Jaysen Tsao
Linear Algebra ›

Introduction to Linear Maps

Linear Maps

Definition 26: Linear Map

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹. A linear map (or linear transformation) from 𝑉 to 𝑊 is a function 𝑇:𝑉→𝑊 such that the following properties hold:

  1. Additivity. ∀𝐮,𝐯∈𝑉,𝑇(𝐮+𝐯)=𝑇(𝐮)+𝑇(𝐯).
  2. Homogeneity. ∀𝐯∈𝑉,∀𝑐∈𝐹,𝑇(𝑐𝐯)=𝑐𝑇(𝐯).

These two properties are known as the linearity conditions or linear map axioms.

Terminology: Domain and Codomain

Let 𝑇:𝑉→𝑊 be a linear map. The domain of 𝑇 is 𝑉, and the codomain of 𝑇 is 𝑊.

Terminology: Preimage and Image of a Vector under a Linear Map

Let 𝑇:𝑉→𝑊 be a linear map such that 𝑇(𝐯)=𝐰 for some 𝐯∈𝑉 and 𝐰∈𝑊.

  • 𝐯 is called the preimage of 𝐰 under 𝑇.
  • 𝐰 is called the image of 𝐯 under 𝑇.

In which case, we say that 𝑇 maps 𝐯 to 𝐰.

Notation

The shorthand notation 𝐯↦𝑇(𝐯) is often used to denote the action of a linear map 𝑇 on a vector 𝐯. For example, if 𝑇(𝐯)=𝐰, we would write 𝐯↦𝐰 to indicate that 𝑇 maps 𝐯 to 𝐰.

Example: Differentiation is a linear map.

Let ℙ𝑛(ℝ) denote the set of polynomials with real coefficients of degree at most 𝑛. Define a function 𝑇:ℙ𝑛(ℝ)→ℙ𝑛−1(ℝ) by 𝑓↦𝑓′, where 𝑓′ is the derivative of 𝑓. Show that 𝑇 is a linear map.

Proof.

To show that 𝑇 is a linear map, we need to verify the two linearity conditions: additivity and homogeneity.

  1. Additivity. Let 𝑓,𝑔∈ℙ𝑛(ℝ). We need to show that 𝑇(𝑓+𝑔)=𝑇(𝑓)+𝑇(𝑔). We have:

    𝑇(𝑓+𝑔)=(𝑓+𝑔)′=𝑓′+𝑔′=𝑇(𝑓)+𝑇(𝑔).

    Thus, additivity holds.

  2. Homogeneity. Let 𝑓∈ℙ𝑛(ℝ) and let 𝑐∈ℝ. We need to show that 𝑇(𝑐𝑓)=𝑐𝑇(𝑓). We have:

    𝑇(𝑐𝑓)=(𝑐𝑓)′=𝑐𝑓′=𝑐𝑇(𝑓).

    Thus, homogeneity holds.

Since both linearity conditions are satisfied, we conclude that 𝑇 is a linear map. ∎

Notation: Set of Linear Maps

The set of all linear maps from 𝑉 to 𝑊 may be denoted by ℒ(𝑉,𝑊) or ℒ𝑉→𝑊. Additionally, the set of all linear maps from 𝑉 to itself may be denoted by ℒ(𝑉) or ℒ𝑉.

Theorem 26: Alternative Characterization of Linearity

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹, and let 𝑇:𝑉→𝑊 be a function. Then 𝑇 is a linear map if and only if:

∀𝐮,𝐯∈𝑉,∀𝑎,𝑏∈𝐹,𝑇(𝑎𝐮+𝑏𝐯)=𝑎𝑇(𝐮)+𝑏𝑇(𝐯).

Corollary 13: General Linearity

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹, and let 𝑇:𝑉→𝑊 be a function. Then 𝑇 is a linear map if and only if for all 𝑐1,𝑐2,…,𝑐𝑘∈𝐹 and all 𝐯1,𝐯2,…,𝐯𝑘∈𝑉:

𝑇(∑𝑖=1𝑘𝑐𝑖𝐯𝑖)=∑𝑖=1𝑘𝑐𝑖𝑇(𝐯𝑖).

