Chapter 4 Linear Transformations
4.2 Matrix Representations of Linear
Transformations
2.3 Additional topics
Theorem
If 𝐿 is a linear transformation mapping ℝ𝑛 into ℝ𝑚 , then there is an 𝑚 × 𝑛
matrix 𝐴, such that
𝐿(𝐱) = 𝐴𝐱
for each x ∈ ℝ𝑛 . In fact, the 𝑗𝑡ℎ column vector of 𝐴 is given by
𝒂𝑗 = 𝐿(𝒆𝑗 ) 𝑗 = 1,2, … , 𝑛
Proof
For 𝑗 = 1,2, … , 𝑛, define 𝒂𝑗 = 𝐿(𝒆𝑗 ), and let 𝐴 = 𝑎𝑖𝑗 = (𝒂1 , 𝒂2 , … , 𝒂𝑛 ).
If 𝐱 = 𝑥1 𝒆1 + 𝑥2 𝒆2 + ⋯ + 𝑥𝑛 𝒆𝑛 is an arbitrary element of ℝ𝑛 , then
𝐿 𝐱 = 𝑥1 𝐿 𝒆1 + 𝑥2 𝐿 𝒆2 + ⋯ + 𝑥𝑛 𝐿 𝒆𝑛 = 𝑥1 𝑎1 + 𝑥2 𝑎2 + ⋯ + 𝑥𝑛 𝑎𝑛
= 𝐴𝐱
Theorem
The theorem tells us how to construct the matrix 𝐴 corresponding to a
linear transformation 𝐿. To get the 𝑗𝑡ℎ column of 𝐴, see what 𝐿 does to
the basis element 𝒆𝑗 of ℝ𝑛 . Since the standard basis elements
𝒆1 , 𝒆2 , … , 𝒆𝑛 (the column vectors of the 𝑛 × 𝑛 identity matrix) are used
for ℝ𝑛 , we refer to 𝐴 as the standard matrix representation of 𝐿.
Example
Find the standard matrix representation for the linear transformation
𝑥+𝑦
𝑥
𝑥−𝑦
2
4
𝐿: ℝ → ℝ given by 𝐿 𝑦 =
.
2𝑥 + 2𝑦
𝑥
Solution

First compute 𝐿 𝒆1 and 𝐿 𝒆2 :
1
= (1,1,2,1)𝑇
0
0
𝐿 𝒆2 = 𝐿
= (1, −1,2,0)𝑇
1
Therefore, we have standard matrix representation:
1 1
𝐴 = 1 −1
2 2
1 0
𝐿 𝒆1 = 𝐿

Example
Rotation Matrix for ℝ2
Let 𝐿: ℝ2 → ℝ2 be the linear operator that rotates each vector by an angle
𝜃 in the counterclockwise direction. Then a matrix representation for 𝐿 is
given by
cos 𝜃 −sin 𝜃
𝐴=
sin 𝜃 cos 𝜃
Matrix Representation Theorem(MRT)
If 𝐸 = {𝐯1 , 𝐯2 , … , 𝐯n } and 𝐹 = {𝐰1 , 𝐰2 , … , 𝐰m } are ordered bases for
vector spaces 𝑉 and 𝑊, respectively. Then corresponding to each
linear transformation 𝐿: 𝑉 → 𝑊, there is an 𝑚 × 𝑛 matrix 𝐴 such that
[𝐿(𝐯)]𝐹 = 𝐴[𝐯]𝐸 for each v ∈ 𝑉
𝐴 is the matrix representing 𝐿 relative to the ordered bases 𝐸 and 𝐹.
