this presentation is about linear algebra of matrix

1. Matrix and linear algebra
Introduced by:
Khalid Jawad Kadhim
2022-2023

2. 1. Propreties of Determinait.
2. Row Echelon Form (REF).
3. Reduced Row Echelon Form (RREF).
4. Rank of the Matrix.
5. Kernel of Matrix.
6. Image of the Matrix.
7. Egienvalue & Egienvector.
8. Gradient.
KEY TERMS

3. 1. If all the elements in any row (column) equal zero,
the determinant equals zero.
2. If all the elements of any row(column) are
multiplied by a constant c, the value of the
determinant is multiplied by c.
14
)]
4
(
2
)
5
(
3
[
2
|
|
where
,
5
4
)
2
(
2
)
3
(
2
5
4
4
6
















 A
A
PROPERTIES OF DETERMINANTS

4. 3. The value of the determinant is not changed by adding any
row (column) multiplied by a constant c to another row
(column).
4. If any two rows (columns) are interchanged, the sign of the
determinant is changed.
7
)]
4
(
2
)
5
(
3
|
|
where
,
5
4
2
3









 A
A
7
)
4
(
3
)
5
(
1
|
|
where
,
5
4
3
-
1
-










 B
B
-7
3(5)
-
2(4)
4
5
3
2
and
7
2(4)
-
3(5)
5
4
2
3




PROPERTIES OF DETERMINANTS

5. 5. The determinant of a matrix equals that of its
transpose; that is, |A| = |AT|.
6. If a matrix A is placed in diagonal form, then the
product of the elements on the diagonal equals the
determinant of A.
7
4(2)
-
3(5)
5
2
4
3
and
7
2(4)
-
3(5)
5
4
2
3




7
)
3
7
(
3
3
7
0
0
3
|
|
3
7
0
0
3
3
7
0
2
3
7
2(4)
3(5)
|
A
|
with
,
5
4
2
3































A
A
A
PROPERTIES OF DETERMINANTS

6. Rank of Matrix : Definition
• Definition :
The rank of a matrix is defined as
(a) the maximum number of linearly independent
column vectors in the matrix or
(b) (b) the maximum number of linearly independent
row vectors in the matrix.
• Theorem : The row-rank of a row-reduced matrix is
the number of nonzero rows in that matrix.

7. Rank of Matrix
• When all of the vectors in a matrix are linearly independent, the
matrix is said to be full rank. Consider the matrices A and B below
• Notice that row 2 of matrix A is a scalar multiple of row 1; that is, row
2 is equal to twice row 1. Therefore, rows 1 and 2 are linearly
dependent. Matrix A has only one linearly independent row, so its
rank is 1. Hence, matrix A is not full rank.
• Now, look at matrix B. All of its rows are linearly independent, so the
rank of matrix B is 3. Matrix B is full rank.

8. How to Find Matrix Rank?
• We describe a method for finding the rank of any
matrix. This method assumes familiarity with echelon
matrices and echelon transformations.
• The maximum number of linearly independent vectors
in a matrix is equal to the number of non-zero rows in
its row echelon matrix. Therefore, to find the rank of
a matrix, we simply transform the matrix to its row
echelon form and count the number of non-zero rows.

9. How to Find Row echelon form?
• Consider matrix A and its row echelon form matrix, Aref. Previously,
we showed how to find the row echelon form for matrix A.
• Because the row echelon form Aref has two non-zero rows, we know
that matrix A has two independent row vectors; and we know that
the rank of matrix A is 2.
• You can verify that this is correct. Row 1 and Row 2 of matrix A are
linearly independent. However, Row 3 is a linear combination of
Rows 1 and 2. Specifically, Row 3 = 3*( Row 1 ) + 2*( Row 2).
Therefore, matrix A has only two independent row vectors.

10. Row Echelon Form(REF)
• A matrix is in row echelon form (ref) when it satisfies
the following conditions.
• The first non-zero element in each row, called the
leading entry, is 1.
• Each leading entry is in a column to the right of
the leading entry in the previous row.
• Rows with all zero elements, if any, are below rows
having a non-zero element.

