The document discusses reducing a quadratic form to canonical form using an orthogonal transformation. It begins by defining quadratic forms and representing them using matrices. It then provides examples of writing the matrix and quadratic form for functions of 2 and 3 variables. The document explains finding the eigenvectors and eigenvalues of the coefficient matrix to form an orthogonal transformation matrix B. Premultiplying the coefficient matrix A by the inverse of B results in the diagonal canonical form matrix D, where the diagonal elements are the eigenvalues of A. The quadratic form is then in canonical (sum of squares) form. An example problem demonstrates reducing a 3D quadratic form to canonical form using this process.

1. Reduction Of
Quadratic Form
To
Canonical Form
1

2. Quadratic Form
2
Definition: The quadratic form in n variables 𝑥1, 𝑥2., … . 𝑥𝑛 is the general homogenous
function of second degree in the variables
i.e. Y=𝑓(𝑥1, 𝑥2., … . 𝑥𝑛 )= σ𝑖,𝑗=1
𝑛
𝑎𝑖𝑗 𝑥𝑖 𝑥𝑗
In terms of matrix notation , the quadratic form is given by
Y=𝑋𝑇
𝐴𝑋 = 𝑥1 𝑥2 … 𝑥𝑛
𝑎11 𝑎12 … 𝑎1𝑛
𝑎21 𝑎22 … 𝑎2𝑛
… … … …
𝑎𝑛1 𝑎𝑛2 … 𝑎𝑛𝑛
𝑥1
𝑥2
…
𝑥𝑛

3. In Matrix Notation
3
In two variables (𝑥,𝑦): 𝑎𝑥2+2ℎ𝑥𝑦 + 𝑏𝑦2
𝑄 𝑥, 𝑦, 𝑧 = 𝑋𝑇𝐴𝑋 = 𝑥 𝑦
𝑎 ℎ
ℎ 𝑏
𝑥
𝑦
Here A is known as matrix of quadratic form
.

4. In three variables
4
𝑄(𝑥, 𝑦, 𝑧)=𝑎𝑥2
+ 𝑏𝑦2
+ 𝑐𝑧2
+ 2ℎ𝑥𝑦 + 2𝑓𝑦𝑧 + 2𝑔𝑧𝑥
In matrix notation
𝑄 𝑥, 𝑦, 𝑧 = 𝑋𝑇
𝐴𝑋= 𝑥 𝑦 𝑧
𝑎 ℎ 𝑔
ℎ 𝑏 𝑓
𝑔 𝑓 𝑐
𝑥
𝑦
𝑧

5. To write matrix of quadratic form in three variables
5
𝑄 𝑥, 𝑦, 𝑧 = 𝑎𝑥2
+ 𝑏𝑦2
+ 𝑐𝑧2
+ 2ℎ𝑥𝑦 + 2𝑓𝑦𝑧 + 2𝑔𝑧𝑥
A=
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑥2
)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑥𝑦)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑥𝑧)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑥𝑦) 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑦2)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑦𝑧)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑥𝑧)
1
2
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑦𝑧) 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡(𝑧2
)

6. 6
Ex. Obtain the matrix of the Quadratic form
Q= 𝑥2
+ 2 𝑦2
− 7𝑧2
− 4𝑥𝑦+8𝑥𝑧 + 5𝑦𝑧
Sol. Compare the given quadratic form by standard form
𝑄 𝑥, 𝑦, 𝑧 = 𝑎𝑥2
+ 𝑏𝑦2
+ 𝑐𝑧2
+ 2ℎ𝑥𝑦 + 2𝑓𝑦𝑧 + 2𝑔𝑧𝑥
Here 𝑎 = 1 , 𝑏 = 2 , 𝑐 = −7; ℎ = −2 ; 𝑔 = 4 ; 𝑓 =
5
2
Therefore coefficient matrix is
𝐴 =
𝑎 ℎ 𝑔
ℎ 𝑏 𝑓
𝑔 𝑓 𝑐
=
1 −2 4
−2 2
5
2
4
5
2
−7
Which is a symmetric matrix.

7. 7
Ex. Obtain the matrix of the Quadratic form
Q= 2𝑥2
+ 3 𝑦2
+ 𝑧2
− 3𝑥𝑦+2𝑥𝑧 + 4𝑦𝑧
𝐴 =
𝑎 ℎ 𝑔
ℎ 𝑏 𝑓
𝑔 𝑓 𝑐
=
2 −
3
2
1
−
3
2
3 2
1 2 1
Ans.

