DEFINITION:- (QUADRATIC FORM)

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.

n n
In n – variables x1,x2,…,xn, the general quadratic form is  b x x , where b
j1 i1
ij i j ij  b ji

In the expansion, the co-efficient of xixj = (bij + bji).

Suppose 2aij  bij  bij where aij  a ji andaii  bii

n n n n
1

j1 i1
b x x
ij i j  
j1 i1
aijx i x j where aij  (bij  b ji ).
2

Hence every quadratic form can be written as

n n

 a x x
j1 i1
ij i j  X' AX,

so that the matrix A is always symmetric,

where A  aij  and X  x1, x 2 ,..., x n .
Now writing the above said examples of quadratic forms
in matrix form, we get

a h  x 
(i) ax 2  2hxy  by 2  [x y]   
h b  y 

a h f   x 
(ii) ax  by  cz  2hxy  2gyz  2fzx  x y z h b g  y 
2 2 2

 f g c   z 

(iii) ax 2  by 2  cz 2  dw2  2hxy  2gyz  2f zx  2lxw  2myw  2nzw

a h f l   x 
h b g m  y 
 x y z w    
f g c n   z 
  
 l m n d  w 
NATURE OF A QUADRATIC FORM

A real quadratic form X’AX in n variables is said to be

i. Positive definite if all the eigen values of A > 0.

ii. Negative definite if all the eigen values of A < 0.

iii. Positive semidefinite if all the eigen values of A are positive and at least one eigen value
= 0.

iv. Negative semidefinite if all the eigen values of A are negative and at least one eigen
value = 0.

v. Indefinite if some of the eigen values of A are + ve and others – ve.

Example:1 Find the nature of the following quadratic forms

i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx

ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy

Solution:-
1 1 3
i. The matrix of the quadratic form is A  1 5 1
 
3 1 1

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.

ii. The matrix of the quadratic form is  3 1 1 

A   1 5  1
 1  1 3 
The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form
is positive definite

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 a h g x

A  h b f , X   y  and X'  [ x y z ],
 
g f c   z 
then X' AX  ax 2  by 2  cz 2  2fyz  2gzx  2hxy .... (i)
which is a quadratic form
 x1  x 2  x 3 
Let λ1, λ2, λ3 be the eigen values of the matrix A and X1  y1 , X2  y 2 , X3   y 3 
   
     
 z1   z 2   z3 

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).

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.

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.

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

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.

Problem: Reduce 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz into canonical form by orthogonal
reduction.  8 6 2 
Solution:- The matrix of the quadratic form is A   6 7  4
 2  4 3 

The characteristic roots of A are given by| A  λ I | 0

8λ 6 2
i.e.,  6 7λ 4  0
2 4 3λ
or λ(λ  3)(λ  15)  0
 λ  0, 3, 15

Characteristic vector for λ = 0 is given by

[A – (0)I] X1 = 0

i.e., 8x1  6x 2  2x3  0

 6x1  7x 2  4x3  0
2x1  4x 2  3x3  0
Solving first two , we get
x1 x 2 x 3
 
1 2 2
giving the eigen vector X1  k1(1, 2, 2)'
When λ = 3, the corresponding characteristic vector is given by [A – 3I] X2 = 0

i.e., 5x1  6x 2  2x3  0

 6x1  4x 2  4x3  0
2x1  4x 2 0
Solving any two equations, we get X2 = k2 (2, 1, -2)’.

Similarly characteristic vector corresponding to λ = 15 is X3 = k3 (2, -2, 1)’.

Now, X1, X2, X3 are pairwise orthogonal

i.e., X1 . X2 = X2 . X3 = X3 . X1 = 0.

 2 1 2 
 3 3 3 
 1 2 2
B   
 3 3 3
 2 2 1 
 
The normalised modal matrix is

Now B is orthogonal matrix and B  1

i.e., B 1  B T andB 1AB  D  diag{3,0,15}

2 1 2  2 1 2 
3 3    3
3 3 3  3 0 0 
1 2   1 2
ie., 
2
A 
2
  0 0 0 
3 3 3   3 3 3  
2  2 1   2 2 1  0 0 15
 3 3 3   3 3 3 

X' AX  Y' (B 1AB)Y  Y' DY

3 0 0   y1 
 y1 y 2 y 3  0 0 0   y 2 
0 0 15  y 3 
 3y12  0.y 22  15y32

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.