Theorem 27: Zero Vector Preservation of Linear Maps

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹, and let 𝑇:𝑉→𝑊 be a linear map. Then 𝑇 preserves the zero vector, i.e. 𝑇(𝟎𝑉)=𝟎𝑊.

Proof: Theorem 27.
Assume 𝑇:𝑉→𝑊 over 𝐹 is a linear map that satisfies the linearity conditions in Definition 26. Then by Theorem 3, 𝑇(𝟎𝑉)=𝑇(0𝐹𝐯) for any 𝐯∈𝑉. By homogeneity, 𝑇(0𝐹𝐯)=0𝐹𝑇(𝐯). Since 𝑇(𝐯)∈𝑊, by Theorem 3 again, 0𝐹𝑇(𝐯)=𝟎𝑊. Thus, by transitivity of equality, 𝑇(𝟎𝑉)=𝟎𝑊. ∎

We can apply multiple linear maps in succession by composing them, and the result is still a linear map. This allows us to build more complex linear maps from simpler ones, and to analyze the structure of linear maps in terms of their compositions.

Theorem 28: Composition of Linear Maps

Define the linear maps 𝑈:𝑉→𝑊 and 𝑇:𝑊→𝑋. Then the function composition of 𝑇 and 𝑈, denoted 𝑇∘𝑈, is a linear map from 𝑉 to 𝑋, where 𝑇∘𝑈 is defined by:

(𝑇∘𝑈)(𝐯)=𝑇(𝑈(𝐯)).

Notation

The composition 𝑇∘𝑈 of two linear maps may be denoted by juxtaposition, i.e. 𝑇𝑈. This should not be confused with the product of two functions. Furthermore, the repeated composition of a linear map 𝑛 times may be denoted 𝑇𝑛, and should not be confused with the 𝑛th power of a function:

𝑇𝑈=𝑇∘𝑈,𝑇𝑛=𝑇∘𝑇∘…∘𝑇⏟𝑛 times.
Proof: Theorem 28.

Let 𝑈:𝑉→𝑊 and 𝑇:𝑊→𝑋 be linear maps. We need to show that 𝑇∘𝑈:𝑉→𝑋 is a linear map, i.e. it satisfies the linearity conditions.

  1. Additivity. Let 𝐮,𝐯∈𝑉. We need to show that (𝑇∘𝑈)(𝐮+𝐯)=(𝑇∘𝑈)(𝐮)+(𝑇∘𝑈)(𝐯). We have:

    (𝑇∘𝑈)(𝐮+𝐯)=𝑇(𝑈(𝐮+𝐯))=𝑇(𝑈(𝐮)+𝑈(𝐯))=𝑇(𝑈(𝐮))+𝑇(𝑈(𝐯))=(𝑇∘𝑈)(𝐮)+(𝑇∘𝑈)(𝐯).

    Thus, additivity holds.

  2. Homogeneity. Let 𝐯∈𝑉 and let 𝑐∈𝐹. We need to show that (𝑇∘𝑈)(𝑐𝐯)=𝑐(𝑇∘𝑈)(𝐯). We have:

    (𝑇∘𝑈)(𝑐𝐯)=𝑇(𝑈(𝑐𝐯))=𝑇(𝑐𝑈(𝐯))=𝑐𝑇(𝑈(𝐯))=𝑐(𝑇∘𝑈)(𝐯).

    Thus, homogeneity holds.

Since both linearity conditions are satisfied, we conclude that 𝑇∘𝑈 is a linear map. ∎

The simplest linear maps are the identity and zero linear maps. The identity linear map on a vector space 𝑉 maps every vector to itself:

Definition 27: Identity Linear Map

Let 𝑉 be a vector space. The identity map or identity transformation on 𝑉, denoted id𝑉, is the linear map from 𝑉 to itself defined by:

id𝑉(𝐯)=𝐯 for all 𝐯∈𝑉.

That is, id𝑉:𝑉→𝑉 is the linear map defined by 𝐯↦𝐯.