8. 𝐴 =
EX. Write down the quadratic form corresponding to the
matrix
𝐴 =
1 2 5
2 0 3
5 3 4
𝑆𝑜𝑙. Quadratic form 𝑖𝑠 𝑋𝑇
𝐴𝑋 𝑤ℎ𝑒𝑟𝑒 𝑋 =
𝑥
𝑦
𝑧
𝑋𝑇
𝐴𝑋 = 𝑥 𝑦 𝑧
1 2 5
2 0 3
5 3 4
𝑥
𝑦
𝑧
or 𝑄 𝑥, 𝑦, 𝑧 = 𝑥2 + 0𝑦2 + 4𝑧2 + 4𝑥𝑦 + 6𝑦𝑧 + 10𝑧𝑥
is the required Quadratic form.
8

9. Sol. Quadratic form = 𝑋𝑇
𝐴𝑋 𝑤ℎ𝑒𝑟𝑒 X =
𝑥
𝑦
𝑧
𝑋𝑇
𝐴𝑋 = 𝑥 𝑦 𝑧
3 2 4
2 0 4
4 4 3
𝑥
𝑦
𝑧
or 𝑄 𝑥, 𝑦, 𝑧 = 3𝑥2
+ 0𝑦2
+ 3𝑧2
+ 4𝑥𝑦 + 8𝑦𝑧 + 8𝑧𝑥
is the required Quadratic form.
EX. Write down the quadratic form corresponding to the matrix
𝐴 =
3 2 4
2 0 4
4 4 3
9

10. CANONICAL FORM OR
SUM OF SQUARE FORM
10
The sum of square form of a real quadratic form 𝑋𝑇
𝐴𝑋 is
𝒀𝑻𝑫𝒀 = 𝝀𝟏𝒚𝟏
𝟐
+ 𝝀𝟐𝒚𝟐
𝟐
+…… +𝝀𝒏𝒚𝒏
𝟐
is known as canonical form which is formed with the help of the
orthogonal transformation 𝑋 = 𝐵𝑌 where B is the modal matrix of
normalized eigen vectors , Y=
𝑦1
𝑦2
…
𝑦𝑛
and D is a diagonal matrix whose
diagonal elements are the eigen values of the matrix A.

11. 11
1. First write the coefficient matrix of the given quadratic
form.
2. Find the eigen values and eigen vector of the coefficient
matrix A.
3. Now check whether eigen vectors are pairwise
orthogonal or not.
Orthogonal Vector; Two vectors 𝑋1, 𝑋2 are said to be
orthogonal if their inner product is zero i.e. 𝑋1
𝑇
. 𝑋2 = 0
Working Method

12. 12
4. Form a modal matrix B of normalized eigen
vector
i.e. B = 𝑋1 𝑋2 𝑋3
If X1 =
𝑥1
𝑥2
𝑥3
then 𝑋1 =
𝑥1
𝑥1
2 + 𝑥2
2 + 𝑥3
2
𝑥2
𝑥1
2 + 𝑥2
2 + 𝑥3
2
𝑥3
𝑥1
2 + 𝑥2
2 + 𝑥3
2
Similarly for 𝑋2 and 𝑋3
Here B is matrix of transformation

13. 13
𝑋
5. Here B matrix is an orthogonal matrix
∴ 𝐵𝐵𝑇
= 𝐼
∵ 𝐵𝑇=𝐵−1
Find 𝐵−1
𝐴𝐵
𝐵−1𝐴𝐵= 𝐷 (diagonal matrix whose diagonal elements are
eigen values of the matrix A)
The required canonical form is
𝑌𝑇(𝐵−1𝐴𝐵) 𝑌= 𝑌𝑇𝐷 𝑌= 𝑦1 𝑦2 𝑦3
𝜆1 0 0
0 𝜆2 0
0 0 𝜆3
𝑦1
𝑦2
𝑦3
i.e. 𝜆1𝑦1
2
+ 𝜆2𝑦2
2
+ 𝜆3𝑦3
2
is the required canonical form and
𝑋=𝐵𝑌 is required orthogonal transformation .

14. Ex. Reduce the quadratic form
Q(𝑥, 𝑦, 𝑧)=𝑥1
2
+ 3𝑥2
2
+ 3𝑥3
2
− 2𝑥2𝑥3 into canonical
form using orthogonal transformation.
Sol. Write the coefficient matrix of given quadratic
form
𝐴 =
1 0 0
0 3 −1
0 −1 3
The eigen values of A are 1,2,4
14