In fact,
𝒂𝑗 = [𝐿(𝐯𝑗 )]𝐹 𝑗 = 1,2, … , 𝑛
Proof of MRT(outline)
Firstly: for ∀ 𝐯 ∈ 𝑉, there is a unique representation under the basis
𝐸 = {𝐯1 , 𝐯2 , ⋯ , 𝐯n }, we can write it as
↔ 𝒙= 𝐯𝐸
𝐯 = 𝑥1 𝐯1 + 𝑥2 𝐯2 + ⋯ + 𝑥𝑛 𝐯𝑛
Similarly, 𝐿 𝐯 ∈ 𝑊,𝐹 = 𝐰1 , 𝐰2 , … , 𝐰m is a basis set of W, we have
L(𝐯) = 𝑦1 𝒘1 + 𝑦2 𝒘2 + ⋯ + 𝑦𝑚 𝒘𝑚
↔
𝒚 = L(𝐯)
𝐹
According to the definition of linear transformation ,
𝐿 𝛼𝐯1 + 𝛽𝐯2 = 𝛼𝐿 𝐯1 + 𝛽𝐿(𝑣2 )
L 𝐯 = 𝐿 𝑥1 𝐯1 + 𝑥2 𝐯2 + ⋯ + 𝑥𝑛 𝐯𝑛 = 𝑥1 𝐿(𝐯1 ) + 𝑥2 𝐿(𝐯2 ) + ⋯ + 𝑥𝑛 𝐿(𝐯𝑛 )
Proof of MRT(outline)
L 𝐯 = 𝐿 𝑥1 𝐯1 + 𝑥2 𝐯2 + ⋯ + 𝑥𝑛 𝐯𝑛 = 𝑥1 𝐿(𝐯1 ) + 𝑥2 𝐿(𝐯2 ) + ⋯ + 𝑥𝑛 𝐿(𝐯𝑛 )
𝐿 𝐯𝑗 ∈ 𝑊,𝐹 = 𝐰1 , 𝐰2 , … , 𝐰m is a basis set of W, we have
L(𝐯j ) = 𝑎1𝑗 𝒘1 + 𝑎2𝑗 𝒘2 + ⋯ + 𝑎𝑚𝑗 𝒘𝑚
↔
𝐚j = L(𝐯j )
𝐹
j=1,2,…,n
L 𝐯 = 𝑥1 𝐿(𝐯1 ) + 𝑥2 𝐿(𝐯2 ) + ⋯ + 𝑥𝑛 𝐿(𝐯𝑛 )
𝑛
𝑛
L 𝐯 =
𝑥𝑗 𝐿(v𝑗 ) =
𝑗=1
𝑚
=
𝑥𝑗
𝑗=1
𝑛
𝑚
𝑎𝑖𝑗 𝐰𝑖 =
𝑖=1
𝑚
𝑛
𝑎𝑖𝑗 𝐰𝑖 𝑥𝑗 =
𝑗=1 𝑖=1
𝑎𝑖𝑗 𝐰𝑖 𝑥𝑗
𝑖=1 𝑗=1
𝑛
𝐰𝑖
𝑖=1
𝑚
𝑎𝑖𝑗 𝑥𝑗
𝑗=1
=
L(𝐯) = 𝑦1 𝒘1 + 𝑦2 𝒘2 + ⋯ + 𝑦𝑚 𝒘𝑚
Proof of MRT(outline)
Because, the coordinate for a given basis set is unique, compare
𝑚
𝑛
𝐰𝑖
𝑖=1
𝑎𝑖𝑗 𝑥𝑗
=
L(𝐯) = 𝑦1 𝒘1 + 𝑦2 𝒘2 + ⋯ + 𝑦𝑚 𝒘𝑚
𝑗=1
𝑛
We have
𝑦𝑖 =
𝑎𝑖𝑗 𝑥𝑗 = 𝑎𝑖1 𝑥1 + 𝑎𝑖2 𝑥2 + ⋯ + 𝑎𝑖𝑛 𝑥𝑛 ,
i = 1,2, … , 𝑚
𝑗=1
That is: 𝒚 = 𝐴𝒙, for 𝐴 ∈ 𝑅𝑚×𝑛
𝐿 𝐯
𝐹
Recall that 𝒙 = 𝐯
=𝐴𝐯
𝒚 = L(𝐯)
𝐸
𝐸
L(𝐯j ) = 𝑎1𝑗 𝒘1 + 𝑎2𝑗 𝒘2 + ⋯ + 𝑎𝑚𝑗 𝒘𝑚
↔
𝐚j = L(𝐯j )
𝐹
j=1,2,…,n
𝐹
Theorem
L(𝐯j ) = 𝑎1𝑗 𝒘1 + 𝑎2𝑗 𝒘2 + ⋯ + 𝑎𝑚𝑗 𝒘𝑚
𝑎1𝑗
𝑎2𝑗
𝐿 𝐯𝑗 = [𝑤1 , 𝑤2 , … , 𝑤𝑚 ]
⋮
𝑎𝑚𝑗
↔
𝐚j = L(𝐯j )
𝐹
j=1,2,…,n
= 𝐹𝐚j
If 𝐸 = {𝐯1 , 𝐯2 , … , 𝐯n } and 𝐹 = {𝐰1 , 𝐰2 , … , 𝐰m } be ordered bases for ℝ𝑛
and ℝ𝑚 , respectively. If 𝐿: ℝ𝑛 → ℝ𝑚 is a linear transformation and 𝐴 is the
matrix representing 𝐿 w.r.t. 𝐸 and 𝐹, then
𝒂𝑗 = 𝐹 −1 𝐿 𝒗𝑗
𝑗 = 1,2, … , 𝑛
