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2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

1 Matrices and determinants
1
...
The numbers
are called the elements of the matrix
...

Unless specifically stated otherwise, we will assume that vectors are column vectors
...
1

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

A general real matrix, A ∈ Rm×n with m × n elements is of
the form


a11 a12 a13
...
a2n 


A =  a31 a32 a33
...


...


...


...


...


...

am1 am2 am3
...

(ii) The second index represents the column
...

Notation/Conventions:
Use lowercase boldface (or underlined) letters for vectors
a b c

(or

a,

b,

c)

Use uppercase boldface (or underlined) letters for matrices
A

B C

(or

A,

B,

C)

Refer to the respective elements by lowercase letters with the
appropriate number of indices e
...

bi
aij

is a vector element
is a matrix element

1
...
2 Special matrices
The unit matrix, I, is a square matrix whose only non-zero
elements are on the diagonal and are equal to one, e
...



1 0 0
...
0 0 




1 0 0


 0 0 1
...


...


...
1 0 
0 0 0
...
g
...
0 0
 0 0 0
...
0 0 
0 =  0 0 0  , 0 = 
...


...



0 0 0
 0 0 0
...
0 0
A diagonal matrix only has non-zero elements on the main
diagonal
...
g
...

0
0


 0 d
...


...


...


...

D =  0 d22 0  , D = 
...


...


...
dn−1,n−1 0 
0 0
...


1
...
3 Matrix algebra
1
...
1

Matrix equality

Two matrices are equal if they have the same size and if their
corresponding elements are identical , i
...

A=B
if and only if
aij = bij

for i = 1,
...
, n

1
...
2 Matrix addition
Two matrices can only be added if they have the same size
...

We add matrices by adding their corresponding elements, i
...

A=B+C
is obtained (element-wise) via
aij = bij + cij

for i = 1,
...
, n

Example
A=

A+B=

1 2 3
5 6 7

B=

1 + 10 2 + 1 3 + 23
5 + 5 6 + 16 7 + 3

1
...
3
...

The result is a matrix of the same size
...
, m; j = 1,
...
3
...

Let a ∈ Rn and b ∈ Rn be two column vectors with n real
elements each:


 
a1
b1
a 
b 
 2
 2


 
a =  a3  ,
b =  b3 
...


...


...


...
5

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

Then the dot product (or, scalar product) of a and b is defined
as
a · b = a1b1 + a2b2 + a3b3 + · · · + anbn
...
an
...


...

an
Then the product of the row vector aT and the column vector
b can be defined as follows:
def

aT b = a · b = a1b1 + a2b2 + a3b3 + · · · + anbn

(2)

Example






2
5
5
a =  −1  , b =  2  ; aT b = 2 −1 3  2 
3
−4
−4

 

5
2
=  −1  ·  2  = 10 − 2 − 12 = −4
...
6

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

Now we can define a matrix – vector multiplication
...


m×1

This implies that the number of columns of the matrix must
be equal to the number of rows in the column vector!
The elements of the vector b are calculated by taking
dot products of the rows of the matrix with the column vector:
if rows of A ∈ Rm×n are vectors aT , aT ,
...


...

aT
m



⇒

aT
1
aT
2



x


x

Ax = 
...


...

i
Element wise this is
n

aij xj = bi

for i = 1,
...


j=1

1
...

3 



4
 ( 5 6 7 ) 2  
4

The vector products are simply the dot products of the respective row aT with the column vector x
i
Ax =

1×3+2×2+3×4
5×3+6×2+7×4

and hence
b=

3 + 4 + 12
15 + 12 + 28

1
...


2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

1
...
5 Matrix – Vector multiplication using Falk’s scheme
For practical computations use Falk’s scheme and evaluate
Ax = b
in tabular form as follows:

x
A b
Example



1 2 3
5 6 7

A=

,



3
x=2
4

Evaluate Ax as follows:
3
2
4
1 2 3
5 6 7
Now place the elements of the resulting vector at the ‘intersection’ of the column vector and the rows of the matrix:

1 2 3
5 6 7

3
2
4
1×3+2×2+3×4
=
5×3+6×2+7×4

1
...
3
...
e
...


m×n n×p

m×p

This implies that the number of columns of the first matrix
must be equal to the number of rows in the second matrix
...
10

AD



2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

The product matrix of two matrices is obtained by taking dot
products of the rows of the left matrix with the columns of the
right matrix
...
, aT ∈ Rn
1
2
m
n×p
n
and the columns of B ∈ R
are b1, b2,
...
Let


T
a1
 T 
a 
A = 
...
bp

...

aT
m
then the matrix matrix product is


aT
1
aT
2

b1
b1

aT
1
aT
2

b2
...


...


...


aT
1
aT
2



bp


bp 

m×p
AB = 
...
∈R

...


