(x,0)P(x,y)=(x,0).
  • Rotation by 90∘90^\circ counterclockwise: R(x,y)=(−y, x)R(x,y)=(-y,\,x).
  • T(x,y)=(2x,  x−y)T(x,y)=(2x,\;x-y) — each coordinate is a combination of xx and yy.
  • The zero map T(v)=0T(v)=\mathbf{0} and the identity map T(v)=vT(v)=v.

Not linear (and the axiom each breaks):

  • T(x)=x+1T(x)=x+1 — affine: the constant term gives T(0)=1≠0T(0)=1\neq 0, so homogeneity/additivity fail.
  • T(x,y)=(x2, y)T(x,y)=(x^2,\,y) — squaring breaks homogeneity: T(2⋅(1,0))=(4,0)T(2\cdot(1,0))=(4,0) but 2 T(1,0)=(2,0)2\,T(1,0)=(2,0).
  • T(x,y)=(x, ∣y∣)T(x,y)=(x,\,|y|) — absolute value breaks homogeneity for negative scalars: T(−1⋅(0,1))=T(0,−1)=(0,1)T(-1\cdot(0,1))=T(0,-1)=(0,1), yet −1⋅T(0,1)=(0,−1)-1\cdot T(0,1)=(0,-1).
Definition of a Linear Map

Let V,WV,W be vector spaces over R\mathbb{R}. A function T:V→WT:V\to W is a linear map if and only if, for all u,v∈Vu,v\in V and all c∈Rc\in\mathbb{R}: (i) T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v) (additivity), and (ii) T(cu)=cT(u)T(cu)=cT(u) (homogeneity).

Intuition. A linear map is exactly a function that commutes with the vector-space operations: you may add and scale either before or after applying TT and get the same result. The two axioms are the minimal requirements that capture "respects the structure."
Preservation of Linear Combinations

A map T:V→WT:V\to W is linear if and only if T(cu+dv)=cT(u)+dT(v)T(cu+dv)=cT(u)+dT(v) for all u,v∈Vu,v\in V and c,d∈Rc,d\in\mathbb{R}. More generally, a linear map satisfies T ⁣(∑i=1ncivi)=∑i=1nciT(vi)T\!\left(\sum_{i=1}^{n} c_i v_i\right)=\sum_{i=1}^{n} c_i T(v_i).

Intuition. Additivity and homogeneity are equivalent to a single statement: linear combinations pass straight through TT. This is the working form you use to compute: break a vector into pieces whose images you know, then reassemble.
Linear Maps Preserve the Zero Vector

If T:V→WT:V\to W is linear, then T(0V)=0WT(\mathbf{0}_V)=\mathbf{0}_W. Equivalently (contrapositive): if T(0)≠0T(\mathbf{0})\neq\mathbf{0}, then TT is not linear.

Intuition. Scaling any vector by 00 gives the zero vector, and homogeneity forces its image to be scaled by 00 too. This is the single fastest test to rule out linearity — it immediately kills translations and any formula with a constant term.

Worked examples

Example 1

Show from the definition that T:R2→R2T:\mathbb{R}^2\to\mathbb{R}^2 given by T(x,y)=(2x,  x−y)T(x,y)=(2x,\;x-y) is linear.

  1. 1

    Name general inputs and a scalar. Take two arbitrary vectors u=(x1,y1)u=(x_1,y_1) and v=(x2,y2)v=(x_2,y_2) in R2\mathbb{R}^2, and an arbitrary scalar c∈Rc\in\mathbb{R}. We must verify both axioms hold for these.

  2. 2

    Check additivity. First add in the input space: u+v=(x1+x2,  y1+y2)u+v=(x_1+x_2,\;y_1+y_2). Apply TT: T(u+v)=(2(x1+x2),  (x1+x2)−(y1+y2))=(2x1+2x2,  (x1−y1)+(x2−y2)).T(u+v)=\big(2(x_1+x_2),\;(x_1+x_2)-(y_1+y_2)\big)=\big(2x_1+2x_2,\;(x_1-y_1)+(x_2-y_2)\big). Now compute the other side by adding the images: T(u)+T(v)=(2x1, x1−y1)+(2x2, x2−y2)=(2x1+2x2,  (x1−y1)+(x2−y2)).T(u)+T(v)=(2x_1,\,x_1-y_1)+(2x_2,\,x_2-y_2)=\big(2x_1+2x_2,\;(x_1-y_1)+(x_2-y_2)\big). The two expressions are identical, so T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v). ✓\checkmark

  3. 3

    Check homogeneity. Scale first: cu=(cx1, cy1)cu=(cx_1,\,cy_1). Apply TT: T(cu)=(2cx1,  cx1−cy1)=c (2x1,  x1−y1)=c T(u). ✓T(cu)=(2cx_1,\;cx_1-cy_1)=c\,(2x_1,\;x_1-y_1)=c\,T(u).\ \checkmark

  4. 4

    Conclude. Both axioms hold for all u,vu,v and all cc, so TT is linear. (Sanity check: T(0,0)=(0,0)T(0,0)=(0,0), consistent with the requirement T(0)=0T(\mathbf{0})=\mathbf{0}.)

Answer. TT is linear.
Example 2

Decide whether T:R→RT:\mathbb{R}\to\mathbb{R}, T(x)=x+1T(x)=x+1, is linear.

Example 3

Suppose T:R2→R2T:\mathbb{R}^2\to\mathbb{R}^2 is linear with T(1,0)=(3,−1)T(1,0)=(3,-1) and T(0,1)=(0,2)T(0,1)=(0,2). Find T(4,−2)T(4,-2).