Where 𝐹 = {𝐰1 , 𝐰2 , … , 𝐰m }
Example
Let 𝐿 be the linear transformation mapping ℝ3 to ℝ2 defined by
𝐿 𝐱 = 𝑥1 𝒃1 + (𝑥2 + 𝑥3 )𝒃2
For each 𝐱 ∈ ℝ3 , where
1
−1
𝒃1 =
and 𝒃2 =
1
1
Find the matrix 𝐴 representing 𝐿 with respect to the ordered bases
{𝒆1 , 𝒆2 , 𝒆3 } and {𝒃1 , 𝒃2 }.
Proof

𝐿 𝒆1 = 1𝒃1 + 0𝒃2 , 𝐿 𝒆2 = 0𝒃1 + 1𝒃2 , 𝐿 𝒆3 = 0𝒃1 + 1𝒃2

The 𝑗𝑡ℎ column of 𝐴 is determined by the coordinates of 𝐿 𝒆𝑗 w.r.t.
{𝒃1 , 𝒃2 }. Thus
1 0 0
𝐴=
0 1 1
Corollary
If 𝐴 is the matrix representing the linear transformation 𝐿: ℝ𝑛 → ℝ𝑚 with
respect to the bases
𝐸 = {𝒖1 , 𝒖2 , … , 𝒖𝑛 } and 𝐹 = {𝒃1 , 𝒃2 , … , 𝒃𝑚 }
then the reduced row echelon form of (𝒃1 , 𝒃2 , … , 𝒃𝑚 |𝐿 𝒖1 , … , 𝐿 𝒖𝑛 ) is
𝐼|𝐴 .
Proof

Let 𝐵 = (𝒃1 , 𝒃2 , … , 𝒃𝑚 ). The matrix 𝐵|𝐿 𝒖1 , … , 𝐿 𝒖𝑛 is row
equivalent to
𝐵−1 𝐵|𝐿 𝒖1 , … , 𝐿 𝒖𝑛 = 𝐼 𝐵 −1 𝐿 𝒖1 , … , 𝐵 −1 𝐿 𝒖𝑛
= 𝐼|𝒂1 , … , 𝒂𝑛 = 𝐼|𝐴
Example
Let 𝐿: ℝ2 → ℝ3 be the linear transformation defined by 𝐿 𝐱 =
(𝑥2 , 𝑥1 + 𝑥2 , 𝑥1 − 𝑥2 )𝑇
Find the matrix representations of 𝐿 with respect to the ordered bases
{𝒖1 , 𝒖2 } and {𝒃1 , 𝒃2 , 𝒃3 }, where
𝒖1 = (1,2)𝑇 , 𝒖2 = (3,1)𝑇
and
𝒃1 = (1,0,0)𝑇 , 𝒃2 = (1,1,0)𝑇 , 𝒃3 = (1,1,1)𝑇
Solution
First compute 𝐿 𝒖1 = (2,3, −1)𝑇 , 𝐿 𝒖2 = (1,4,2)𝑇
 Transform matrix (𝒃1 , 𝒃2 , 𝒃3 |𝐿 𝒖1 , 𝐿 𝒖2 ) to the row echelon
form

1 1 1 2 1
1 0
0 1 1 3 4 → 0 1
0 0 1 −1 2
0 0
0 −1 −3
0 4
2
1 −1 2
Revision Exercises
Computer Graphics and Animation
Computer Graphics and Animation
• Four Basic Transformation Functions
1. Rotation
2. Dilation and contraction (Scaling)
3. Reflection
•
•
w.r.t. y-axis
w.r.t. x-axis
4. Translation (transposition)
What are the transformation functions?