Corollary 14

For any linear map 𝑇:𝑉→𝑊, id𝑊∘𝑇=𝑇∘id𝑉=𝑇.

Definition 28: Zero Linear Map

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹. The zero map or zero transformation from 𝑉 to 𝑊, denoted 0𝑉→𝑊, is the linear map defined by:

0𝑉→𝑊(𝐯)=𝟎𝑊 for all 𝐯∈𝑉.

That is, 0𝑉→𝑊:𝑉→𝑊 is the linear map defined by 𝐯↦𝟎𝑊.

Theorem 29: Linearity of the Basis Transformation

Let 𝑉 be an 𝑛-dimensional vector space over a field 𝐹, and suppose ℬ is a basis for 𝑉. Then the basis transformation [⋅]ℬ:𝑉→𝐹𝑛 defined by 𝐯↦[𝐯]ℬ is a linear map.

Proof: Theorem 29.

Let 𝑉 be an 𝑛-dimensional vector space over a field 𝐹, and suppose ℬ={𝐛1,…,𝐛𝑛} is an ordered basis for 𝑉. Let 𝐮,𝐯∈𝑉 and let 𝑐∈𝐹, and suppose [𝐮]ℬ=(𝛼1,…,𝛼𝑛) and [𝐯]ℬ=(𝛽1,…,𝛽𝑛). Then we can write:

𝐮=∑𝑖=1𝑛𝛼𝑖𝐛𝑖,𝐯=∑𝑖=1𝑛𝛽𝑖𝐛𝑖.

We need to show that the basis transformation [⋅]ℬ:𝑉→𝐹𝑛 defined by 𝐯↦[𝐯]ℬ satisfies linearity conditions.

  1. Additivity. We have:

    [𝐮+𝐯]ℬ=[∑𝑖=1𝑛𝛼𝑖𝐛𝑖+∑𝑖=1𝑛𝛽𝑖𝐛𝑖]ℬ by substitution =[∑𝑖=1𝑛(𝛼𝑖+𝛽𝑖)𝐛𝑖]ℬ by distributivity =(𝛼1+𝛽1,…,𝛼𝑛+𝛽𝑛) by definition of basis coordinates =(𝛼1,…,𝛼𝑛)+(𝛽1,…,𝛽𝑛) by vector addition in 𝐹𝑛=[𝐮]ℬ+[𝐯]ℬ by substitution.
  2. Homogeneity. We have:

    [𝑐𝐮]ℬ=[𝑐∑𝑖=1𝑛𝛼𝑖𝐛𝑖]ℬ by substitution =[∑𝑖=1𝑛(𝑐𝛼𝑖)𝐛𝑖]ℬ by distributivity =(𝑐𝛼1,…,𝑐𝛼𝑛) by definition of basis coordinates =𝑐(𝛼1,…,𝛼𝑛) by scalar multiplication in 𝐹𝑛=𝑐[𝐮]ℬ by substitution.

Since both linearity conditions are satisfied, [⋅]ℬ is a linear map. ∎

Theorem 30: Sets of Linear Maps are Vector Spaces

Let 𝑉 and 𝑊 be vector spaces over the same field 𝐹. Then the set of all linear maps from 𝑉 to 𝑊, denoted ℒ(𝑉,𝑊), is a vector space over 𝐹 under the following operations:

  • Pointwise addition. For 𝑇,𝑆∈ℒ(𝑉,𝑊), define (𝑇+𝑆)(𝐯)=𝑇(𝐯)+𝑆(𝐯) for all 𝐯∈𝑉.
  • Scalar multiplication. For 𝑇∈ℒ(𝑉,𝑊) and 𝑐∈𝐹, define (𝑐𝑇)(𝐯)=𝑐𝑇(𝐯) for all 𝐯∈𝑉.

Furthermore, dim(ℒ(𝑉,𝑊))=dim(𝑉)dim(𝑊).