...

aT b1 aT b2
...
, m; k = 1,
...


j=1

1
...
That is


1×4+2×2 1×3+2×1 1×2+2×1

AB = 
3×4+4×2 3×3+4×1 3×2+4×1
Giving the final result



C=

8

5

4

20

13

10

1
...
3
...
13

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

1
...
8 Differences from multiplication with numbers
(i) Matrix multiplication is not commutative
AB = BA
Example
1 2
4 1

2 2
3 1

1×2+2×3 1×2+2×1
4×2+1×3 4×2+1×1

=
=

2 2
3 1

1 2
4 1

8 4
11 9
2×1+2×4 2×2+2×1
3×1+1×4 3×2+1×1

=
=

10 6
7 7

We must be careful how we multiply out!
(ii) AB = 0 does not imply A = 0, B = 0 or BA = 0
Example
1 1
2 2

−1 1
1 −1

−1 1
1 −1

1 1
2 2

1
...
15

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

1
...
, m; j = 1,
...
am1
 a a a
...
am3 

...


...


...


...


...


...
amn
Example



A=


1 5
AT =  2 6 
3 7

1 2 3
5 6 7

If A = AT then A is a symmetric matrix, e
...



3 2 −1
A= 2 7 0
−1 0 8
The matrix transpose also satisfies the following rules:
i) (AT )T = A for any matrix A;
ii) (A + B)T = AT + BT and (AB)T = BT AT ,
provided that matrices A and B have compatible dimensions
...
16

2E1: Linear Algebra | Lecture Notes

<

§1 Matrices and matrix algebra

Proof of the identity (AB)T = BT AT ∗

If the rows of A ∈ Rm×n are the vectors aT , aT ,
...
, bp ∈ R
...


...


b1 aT
2
b2 aT
2

b1
...


...


...


a T b1
m
T
a m bp

...


...
aT bp
1
2
m
For column vectors ai and bj vector multiplication aT bj is
i
defined as the dot product between ai and bj
...


...


...


...
bT am
p
p
p

since



bT
1
bT
2




 = B T AT








B =
...


...

a2
...


...


...
am
...

1
...
5 Determinant of a matrix
The determinant of a 2 × 2 matrix
A=

a11 a12
a21 a22

is written det A or |A| or
a11 a12
= a11a22 − a12a21
a21 a22
Example
A=

1 2
4 −7

,

det A = −7 − 8 = −15
...
These are called the minors
of A
...
18

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

Example
a11 a12 a13
m12 = a21 a22 a23
a31 a32 a33

=

a21 a23
a31 a33

is obtained by suppressing the elements in row 1 and column
2 of matrix A
...
If is given by the formula
cij = (−1)i+j mij
where the minor is the determinant of order (n − 1) × (n − 1)
formed by deleting the column and row containing aij
...
19

= a22a33 − a32a23

= −a11a32 + a31a12

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

General determinant
The value of an n × n determinant equals the sum of the products of the elements in any row (or column) and their cofactors, i
...

n

|A| =

aij cij ,

for

i = 1,
...
, n − 1, or n

j=1

or
n

|A| =
i=1

Example
For a 3 × 3 matrix
det A = a11c11 + a12c12 + a13c13

(1st row)

or
det A = a12c12 + a22c22 + a32c32

(2nd column)

Points to note:
• the determinant det A is equal to zero if
(i) rows or columns of A are multiples of each other,
(ii) rows or columns are linear combinations of each other,
(iii) entire rows or columns are zero;
if det A = 0 the matrix A is called a singular matrix;
• for any square matrices A and B there holds
det A = det(AT ),

det(AB) = det(A) det(B)
...

1
...
6 The matrix inverse
The inverse a−1 of a scalar (=a number) a is defined by
a a−1 = 1
...

Note: if A−1 exists then
det(A) det(A−1) = det(AA−1) = det I = 1
...

Example
The inverse of

3 2
7 5

A=
is given by

5 −2
−7 3

B = A−1 =
since

5
−2
−7
3
AB =
3 2 (3 × 5 − 2 × 7) (−3 × 2 + 2 × 3)
7 5 (7 × 5 − 5 × 7) (−7 × 2 + 5 × 3)
which gives
AB =

1 0
0 1

as required
...
21

=I

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

The matrix inverse can be computed as follows
1
...
Find the cofactors of all elements in A and form a new
matrix C of cofactors, where each element is replaced by
its cofactor
...
The inverse of A is now given as
A

−1

CT
=
det A

Note: the inverse A−1 exists if (and only if) det A = 0
...

3 −2 −1
det A = 1

1 2
−3 2
−3 1
− (−1)
+2
−2 −1
3 −1
3 −2

= 1 × 3 + 1 × (−3) + 2 × 3
=6
Since the determinant is nonzero an inverse exists
...
22

2E1: Linear Algebra | Lecture Notes

§1 Matrices and matrix algebra

Calculate the matrix of minors

1 2
−3 2
 −2 −1
3 −1


 −1 2
1 2
M=
 −2 −1
3 −1


 −1 2
1 2
1 2
−3 2


3 −3 3
=  5 −7 1 
−4 8 −2


−3 1
3 −2 


1 −1 

3 −2 


1 −1 
−3 1

Modify the signs according to whether i + j is even or odd to
calculate the matrix of cofactors


3 3 3
C = −5 −7 −1
...

6
6
3 −1 −2

To check that we have made no mistake we can compute

 


1 0 0
1 −1 2
3 −5 −4
1
A−1A = 3 −7 −8 −3 1 2  = 0 1 0
...


1

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