15. Eigen Vectors for 𝜆 = 1
𝐴 − 𝜆1𝐼 𝑋1 = 𝟎 𝑜𝑟 𝐴 − 𝐼 𝑋1 = 𝟎
or
0 0 0
0 2 −1
0 −1 2
𝑥1
𝑥2
𝑥3
=
0
0
0
or
0 0 0
0 2 −1
0 0
3
2
𝑥1
𝑥2
𝑥3
=
0
0
0
Applying 𝑅3 → 𝑅3 +
1
2
𝑅2
2𝑥2 − 𝑥3 = 0. . . . . . . (1)
𝑥3 = 0. . . . . . . . . . . . . (2)
Here no. of equations (m) are 2 and no. of
variables (n) are 3
n−m = 3−2 = 1 variable is independent.
Let 𝑥1 = k
𝑥3 = 0; 𝑥2 = 0
𝑥1 = 𝑘
15

16. 𝑋1 =
𝑥1
𝑥2
𝑥3
=
𝑘
0
0
If 𝑘 = 1 𝑡ℎ𝑒𝑛 𝑋1
=
1
0
0
is the eigen vector
corresponding to 𝜆 = 1
16

17. Eigen vector for 𝜆 = 2
𝐴 − 𝜆2𝐼 𝑋2 = 𝟎 or 𝐴 − 2𝐼 𝑋2 = 𝟎
−1 0 0
0 1 −1
0 −1 1
𝑥1
𝑥2
𝑥3
=
0
0
0
Applying 𝑅3 → 𝑅3 + 𝑅2
−1 0 0
0 1 −1
0 0 0
𝑥1
𝑥2
𝑥3
=
0
0
0
𝑥1 = 0. . . . . . . (1)
𝑥2 − 𝑥3 = 0. . . . . . . . . . . . . (2)
Here no. of equations (m) are 2 and no. of
variables (n) are 3
n−m = 3−2 = 1 variable is independent.
Let 𝑥3 = k
𝑥2 = 𝑥3,
𝑥3 = 𝑘 = 𝑥2
17

18. 𝑋2 =
𝑥1
𝑥2
𝑥3
=
0
𝑘
𝑘
If 𝑘 = 1 𝑡ℎ𝑒𝑛 𝑋2 =
0
1
1
is the eigen vector
corresponding to 𝜆 = 2
18

19. Eigen Vector for 𝜆 = 4
𝐴 − 𝜆3𝐼 𝑋3 = 𝟎 or 𝐴 − 4𝐼 𝑋3 = 𝟎
or
−3 0 0
0 −1 −1
0 −1 −1
𝑥1
𝑥2
𝑥3
=
0
0
0
𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 R3 → R3 − R2
−3 0 0
0 −1 −1
0 0 0
𝑥1
𝑥2
𝑥3
=
0
0
0
𝑥1 = 0; 𝑥2 + 𝑥3 = 0
Here no. of equations are (m) 2 and
no. of variables (n) are 3
n−m = 3−2 = 1 variable is independent.
∴ Let 𝑥3 = k
𝑥2 = −𝑥3
𝑥2 = −𝑘; 𝑥1 = 0
19

20. 𝑋3 =
𝑥1
𝑥2
𝑥3
=
0
−𝑘
𝑘
𝐼𝑓 𝑘 = 1 𝑡ℎ𝑒𝑛 𝑋3 = −
0
1
1
is the eigen vector
corresponding to 𝜆 = 4
20

21. Continued.
21
Eigen vector corresponding to matrix A
𝑋1 =
1
0
0
, 𝑋2=
0
1
1
, 𝑋3 =
0
−1
1
Now to check whether 𝑋1, 𝑋2, 𝑋3 are pairwise orthogonal.
Pairwise Orthogonal –Two vectors 𝑋1, 𝑋2 are said to be orthogonal
if their inner product i.e. 𝑋1
𝑇
. 𝑋2=0
 
orthogonal
pairwise
are
,
,
Here
also.
vectors
others
for
check
can
we
Similarly
.
orthogonal
pairwise
are
,
Therefore
0
1
1
0
0
0
1
1
1
0
0
0
1
.
3
2
1
2
1
2
1
X
X
X
X
X
X
X
T
T
=










=




















=

22. Now we form normalized modal matrix
𝐵 =
1 0 0
0
1
2
−
1
2
0
1
2
1
2
, 𝐵−1 =
1 0 0
0
1
2
1
2
0 −
1
2
1
2
𝐵−1
𝐴𝐵 =
1 0 0
0
1
2
1
2
0 −
1
2
1
2
1 0 0
0 3 −1
0 −1 3
1 0 0
0
1
2
−
1
2
0
1
2
1
2
22

23. 23
𝐵−1𝐴𝐵 =
1 0 0
0 2 0
0 0 4
i.e. 𝐵−1𝐴𝐵=D
where D is a diagonal matrix whose diagonal
elements are the eigen values of the matrix of
quadratic form.