Kernel, Nullity, and Injectivity

Definition 29: Kernel of a Linear Map

Let 𝑇:𝑉→𝑊 be a linear map. The kernel or null space of 𝑇, denoted ker(𝑇), is the set of all vectors in 𝑉 that are mapped to the zero vector in 𝑊. Formally:

ker(𝑇)={𝐯∈𝑉|𝑇(𝐯)=𝟎𝑊}.

Theorem 31: Kernel is a Subspace of the Domain

Let 𝑇:𝑉→𝑊 be a linear map. Then ker(𝑇) is a subspace of 𝑉.

Proof: Kernel is a Subspace of the Domain.

Let 𝑇:𝑉→𝑊 be a linear map. Checking the Subspace Criteria:

  1. Existence of vector additive identity. Since 𝑇(𝟎𝑉)=𝟎𝑊 by Theorem 27, 𝟎𝑉∈ker(𝑇).
  2. Closure under vector addition. Let 𝐮,𝐯∈ker(𝑇). Then 𝑇(𝐮)=𝑇(𝐯)=𝟎𝑊. By additivity, 𝑇(𝐮+𝐯)=𝑇(𝐮)+𝑇(𝐯)=𝟎𝑊+𝟎𝑊=𝟎𝑊, so 𝐮+𝐯∈ker(𝑇).
  3. Closure under scalar multiplication. Let 𝐯∈ker(𝑇) and let 𝑐∈𝐹. Then 𝑇(𝐯)=𝟎𝑊. By homogeneity, 𝑇(𝑐𝐯)=𝑐𝑇(𝐯)=𝑐𝟎𝑊=𝟎𝑊, so 𝑐𝐯∈ker(𝑇). ∎

Definition 30: Nullity of a Linear Map

Let 𝑇:𝑉→𝑊 be a linear map. The nullity of 𝑇 is the dimension of the kernel of 𝑇:

nullity(𝑇)=dim(ker(𝑇)).

Theorem 32: Nullity Theorem

Let 𝑇:𝑉→𝑊 be a linear map. Then 𝑇 is injective if and only if nullity(𝑇)=0; that is, 𝑇 is injective if and only if ker(𝑇)={𝟎𝑉}.

Image, Rank, and Surjectivity

Definition 31: Image of a Linear Map

Let 𝑇:𝑉→𝑊 be a linear map. The image or range of 𝑇, denoted im(𝑇), is the set of all vectors in 𝑊 that are the image of some vector in 𝑉 under 𝑇. Formally:

im(𝑇)={𝐰∈𝑊|∃𝐯∈𝑉 s.t. 𝑇(𝐯)=𝐰}.

Theorem 33: Image is a Subspace of the Codomain

Let 𝑇:𝑉→𝑊 be a linear map. Then im(𝑇) is a subspace of 𝑊.

Proof: Image is a Subspace of the Codomain.

Let 𝑇:𝑉→𝑊 be a linear map. Checking the Subspace Criteria:

  1. Existence of vector additive identity. Since 𝑇(𝟎𝑉)=𝟎𝑊 by Theorem 27, 𝟎𝑊∈im(𝑇).
  2. Closure under vector addition. Let 𝐰1,𝐰2∈im(𝑇). Then there exist 𝐯1,𝐯2∈𝑉 such that 𝑇(𝐯1)=𝐰1 and 𝑇(𝐯2)=𝐰2. By additivity, 𝑇(𝐯1+𝐯2)=𝑇(𝐯1)+𝑇(𝐯2)=𝐰1+𝐰2. Since 𝐯1+𝐯2∈𝑉, it is the preimage of 𝐰1+𝐰2, so 𝐰1+𝐰2∈im(𝑇).
  3. Closure under scalar multiplication. Let 𝐰∈im(𝑇) and let 𝑐∈𝐹. Then there exists 𝐯∈𝑉 such that 𝑇(𝐯)=𝐰. By homogeneity, 𝑇(𝑐𝐯)=𝑐𝑇(𝐯)=𝑐𝐰, so 𝑐𝐰∈im(𝑇). ∎

Notation

Let 𝑓:𝑈→𝑉 be a function. Then for any subset 𝑆⊆𝑈, 𝑓(𝑆) (called the image of 𝑆 under 𝑓) denotes the set of all images of elements in 𝑆 under 𝑓:

𝑓(𝑆)={𝑓(𝑠)∈𝑉|𝑠∈𝑆}⊆im(𝑓).

