Chapter 5

Linear Transformations and Matrices

5.1 Linear transformations from Rn to Rm

When A is an m × n matrix and v is a vector of dimension n then Av is a vector of

dimension m. Since matrix multiplication of A by a vector is a linear operation it

follows that the matrix product Av produces a natural linear transformation from Rn

to Rm.

Theorem:. If A is an m × n matrix (m rows and n columns), and T : Rn → Rm is

defined by T(v) = Av, for every v ∈ V ; then T is a linear transformation.

Proof. Since A has size m × n, for any vector v ∈ Rn we have that the product Av

exists and is anm-dimensional vector, that is, T(v) = Av ∈ Rm. We have to prove that

T is a linear transformation. Let v1, v2 ∈ Rn, and λ be a scalar. We have:

T(v1 + v2) = A(v1 + v2) = A(v1) +A(v2) ⇒ T(v1 + v2) = T(v1) + T(v2),

and:T(λ v1) = A(λ v1) = λAv1 ⇒ T(λ v1) = λ T(v1).

�

The matrix-based linear transformation T , defined by T(v) = Av is often denoted as

TA.

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There is a reverse connection: for every linear transformation T : Rn → Rm, there is a

matrix A such that T(v) = Av.

Theorem: Suppose that T : Rn → Rm is a linear transformation. Let {e1, . . . , en}

denote the standard basis of Rn and letA be the matrix whose columns are the vectors

T(e1), . . . , T(en). Then for every u ∈ Rn,

T(u) = Au.

Proof: Let u = (u1, . . . , un)′ ∈ Rn. Then

u = u1e1 + · · ·+ unen.

By the linearity properties of T we have

T(u) = u1T(e1) + · · ·+ unT(en),

which is a linear combination of the columns of A, so we have

T(u) = Au,

as we wanted. �

Thus, to each matrix A there corresponds a linear transformation TA, and to each

linear transformation T there corresponds a matrix, that we often denote as AT . Note

that it is enough to know the image of the vectors in the basis to compute the matrix

of the transformation and thus to allow determining the image of any vector.

Example: Let T : R3 → R3 be the linear transformation given by

T((u1, u2, u3)′) = (u1 + u2 + u3, u1 − u3, u1 + 2u2 − 3u3)

′ .

The image of any vector is easily computed by substituting its components into the

definition of T . The matrix of this transformation is found by computing the image of

the standard basis vectors:

T(e1) = (1, 1, 1) ′; T(e2) = (1,−1, 2) ′; T(e3) = (1, 0,−3) ′,

2

and thus

A =

1 1 1

1 0 −1

1 2 −3

.Notice that the entries of the matrix A are just the coefficients of u1, u2, u3 in the

definition of T . �

Example: If A is them× n zero matrix then TA is the zero linear transformation that

maps every v ∈ Rn into the zero vector in Rm. �

Example: If n = m and In is the n× n identity matrix, then TIn is the linear transfor-

mation, Rn → Rn that maps every vector into itself and is called the identity linear

transformation. �

Example: Consider the matrix

A =

2 0 −2

4 1 −1

,and the linear transformation associated to it, TA : R3 → R2.

The kernel of TA is given by

ker(TA) = {u = (u1, u2, u3)′ ∈ R3 : Au = 0}.

This set includes all vectors satisfying 2u1−2u3 = 0 and 4u1+u2−u3 = 0. By solving

this system, we get that the kernel is given by

ker(TA) = {α(1,−3, 1) ′ : α ∈ R},

which is, obviously, a vector space of dimension 1, spanned by vector (1,−3, 1) ′.

The range of TA is given by

R(TA) = {v = (v1, v2)′ ∈ R2 : v1 = 2u1 − 2u3, v2 = 4u1 + u2 − u3, u1, u2, u3 ∈ R},

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and its dimension is easily computed as

rank(TA) = n− nullity(TA) = 3− 1 = 2.

