1. Mathematics-I J.Baskar Babujee Department of Mathematics
Anna University, Chennai

2. Content 1.1 INTRODUCTION 1.2 TYPES OF MATRICES
1.3 CHARACTERISTIC EQUATION
1.4 EIGEN VECTORS
1.5 PROBLEMS
1.6 CAYLEY HAMILTON THEOREM

3. 1.7 DIAGONALISATION OF A MATRIX
1.8 REDUCTION OF A MATRIX TO DIAGONAL FORM
1.9 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM
1.10 QUADRATIC FORMS
1.11 NATURE OF A QUADRATIC FORM

4. 1. 12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM 1
1.12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM 1.13 INDEX AND SIGNATURE OF THE QUADRATIC FORM 1.14 LINEAR TRANSFORMATION OF A QUADRATIC FORM REDUCTION OF QUADRATIC FORM TO CANANONICAL FORM BY ORTHOGONAL TRANSFORMATION

5. Unit 1 MATRICES 1.1 INTRODUCTION:-
A matrix is defined as a rectangular array (or arrangement in rows or columns) of scalar subject to certain rules of operations.
If mn numbers (real or complex) or functions are arranged in the columns (vertical lines) then A is called an m n matrix. Each of the mn numbers is called an element of the matrix.

6. An m n matrix is usually written as
A matrix is usually denoted by a single capital letter A, B or C etc.

7. Thus, an m n matrix A may be written as
A = , where i = 1, 2, 3, … , m ;
j = 1, 2, 3, … , n
In Algebra, the matrices have their largest
application in the theory of simultaneous equations and linear transformations.

8. E.g., The set of simultaneous equations

9. may be symbolically represented by the equation Where A = , X = , B =

10. 1.2 TYPES OF MATRICES (1) Real Matrix :- A matrix is said to be
real if all its elements are real numbers.
E.g., is a real matrix.

11. (2) Square Matrix:- A matrix in which the number of rows is equal to the number of columns is called a square matrix, otherwise, it is said to be rectangular matrix.
i.e., a matrix A = is a square matrix if m = n
a rectangular matrix if m n

12. A square matrix having n rows and n columns is called “ n – rowed square matrix”,
is a 3 – rowed square matrix
The elements of a square matrix are called its diagonal elements and the diagonal along with these elements are called principal or leading diagonal.

13. The sum of the diagonal elements of a square matrix is called its trace or spur.
Thus, trace of the n rowed square matrix
A= is

14. (3) Row Matrix :-
A matrix having only one row and any number of columns,
i.e., a matrix of order 1 n is called a row matrix.
[ ] is a row matrix.
Example:-

15. (4) Column Matrix:-
A matrix having only one column and any number of rows,
ii.e., a matrix of order m 1 is called a column matrix.
is a column matrix.
Example:-

16. (5) Null Matrix:-
A matrix in which each element is zero is called a null matrix or void matrix or zero matrix.
A null matrix of order m n is denoted by =
Example :-

17. (6) Sub - matrix :-
A matrix obtained from a given matrix A by deleting some of its rows or columns or both is called a sub – matrix of A.
is a sub – matrix of
Example:-

18. (7) Diagonal Matrix :-
A square matrix in which all non – diagonal elements are zero is called a diagonal matrix.
i.e., A = [a ] is a diagonal matrix if a = 0 for i j.
is a diagonal matrix.
Example:-

19. (8) Scalar Matrix:-
A diagonal matrix in which all the diagonal elements are equal to a scalar, say k, is called a scalar matrix.
i.e., A = [a ] is a scalar matrix if
is a scalar matrix.
Example :-

20. (9) Unit Matrix or Identity Matrix:-
A scalar matrix in which each diagonal element is 1 is called a unit or identity matrix. It is denoted by
i.e., A = [a ] is a unit matrix if
is a unit matrix.
Example

21. i.e., A = [a ] is an upper triangular matrix if a = 0 for i > j
A square matrix in which all the elements below the principal diagonal are zero is called an upper triangular matrix.
i.e., A = [a ] is an upper triangular matrix if a = for i > j
is an upper triangular matrix
Example:-

22. (11) Lower Triangular Matrix.
A square matrix in which all the elements above the principal diagonal are zero is called a lower triangular matrix.
i.e., A = [a ] is a lower triangular matrix if a = for i < j
is a lower triangular
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Example:-

23. (12) Triangular Matrix:-
A triangular matrix is either upper triangular or lower triangular.
(13) Single Element Matrix:-
A matrix having only one element is called a single element matrix.
i.e., any matrix [3] is a single element matrix.