Definition 32: Rank of a Linear Map

Let 𝑇:𝑉→𝑊 be a linear map. The rank of 𝑇 is the dimension of the image of 𝑇:

rank(𝑇)=dim(im(𝑇)).

Theorem 34: Rank-Nullity Theorem

Let 𝑇:𝑉→𝑊 be a linear map. If 𝑉 is finite-dimensional, then 𝑊 is also finite-dimensional, and the following equation holds:

rank(𝑇)+nullity(𝑇)=dim(𝑉).

Terminology

A transformation 𝑇 is bijective (sometimes also called a one-to-one correspondence) if 𝑇 is both injective and surjective.

Exercises

Exercise 51.
Exercise 52.
Show that the transformation 𝑇:ℝ2→ℝ defined by 𝑇(𝑥,𝑦)=𝑥𝑦 is not linear.
Exercise 53.

Let 𝑇:𝐹∞→𝐹∞ be the transformation defined by:

𝑇(𝑎1,𝑎2,𝑎3,…)=(0,𝑎1,𝑎2,𝑎3,…) for all 𝑎1,𝑎2,𝑎3,…∈𝐹.

Determine, with proof, whether 𝑇 is a linear map.

Exercise 54.
Suppose 𝑇 is a linear map. Show that if the set {𝐯1,𝐯2,…,𝐯𝑘} is linearly independent, then the set {𝑇(𝐯1),𝑇(𝐯2),…,𝑇(𝐯𝑘)} is also linearly independent.
Exercise 55.
Let 𝑉 and 𝑊 be finite-dimensional vector spaces such that dim(𝑉)>dim(𝑊). Show that there does not exist an injective linear map 𝑇∈ℒ(𝑉,𝑊).
Exercise 56.

Let 𝑇,𝑆∈ℒ(𝑉,𝑊) be linear maps. Show that:

ker(𝑇)∩ker(𝑆)⊆ker(𝑇+𝑆).
Exercise 57.

For linear maps 𝑇:𝑈→𝑉 and 𝑆:𝑉→𝑊, both with finite rank, show that:

rank(𝑆∘𝑇)≤min(rank(𝑆),rank(𝑇)).
Exercise 58.
Suppose 𝑉 is a finite-dimensional vector space. Show that there exists a linear map 𝑇:𝑉→𝑉 such that ker(𝑇)=im(𝑇) if and only if dim(𝑉) is even.
Exercise 59.
A linear map 𝑇 is called idempotent iff 𝑇2=𝑇. Suppose that 𝑇:𝑉→𝑉 is a linear map, where 𝑉 is finite-dimensional. Show that if 𝑇 is idempotent, then 𝑉=ker(𝑇)⊕im(𝑇).
Exercise 60.

Suppose 𝑉 is a finite-dimensional vector space, and 𝑇∈ℒ(𝑉). Show that:

  1. ker(𝑇)⊆ker(𝑇2)⊆ker(𝑇3)⊆⋯
  2. If for some 𝑘, ker(𝑇𝑘)=ker(𝑇𝑘+1), then ker(𝑇𝑘)=ker(𝑇𝑚) for all 𝑚≥𝑘
Exercise 61.

Let 𝑉 and 𝑊 be finite-dimensional vector spaces. The set of all linear maps from 𝑉 to 𝑊, ℒ(𝑉,𝑊), is itself a vector space under pointwise addition and scalar multiplication.

Fix some vector 𝐯0∈𝑉 to define the following function Φ:ℒ(𝑉,𝑊)→𝑊:

Φ(𝑇)=𝑇(𝐯0).
  1. Show that Φ is a linear map.
  2. Find nullity(Φ) and rank(Φ) in terms of dim(𝑉) and dim(𝑊).
  3. Find dim(ℒ(𝑉,𝑊)) in terms of dim(𝑉) and dim(𝑊).