11. Reduced Row Echelon Form(RREF)
• A matrix is in reduced row echelon form (rref)
when it satisfies the following conditions.
• The matrix satisfies conditions for a row echelon
form.
• The leading entry in each row is the only non-
zero entry in its column.
• Each of the matrices shown below are examples of
matrices in reduced row echelon form.

12. Example Finding a Row Echelon
Form
21 1 2
Apply elementary row operations to find a row echelon form of the augmented matrix.
2 3 1 1 1 5 3 10 1 5 3 10
1 5 3 10 2 3 1 1 2 0 13 5 21 3
3 1 6 5 3 1 6 5 3 1 6 5
R R R R
     
     
     
        
     
     
     
     
1 3
2 2 3 3
1 5 3 10 1 5 3 10
1 5 3 10
1 5 21 5 21 13
0 13 5 21 0 1 14 0 1
13 13 13 13 13 31
0 14 3 25
0 14 3 25 31 31
0 0
13 13
1 5 3 10
5 21
0 1
13 13
0 0 1 1
R
R R R R

 
 
   
 
 
   
 
  
   
 
  
   
 

   
   
  
   
 
 
 
 



 
 
Convert the matrix to equations and solve by substitution.
1; 5/13 21/13 so 2; 10 3 10 so 3.
The solution is 3, 2,1 .
z y y x x




         


13. EXAMPLE1: RANK OF A MATRIX
• the matrix is 4 x 3
• Since there are 3 nonzero rows remaining in this echelon
form of B, Rank =3.

14. 3 square submatrices:
Each of these has a determinant of 0, so the rank is less than 2.
Thus the rank of R is 1.
Example 2: Rank of Matrix








8
4
2
4
2
1
matrix,
order
3
2 R
8
4
4
2
,
8
2
4
1
,
4
2
2
1
3
2
1 



















 R
R
R

15. Example 3: Rank of Matrix
Since |A|=0, the rank is not 3.
The following submatrix has a nonzero determinant:
Thus, the rank of A is 2.











11
10
9
5
3
1
6
4
2
A
2
)
1
(
4
)
3
(
2
3
1
4
2




16. Example 4: Rank of Matrix

17. Example 4: Rank of Matrix

18. Kernel of matrix
o The Kernel of a linear transformation consists of all of the
solutions to the system Ax=0 We denote Kernel of L by Ker(L).
 Learn about functions that map a vector space V into a
vector space W --- L: V  W
ker(L) is a subspace of V
Domain
Codomain
Kernel
W: codomain of L
V: domain of L

19. Kernel : Example
Find the kernel of a linear transformation

20. KERNEL : EXAMPLE

21. KERNEL : EXAMPLE
o The following is a simple illustration of the computation of the
kernel of a matrix
o The kernel of this matrix consists of all vectors (x, y, z) ∈ R3 for
which
o The same linear equations can also be written in matrix form as:

22. KERNEL : EXAMPLE
o the matrix can be reduced to:
o The elements of the kernel can be further expressed in
parametric form, as follows:
o Rewriting the matrix in equation form yields:

23. The concept of image in linear algebra
• The image of a linear transformation or matrix is the
span of the vectors of the linear transformation. It can
be written as Im(A).
• For example, consider the matrix (call it A)
 The vectors that are possible belong to the span of A. In
this case, the span can be represented by a
"parametrized" matrix, where t and s can be any
number:

24. Example: Image

25. Image :Example

26. Eigenvalue & Eigenvector
• Let A be an n  n matrix. The scalar  is
called an eigenvalue of A if there is a
nonzero vector x s.t. Ax = x
The vector x is called an eigenvector of A
corresponding to .
• An eigenvector cannot be zero.
• An eigenvalue of  = 0 is possible.
• Ax = x  (I – A)x = 0
• det(I – A) = 0.

27. How to find the eigenvalues and
eigenvectors
• Set up the characteristic equation, using |A − λI| = 0.
• Solve the characteristic equation, giving us the
eigenvalues.
• Substitute the eigenvalues into the two equations given
by A − λI.
• Choose a convenient value for x1, then find x2.
• The resulting values form the corresponding
eigenvectors of A.