24. (14) Equal Matrices:-
Two matrices A and B are said to be equal iff they have the same order and their corresponding elements are equal.
i.e., if A = and B = , then A = B
iff a) m = p and n = q
b) a = b for all i and j.

25. (15) Singular and Non – Singular Matrices:-
A square matrix A is said to be singular if |A| = 0 and non – singular if |A| 0.
A = is a singular
matrix since |A| = 0.
Example :-

26. 1.3 CHARACTERISTIC EQUATION
If A is any square matrix of order n, we can form the matrix , where is the nth order unit matrix.
The determinant of this matrix equated to zero,
i.e.,

27. is called the characteristic equation of A.
On expanding the determinant, we get
where k’s are expressible in terms of the elements a
The roots of this equation are called Characteristic roots or latent roots or eigen values of the matrix A.

28. 1.4 EIGEN VECTORS Consider the linear transformation Y = AX ...(1)
which transforms the column vector X into the column vector Y. We often required to find those vectors X which transform into scalar multiples of themselves.
Let X be such a vector which transforms into X by the transformation (1).

29. Then Y = X ... (2)
From (1) and (2), AX = X AX- X = 0
(A )X = 0 ...(3)
This matrix equation gives n homogeneous linear equations
... (4)

30. These equations will have a non-trivial solution only if the co-efficient matrix A - is singular
i.e., if |A | = (5)
Corresponding to each root of (5), the
homogeneous system (3) has a non-zero solution
X = is called an eigen vector or latent vector

31. Properties of Eigen Values:-
The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal.
The product of the eigen values of a matrix A is equal to its determinant.
If is an eigen value of a matrix A, then 1/ is the eigen value of A-1 .
If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value.

32. PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of A, then
k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar.
ii are the eigen values of the inverse
matrix A-1.
iii are the eigen values of Ap, where p is any positive integer.

33. Proof:-
i. Let λr be an eigen value of A and Xr the corresponding eigen vector.
Then, by definition,
Multiplying both sides by k,
(kA)Xr = (kλr)Xr
Then k λr is an eigen value of kA and the corresponding eigen vector is the same as that of λr, namely Xr.
AXr = λrXr

34. ii. Pre multiplying both sides of AXr = λrXr by A-1
A-1 (A Xr) = A-1 (λr Xr )
Xr = λr (A-1 Xr )
=> A-1 Xr = (Xr)
Hence is an eigen value of A-1 and the corresponding eigen vector is the same as that of
λr , namely Xr.

35. iii. Pre multiplying both sides of AXr = λrXr by A
A (A Xr) = A (λr Xr )
A2 Xr = λr (A Xr )
= λr (λr xr)
= λr2 Xr.
Similarly, we can prove that A3Xr = λr3 Xr, …,
ApXr = λrp Xr, where p is any positive integer. Hence λrp is any eigen value of Ap and the corresponding eigen vector is the same as that of λr, namely Xr.

36. THEOREM :-
A matrix A is singular if and only if 0 is an eigen value of A.
1.5 PROBLEMS
1. Find the sum and product of the eigen values of the matrix
without finding the eigen values.

37. Solution:-
Sum of the eigen values of A = sum of its diagonal elements. =
= -1.
Product of the eigen values of A = | A | =
= 45.

38. 2. Two eigen values of the matrix
are equal to 1 each. Find the third eigen value.
Solution:- Let a be the third eigen value of A.
Since sum of the eigen values = sum of the diagonal
elements,
a =
a = 5
Therefore, the third eigen value of A is 5.

39. 3. The product of two eigen values of the matrix
is 16. Find the third eigen value.
Solution:-
Let a be the third eigen value of A.
Since product of the eigen values = | A |
16a =
Therefore, a = 2.

40. 4. Find the sum of the eigen values of the inverse of
Solution:-
The eigen values of the lower triangular matrix A is 1, -3, 2. Then the eigen values of A-1 are
Sum of the eigen values of A-1 = =

41. 5. If , find the eigen values of 3A, A-1
and – 2A-1.
Solution:-
The eigen values of A are 2, -1, 4.
The eigen values of 3A are 3×2, 3×(-1), 3×4
i.e., , -3, 12

42. The eigen values of A-1 are

43. 6. Find the eigen values and eigen vectors of the matrix
Solution:- The characteristic equation of the given matrix is

44. the eigen values of A are 6, -1.or
Thus,
Corresponding to =6, the eigen vectors are given by (A – 6 ) X1 = 0
the eigen values of A are 6, -1.