Finally, notice that the rank of A is also 2. �

Theorem: If TA : Rn → Rm is the linear transformation associated to them×nmatrix

A then:

1. rank(TA) = rank(A).

2. The maximum possible value for rank(TA) is equal to the min(m,n).

Theorem: If A is an n × n matrix and TA : Rn → Rn is the corresponding linear

transformation then each of the following statements is equivalent (if one is true so

are all of the others):

1. A is non-singular (it has an inverse).

2. The range of TA is Rn.

3. TA is one-to-one.

4. The linear transformation T−1A exists and its matrix is A−1.

Notice that a one-to-one linear transformation corresponds exactly to a matrix A

that is non-singular.

5.2 Change of basis as a linear transformation

Recall from chapter 3 that if B = {v1, . . . , vn} is a basis of Rn, the matrix (v1| . . . | vn),

whose columns are the basis vectors in B is the transition matrix from coordinates

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with respect to B to the standard coordinates. Also, the inverse of this matrix is the

transition matrix from standard coordinates to coordinates with respect to the basis B.

If P is the transition matrix from coordinates in a basis B of Rn to standard coordi-

nates, then considered as the matrix of a linear transformation TP(x) = Px, the linear

transformation actually maps the standard basis vectors ei to the new basis vectors

vi, that is, TP(ei) = vi.

Example: Suppose we want to change the basis in R2 by a rotation of the axes π4

radians counterclockwise. What are the coordinates of a vector with respect to this

new basis B = {v1, v2}? Consider the linear transformation T that transforms the

vectors in the standard basis e1, e2 into the new basis vectors:

v1 =(

cosπ

4, sin

π

4

) ′=

(1√2,1√2

) ′

and

v2 =(− sin

π

4, cos

π

4

) ′=

(−1√2,1√2

) ′

that is, T(e1) = v1 and T(e2) = v2. The matrix associated to the linear transformation

is the transition matrix

P =

1√2

− 1√2

1√2

1√2

which corresponds to the transition matrix from coordinates with respect to B to stan-

dard coordinates. Then the coordinates of a vector vwith respect to the new basis are

given by

P−1v .

The inverse of a counterclockwise rotation is a clockwise rotation, so we have:

P−1 =

cos(−π4

)− sin

(−π4

)sin(−π4

)cos(−π4

) =

1√2

1√2

− 1√2

1√2

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and we can compute the new coordinates of any vector v.

Consider the vector v = (1, 1) ′; its coordinates with respect to the basis B is given by

the product

P−1x =

1√2

1√2

− 1√2

1√2

1

1

=

√20

Alternatively we could have noticed that v =

√2v1, and we get that the coordinates

of vwith respect to B are (√2, 0) ′. �

5.2.1 Change of coordinates from basis B to basis B ′

Given a basis B of Rn with transition matrix PB, and another basis B ′ with transition

matrix PB ′ , we can obtain the transition matrix to change from coordinates in the basis

B to coordinates in the basis B ′ directly from these two matrices, instead of computing

the coordinates of one set of vectors with respect to the other.

To do that, we first change from coordinates with respect to B to standard coordinates

using PB, and then change from standard coordinates to coordinates with respect to

B ′ using P−1B ′ . The matrix M = P−1B ′ · PB is the transition matrix from coordinates with

respect to B to coordinates with respect to B ′.

Example: Consider the two basis of R2 given by B = {(1, 2) ′, (−1, 1) ′} and B ′ =

{(3, 1) ′, (5, 2) ′}. The transition matrices from coordinates with respect to B and B ′ to

standard coordinates are given by PB =

1 −1

2 1

and PB ′ =

3 5

1 2

, respec-

tively. If we want to find the coordinates of vector u with respect to B ′ and we know

that its coordinates with respect to B are given by (4,−1) ′, we can simply compute

TM(u) =Mu = P−1B ′ PB u =

2 −5

−1 3

1 −1

2 1

4

−1

=

−25

16

,to obtain those coordinates. �

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