28. Example 1
Find the eigenvalues and corresponding eigenvectors of
Sol:
two eigenvalues: 1,  2









5
1
12
2
A
)
2
)(
1
(
2
3
12
)
5
)(
2
(
5
1
12
2
2























 A
I

29. Example 1 (cont.)
• (I – A)x = 0





 














0
0
4
1
4
1
12
3
)
1
(
:
1 A
I

0
,
1
4
,
4
0
4
2
1
1
2
1
2
1




















t
t
x
x
t
x
t
x
x
x
x

30. Example 1 (cont.)
• (I – A)x = 0
• eigenvectors:
0
,
0
,
1
3
,
1
4




















s
t
s
t





 














0
0
3
1
3
1
12
4
)
2
(
:
2 A
I

0
,
1
3
2
1
2 













 s
s
x
x
x

31. Example 2
Find the eigenvalues and corresponding eigenvectors for
Sol:
eigenvalues  = 2, 2, 2











2
0
0
0
2
0
0
1
2
A
0
)
2
(
2
0
0
0
2
0
0
1
2
3








 



 A
I

32. Example 3
Find the eigenvalues of
Sol: The characteristic equation of A is
Thus the eigenvalues are , , and .














3
0
0
1
0
2
0
1
10
5
1
0
0
0
0
1
A
0
)
3
)(
2
(
)
1
( 2





 


 A
I
1
1 
 2
2 
 3
3 


33. 




 


5
3
6
4
A
Solution























 







5
3
6
4
1
0
0
1
5
3
6
4
2
I
A
2
18
)
5
)(
4
( 2
2 







 



I
A
We now solve the characteristic equation of A.
The eigenvalues of A are 2 and –1.
The corresponding eigenvectors are found by using these values
of  in the equation(A – I2)x = 0. There are many eigenvectors
corresponding to each eigenvalue.
1
or
2
0
)
1
)(
2
(
0
2
2









 




EXAMPLE 4

34. • For  = 2
We solve the equation (A – 2I2)x = 0 for x.
The matrix (A – 2I2) is obtained by subtracting 2 from the
diagonal elements of A. We get
0












 

2
1
3
3
6
6
x
x
This leads to the system of equations
giving x1 = –x2. The solutions to this system of equations are x1
= –r, x2 = r, where r is a scalar. Thus the eigenvectors of A
corresponding to  = 2 are nonzero vectors of the form
0
3
3
0
6
6
2
1
2
1





x
x
x
x
1
1 2
2
1 1
v
1 1
x
x r
x
 
     
  
     
   
 
EXAMPLE 4 (CONT.)

35. • For  = –1
We solve the equation (A + 1I2)x = 0 for x.
The matrix (A + 1I2) is obtained by adding 1 to the diagonal
elements of A. We get
0












 

2
1
6
3
6
3
x
x
This leads to the system of equations
Thus x1 = –2x2. The solutions to this system of equations are x1
= –2s and x2 = s, where s is a scalar. Thus the eigenvectors of A
corresponding to  = –1 are nonzero vectors of the form
0
6
3
0
6
3
2
1
2
1





x
x
x
x
1
2 2
2
2 2
v
1 1
x
x s
x
 
     
  
     
   
 
EXAMPLE 4 (CONT.)

36. The Gradient of a Function of Two
Variables
The gradient of a function of two variables is a vector-
valued function of two variables.

37. Example 1 – Finding the Gradient of a
Function
Find the gradient of f(x, y) = y ln x + xy2 at the point
(1, 2).
Solution:
Using
and
you have
At the point (1, 2), the gradient is

38. Example 2 – Finding the
Gradient of a Function
• Calculate the gradient to determine the direction of the
steepest slope at point (2, 1) for the function
Solution: To calculate the gradient we would need to
calculate
• which are used to determine the gradient at point (2,1) as

39. Example 3 – Finding the Gradient of a
Function
• Calculate the gradient of the vector
• Solution

40. Example 4 – Finding the Gradient of a
Function
• What is the the gradient vector of the following
function?
• Solution