45. We get only one independent equation - 5x1 - 2x2 = 0

46. Corresponding to = -1, the eigen vectos are given by ( A + ) X2 = 0

47. 7. Find the eigen values and eigen vectors of the matrix A =
Solution:- The characteristic equation of the given matrix is

48. the eigen values of A are -3, -3, 5.
Thus,
the eigen values of A are -3, -3, 5.

49. Corresponding to = - 3, the eigen vectors are given by

50. Corresponding to λ = 5, the eigen vectors are given by (A – 5 I )X2 = 0.

52. 8. Find the eigen values and eigen vectors of the matrix
Solution:- The characteristic equation is

53. Eigen vector X1 corresponding to λ1= 0 is given by

54. Solving equations (1) and (2) by cross-multiplication, we get
which satisfy equation (3) also.
Required eigen vector corresponding to λ1 = 0 is

55. Eigen vector X2 corresponding to λ2= 3 is given by

56. Solving equations (4) and (5) by cross-multiplication, we get
which satisfy equation (6) also.
Required eigen vector corresponding to λ2 = 3 is

57. Eigen vector X3 corresponding to λ3= 15 is given by

58. Solving equations (7) and (8) by cross-multiplication, we get
which satisfy equation (9) also.
Required eigen vector corresponding to λ3 = 15 is

59. 9. Find the eigen values and eigen vectors of the matrix
Solution:- The characteristic equation is

60. Eigen vector corresponding to is given by

61. These equations are equivalent to a single equation … (1)
Let x3 = 2 k3 and x2 = 2 k2 then from (1)
2x1 – 2 k2 + 2 k3 = 0
=> x1 = k2 – k3
Required eigen vector corresponding to is

62. Similarly, the eigen vector corresponding to = 8 is given by,

63. Solving equations (2) and (3) by cross-multiplication, we get
which satisfy equation (4) also.
Required eigen vector corresponding to λ3 = 8 is

64. Show that if λ1,λ2, … λn are the latent roots of the matrix A, then A3 has the latent roots
Solution:- Let λ be a latent root of the matrix A. Then there exists a non – zero vector X such that
Example 1:-

65. A X = λ X … (1) => A2 (AX) = A2 (λ X)
=> A3 X = λ (A2 X) [using (1)]
But A2 X = A ( A X) = A (λ X)
= λ (AX) = λ (λX) = λ2 X
Therefore, A3 X = λ (λ2 X) = λ3X
=> λ3 is a latent root of A3.

66. Therefore, If λ1,λ2, … λn are the latent roots of the matrix A, then are the latent roots of A3.
If λ1,λ2, … λn are eigen values of A then find eigen values of the matrix (A – λI)2 .
Solution:- (A – λI)2 = A2 – 2 λAI + λ2I2
= A2 – 2 λA + λ2I
Example 2:-

67. Eigen values of A2 are
Eigen values of 2 λA are 2 λ λ1,2 λ λ2, …, 2 λ λn.
Eigen values of λ2I are λ2
Eigen values of ( A – λI)2 are

68. Example 3:-
Find the eigen values and eigen vectors of the matrix
Solution:- The characteristic equation is

69. Eigen vector X1 corresponding to λ1= 3 is given by

70. The characteristic vector corresponding to eigen value λ1 = 3 is given by
When λ2 = 2, let X2 be the eigen vector then
(A – 2I) X2 = 0 where X2 = [x1 x2 x3]’

72. When λ3 = 5, let X3 be the eigen vector then
(A – 5I) X3 = 0 where X3 = [x1 x2 x3]’
Solving above equations by cross – multiplication, we get

73. Required eigen vector is

74. 1.6 CAYLEY HAMILTON THEOREM
Every square matrix satisfies its own characteristic equation.
Let A = [aij]n×n be a square matrix
then,

75. Let the characteristic polynomial of A be  (λ)
Then,
The characteristic equation is

76. Note 1:- Premultiplying equation (1) by A-1 , we have

77. This result gives the inverse of A in terms of (n-1) powers of A and is considered as a practical method for the computation of the inverse of the large matrices.
Note 2:- If m is a positive integer such that m > n then any positive integral power Am of A is linearly expressible in terms of those of lower degree.