Are they ALL linear transformation?
1. Rotation (Already mentioned)
Rotation Matrix for ℝ2
Let 𝐿R: ℝ2 → ℝ2 be the linear operator that rotates each vector by an angle
𝜃 in the counterclockwise direction.
Then a matrix representation for 𝐿R is given by
cos 𝜃 −sin 𝜃
A𝑅 =
sin 𝜃 cos 𝜃
e2
e1
2. Dilation and contraction (Scaling)
Scaling Matrix for ℝ2
Let 𝐿𝑆: ℝ2 → ℝ2 be the linear operator that dilate or contract the any
vector by c (scalar in ℝ ).
To calculate A:
1) We evaluate 𝐿𝑆(e1) and 𝐿𝑆(e2)
𝐿𝑆 𝑒1 = 𝐿𝑆
1
0
=
𝑐
0
2) Group the result vectors to form A.
A𝑆 =
3) Check linearity
𝑐
0
𝐿𝑆 𝑒2 = 𝐿𝑆
0
1
=
0
𝑐
𝐿𝑆 𝑥 + 𝑦 = 𝑐 𝑥 + 𝑦 = 𝑐𝑥 + 𝑐𝑦 = 𝐿𝑆 𝑥 + 𝐿𝑆 𝑦
𝐿𝑆 𝑘𝑥 = 𝑘𝑐 𝑥 = 𝑘𝐿𝑆 𝑥
i.e. Linear Transformation
0
𝑐
3. Reflection (Mirroring)
3A) Reflection Matrix for ℝ2 (y-axis)
Let 𝐿𝑅𝑦: ℝ2 → ℝ2 be the linear operator Reflect any vector
w.r.t. y-axis
To calculate A:
1) We evaluate 𝐿𝑅𝑦(e1) and 𝐿𝑅𝑦(e2)
𝐿𝑅𝑦 𝑒1 = 𝐿𝑅𝑦
1
0
=
−1
0
𝐿𝑅𝑦 𝑒2 = 𝐿𝑅𝑦
2) Group the result vectors to form A.
A𝑅𝑦 =
3) Check linearity
0
1
=
0
1
−1 0
0 1
𝑥 +𝑦
−1
𝐿𝑅𝑦 𝑥 + 𝑦 = 𝐴𝑅𝑦 𝑥1 + 𝑦1 =
0
2
2
−𝑥1 −𝑦1
= 𝑥 + 𝑦 = 𝐿𝑅𝑦 𝑥 + 𝐿𝑅𝑦 𝑦
1
𝐿𝑅𝑦 𝑐𝑥 = 𝐴𝑅𝑦
−𝑥1 − 𝑦1
0 𝑥1 + 𝑦1
=
𝑥2 + 𝑦2
1 𝑥2 + 𝑦2
𝑐𝑥1
𝑥1
𝑐𝑥2 = 𝑐𝐴𝑅𝑦 𝑥2 = 𝑐𝐿𝑅𝑦 𝑥
i.e. Linear Transformation
4. Translation (Transposition)
4) Translation Matrix for ℝ2
Let 𝐿𝑇: ℝ2 → ℝ2 be the linear operator Translation of any
vector to 𝒙′ = (𝑥1′ , 𝑥2′ )
To calculate A:
1) We evaluate 𝐿𝑇(e1) and 𝐿𝑇(e2)
𝐿 𝑇 𝑒1 = 𝐿 𝑇
1
0
1 + 𝑥1′
=
𝑥2′
2) Group the result vectors to form A.