78. Example 1:-
Verify Cayley – Hamilton theorem for the matrix
A = Hence compute A-1 .
Solution:- The characteristic equation of A is

79. To verify Cayley – Hamilton theorem, we have to
show that A3 – 6A2 +9A – 4I = 0 … (1)
Now,

80. A3 -6A2 +9A – 4I = 0 = -4 =
This verifies Cayley – Hamilton theorem.

81. Now, pre – multiplying both sides of (1) by A-1 , we have
A2 – 6A +9I – 4 A-1 = 0
=> 4 A-1 = A2 – 6 A +9I

82. Example 2:-
Given find Adj A by using Cayley –
Hamilton theorem.
Solution:- The characteristic equation of the given matrix A is

83. By Cayley – Hamilton theorem, A should satisfy
A3 – 3A2 + 5A + 3I = 0
Pre – multiplying by A-1 , we get
A2 – 3A +5I +3A-1 = 0

86. 1.7 DIAGONALISATION OF A MATRIX
Diagonalisation of a matrix A is the process of reduction A to a diagonal form.
If A is related to D by a similarity transformation, such that D = M-1AM then A is reduced to the diagonal matrix D through modal matrix M. D is also called spectral matrix of A.

87. 1.8 REDUCTION OF A MATRIX TO DIAGONAL FORM
If a square matrix A of order n has n linearly independent eigen vectors then a matrix B can be found such that B-1AB is a diagonal matrix.
Note:- The matrix B which diagonalises A is called the modal matrix of A and is obtained by grouping the eigen vectors of A into a square matrix.

88. Similarity of matrices:-
A square matrix B of order n is said to be a similar to a square matrix A of order n if
B = M-1AM for some non singular matrix M.
This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation.
Note:- If the matrix B is similar to matrix A, then B has the same eigen values as A.

89. Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3.
Solution:- Characteristic equation is
=> λ = 1, 2, 3
Hence eigen values of A are 1, 2, 3.
Example:-

90. Corresponding to λ = 1, let X1 = be the eigen
vector then

91. Corresponding to λ = 2, let X2 = be the eigen
vector then,

92. Corresponding to λ = 3, let X3 = be the eigen
vector then,

93. Hence modal matrix is

94. = MD2M-1 [since M-1M = I] Now, since D = M-1AM => A = MDM-1
A2 = (MDM-1) (MDM-1)
= MD2M [since M-1M = I]

95. Similarly, A3 = MD3M-1 =
A3 =

96. 1.9 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM
A square matrix A with real elements is said to be orthogonal if AA’ = I = A’A.
But AA-1 = I = A-1A, it follows that A is orthogonal if
A’ = A-1.
Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.

97. If A is a real symmetric matrix then eigen vectors of A will be not only linearly independent but also pairwise orthogonal.
If we normalise each eigen vector and use them to form the normalised modal matrix N then it can be proved that N is an orthogonal matrix.

98. The similarity transformation M-1AM = D takes the form N’AN = D since N-1 = N’ by a property of orthogonal matrix.
Transforming A into D by means of the transformation N’AN = D is called as orthogonal reduction or othogonal transformation.
Note:- To normalise eigen vector Xr, divide each element of Xr, by the square root of the sum of the squares of all the elements of Xr.

99. Diagonalise the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
Example :-

104. 1.10 QUADRATIC FORMS DEFINITION:-
A homogeneous polynomial of second degree in any number of variables is called a quadratic form.
For example,
ax2 + 2hxy +by2
ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and
ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw
are quadratic forms in two, three and four variables.

105. In n – variables x1,x2,…,xn, the general quadratic formis
In the expansion, the co-efficient of xixj = (bij + bji).

106. Hence every quadratic form can be written as

108. 1.11 NATURE OF A QUADRATIC FORM
A real quadratic form X’AX in n variables is said to be
Positive definite if all the eigen values of A > 0.
Negative definite if all the eigen values of A < 0.
Positive semidefinite if all the eigen values of A 0 and at least one eigen value = 0.
Negative semidefinite if all the eigen values of
A 0 and at least one eigen value = 0.
v. Indefinite if some of the eigen values of A are + ve and others – ve.

109. Example :-
Find the nature of the following quadratic forms
x2 + 5y2 + z2 + 2xy + 2yz + 6zx
3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy
Solution:-
The matrix of the quadratic form is

110. The eigen values of A are -2, 3, 6.
Two of these eigen values being positive and one being negative, the given quadratric form is indefinite.
The matrix of the quadratic form is
The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.

111. 1.12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM
A homogeneous expression of the second degree in any number of variables is called a quadratic form.
For instance, if
which is a quadratic form.

112. Let λ1, λ2, λ3 be the eigen values of the matrix A and
be its corresponding eigen vectors in the normalized form (i.e., each element is divided by square root of sum of the squares of all the three elements in the eigen vector).

113. Then B-1AB = D, a diagonal matrix.
Hence the quadratic form (i) is reduced to a sum of squares (i.e., canonical form).
λ1x2 + λ2y2 + λ3z2
And B is the matrix of transformation which is an orthogonal matrix.
Note:-
1. Here some of λi may be positive or negative or zero
2. If ρ(A) = r, then the quadratic form X’AX will contain only r terms.

114. 1.13 INDEX AND SIGNATURE OF THE QUADRATIC FORM
The number p of positive terms in the canonical form is called the index of the quadratic form.
(The number of positive terms) – ( the number of negative terms)
i.e., p – (r – p) = 2p – r is called signature of the quadratic form, where ρ(A) = r.

115. 1.14 LINEAR TRANSFORMATION OF A QUADRATIC FORM.
Let X’AX be a quadratic form in n- variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation.
From (1), X’ = (PY)’ = Y’P’ and hence
X’AX = Y’P’APY = Y’(P’AP)Y
= Y’BY …. (2)
where B = P’AP.

116. Therefore, Y’BY is also a quadratic form in n- variables
Therefore, Y’BY is also a quadratic form in n- variables. Hence it is a linear transformation of the quadratic form X’AX under the linear transformation X = PY and B = P’AP.
Note. (i) Here B = (P’AP)’ = P’AP = B
(ii) ρ(B) = ρ(A)
Therefore, A and B are congruent matrices.

117. Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form.
Diagonalise the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation.
Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares.
Example:-

118. Solution:- The given quadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix
A =
Let us reduce A into diagonal matrix. We know tat A = I3AI3.

121. The canonical form of the given quadratic form is
Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1.
Note:- In this problem the non-singular transformation which reduces the given quadratic form into the canonical form is X = PY.
i.e.,

122. 1.15 REDUCTION OF QUADRATIC FORM TO CANANONICAL FORM BY ORTHOGONAL TRANSFORMATION
Let X’AX be a given quadratic form. The modal matrix B of A is the matrix whose columns are characteristic vectors of A. If B represents the orthogonal matrix of A,

123. then X = BY will reduce X’AX to Y’ diag(λ1, λ2,…, λn) Y, where λ1, λ2,…, λn are characteristic values of A.
Note. This method works successfully if the characteristic vectors of A are linearly independent which are pairwise orthogonal.

124. Example 1:-
Reduce 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz into canonical form by orthogonal reduction.
Solution:- The matrix of the quadratic form is

125. The characteristic roots of A are given by

126. Characteristic vector for λ = 0 is given by
[A – (0)I] X1 = 0

127. When λ = 3, the corresponding characteristic vector is given by [A – 3I] X2 = 0
i.e.,
Solving any two equations, we get X2 = k2 (2, 1, -2)’.
Similarly characteristic vector corresponding to
λ = 15 is X3 = k3 (2, -2, 1)’.

128. Now, X1, X2, X3 are pairwise orthogonal
i.e., X1 . X2 = X2 . X3 = X3 . X1 = 0.
The normalised modal matrix is

129. Now B is orthogonal matrix and

130. which is the required canonical form.
Note. Here the orthogonal transformation is X =BY, rank of the quadratic form = 2; index = 2, signature = 2. It is positive definite.

131. TEST YOUR KNOWLEDGE If then find the eigen value of
2. Write the matrix of the Quadratic form
3. Obtain the characteristic equation of the matrix whose eigen values are 1,-2 and 3.
4. A = , then find the eigen values of
3A3+5A2-6A+2I. .

132. 5. Write the quadratic form corresponding to the following symmetric matrix
6. Find the sum of the eigen values of the inverse of.
7. Obtain the latent roots of A4 where A =
8. If A is idempotent matrix then A2=A. What will be the eigen values of A.

133. 9. If are the eigen values of the matrix A, whose characteristic equation is
Obtain using the property.
10. (i) Using Cayley-Hamilton Theorem find the inverse of the matrix
(ii) Find the Characteristic roots and Characteristic vectors of the matrix

134. 11. Reduce the quadratic form to the canonical form by orthogonal transformation. Also specify the matrix of transformation. Obtain its index, signature and nature of the quadratic form.
12. (i) Find the eigen value and eigen vector of the matrix
(ii) Using Cayley Hamiltonian find the inverse of

135. 13. Discuss the nature, index and signature of the quadratic form
14. Diagonalise the matrix by
orthogonal reduction and provide the normalized modal matrix.
15. Reduce the quadratic form 2x1x2+2x1x3-2x2x3 to the canonical form by an Orthogonal transformation.

136. 16. (i) Find the eigen value and eigen vector of the matrix
(ii) For A = , compute the value of
, Using Cayley-Hamilton theorem.
17. Reduce the quadratic form 3x12 + 5x22 + 3x32
-2x2x3+2x3x1-2x1x2 to a Canonical form by orthogonal reduction.

137. THANK YOU