1 + 𝑥1′
A𝑇 =
𝑥2′
3) Check linearity
𝐿 𝑇 𝑒2 = 𝐿 𝑇
0
1
𝑥1′
1 + 𝑥2′
𝐿 𝑇 𝑥 + 𝑦 = (𝑥 + 𝑦) + 𝑥 ′ = 𝑥 + 𝑦 + 𝑥 ′
𝐿 𝑇 𝑥 + 𝐿 𝑇 𝑦 = 𝑥 + 𝑥 ′ + 𝑦 + 𝑥 ′ = 𝑥 + 𝑦 + 2𝑥 ′
i.e. non-Linear Transformation
𝑥1′
=
1 + 𝑥2′
Q 7. p188
Let
1
1
1
𝑦1 = 1 , 𝑦2 = 1 , 𝑦2 = 0
1
0
0
and let I be the identity operator on ℝ3 .
a) Find the coordinates of I(e1), I(e2), I(e3) w.r.t. {y1,y2,y3}
b) Find a matrix A such that Ax is the coordinate vector of x w.r.t.
{y1,y2,y3}.
Solution:
 In order to simplify the process for the evaluation of the coordinates
of I(e1), I(e2), I(e3) w.r.t. {y1,y2,y3}.
 It is wiser to calculate the transformation matrix A first (i.e. solution of
b) and base on it to calculate these coordinates w.r.t. y.
i. Identity operator is given by:
𝛪 𝑣 =𝑣
So we have:
1
0
0
𝛪 𝑒1 = 𝑒1 = 0 , 𝛪 𝑒2 = 𝑒2 = 1 , 𝛪 𝑒3 = 𝑒3 = 0
0
0
1
By using corollary for the reduced row echelon form of :
[B | L(u)]  [ I | A] to find A. So we have:
1 1 1 1 0 0
1 1 0 0 1 0
1 0 0 0 0 1
𝑅1 ↔𝑅3
1 0 0 0 0 1
1 1 0 0 1 0
1 1 1 1 0 0
1 0 0 0 0
1
0 1 0 0 1 −1 . So. 𝐴 =
0 0 1 1 −1 0
𝑅2 −𝑅1
𝑅3 −𝑅1
0 0
0 1
1 −1
1
0
0
1
−1
0
0 0 0 0 1
1 0 0 1 −1
1 1 1 0 −1
𝑅3 −𝑅2
Solution:
ii) And hence the coordinates are:
𝛪 𝑒1
0
= 𝐴𝑒1 = 0
1
0
1
−1
1
−1
0
1
0
0 = 0
0
1
𝛪 𝑒2
0
= 𝐴𝑒2 = 0
1
0
1
−1
1
−1
0
0
0
1 = 1
0
−1
0 0
1 0
1
𝛪 𝑒3 = 𝐴𝑒3 = 0 1 −1 0 = −1
1 −1 0 1
0
To verify the result, you can check:
w.r.t. {y1,y2,y3}.
1
1
1
1
= 0 1 +0 1 +1 0 = 0 = 𝑒1
1 𝑒
0 𝑒
0 𝑒 0 𝑒
𝑦
1
1
0
0
1
1 = 0 1 +1 1 −1 0 = 1 = 𝑒2
−1 𝑦
1 𝑒
0 𝑒
0 𝑒 0 𝑒
1
1
1
1
0
−1 = 1 1 −1 1 + 0 0 = 0 = 𝑒3
0 𝑦
1 𝑒
0 𝑒
0 𝑒 1 𝑒
0
0
1
Homework
Ex. 4.2
3(a, c), 5(a, d), 9(a, b), 15.
Extra: 13, 19.