24.  The canonical Form
𝑌𝑇(𝐵−1𝐴𝐵)𝑌 = 𝑌𝑇𝐷𝑌
= 𝑦1 𝑦2 𝑦3
1 0 0
0 2 0
0 0 4
𝑦1
𝑦2
𝑦3
= 𝑦1
2 + 2𝑦2
2 + 4𝑦3
2
𝑦1
2 + 2𝑦2
2 + 4𝑦3
2
is the required canonical form.
24

25. Orthogonal
Transformation
𝑋 = 𝐵𝑌
𝑥1
𝑥2
𝑥3
=
𝑋 = 𝐵𝑌
𝑥1
𝑥2
𝑥3
=
1 0 0
0
1
2
−
1
2
0
1
2
1
2
𝑦1
𝑦2
𝑦3
𝑥1 = 𝑦1,
𝑥2 =
1
2
𝑦2 −
1
2
𝑦3,
𝑥3 =
1
2
𝑦2 +
1
2
𝑦3
is the required
orthogonal transformation.
25

26. Reduce 6𝑥2 + 3𝑦2 + 3𝑧2 − 4𝑥𝑦 − 2𝑦𝑧 + 4𝑥𝑧 into
canonical form by orthogonal transformation.
Sol. Write the coefficient matrix of given quadratic
form
𝐴 =
6 −2 2
−2 3 −1
2 −1 3
The eigen values of A are 2,2,8
The eigen vectors corresponding to 2, 2, 8 are
−1
0
2
,
1
2
0
and
2
−1
1
respectively.
26

27. 27
𝑋1=
−1
0
2
, 𝑋2=
1
2
0
, 𝑋3=
2
−1
1
Here 𝑋1 , 𝑋2, are not pairwise orthogonal
Let 𝑋1 is a vector which is orthogonal to 𝑋2 and 𝑋3
Let 𝑋1=
𝑎
𝑏
𝑐
than 𝑋1
𝑇
. 𝑋2= 0 or 𝑎 𝑏 𝑐
1
2
0
=0
or a+2b=0……(1)
similarly 𝑋1
𝑇
. 𝑋3= 0 or 𝑎 𝑏 𝑐
2
−1
1
=0 or 2a - b + c=0…..(2)
Assuming b =𝑘 , 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡
𝑎 = −2𝑘 𝑎𝑛𝑑 𝑐 = 5𝑘 or 𝑋1=
−2𝑘
𝑘
5𝑘
or if 𝑘 =1 than 𝑋1=
−2
1
5
Now 𝑋1 , 𝑋2, and 𝑋3 are pairwise orthogonal .
Now rest of the process is same as in previous example.

28. INDEX ,SIGNATURE,RANK
Let 𝑄 = 𝑋𝑇
𝐴𝑋
be a quadratic form in n variables 𝑥1, 𝑥2, . . . . . . . . . . . 𝑥𝑛 .
INDEX ∶ The no. of positive terms (p) in the canonical form.
Signature: The difference between positive and negative terms in the canonical form.
Rank: A matrix is said to be of rank r when
(i) it has at least one non-zero minor of order r,
and (ii) every minor of order higher than r vanishes.
Briefly, the rank of a matrix is the largest order of any non-vanishing minor of
the matrix.
28

29. 29
Determine the rank of the following matrices:
(i)A=
𝟏 𝟐 𝟑
𝟏 𝟒 𝟐
𝟐 𝟔 𝟓
Sol. (i)Here A is a 3 x3 matrix
∴ 𝜌(𝐴) ≤ 3
After operating 𝑅2 → 𝑅2- 𝑅1 , 𝑅3 → 𝑅3−2𝑅1
A~
𝟏 𝟐 𝟑
𝟎 𝟐 −𝟏
𝟎 𝟐 −𝟏
Operating 𝑅3 → 𝑅3- 𝑅2
A~
𝟏 𝟐 𝟑
𝟎 𝟐 −𝟏
𝟎 𝟎 𝟎
The only third order minor is zero, but the second order minor
1 2
0 2
= 2 ≠0
∴ 𝜌 𝐴 = 2

30. 30
Sol. (ii)Here B is a 2 x 4 matrix.
∴ 𝜌(𝐴) ≤ 2, the smaller of 2 and 4.
The second order minor
1 2
−2 0
=4≠ 0
∴ 𝜌 𝐴 = 2
(ii) B=
𝟏 𝟐 𝟑 𝟒
−𝟐 𝟎 𝟓 𝟕

31.  Let 𝑋𝑇
𝐴𝑋 be a real quadratic form in n
variables 𝑥1,𝑥2, …. 𝑥𝑛 with rank r and signature s.
 Then we say that the quadratic form is
(i) positive definite if r = n, s = n
(ii) negative definite if r = n, s = 0
(iii) positive semidefinite if r<n and s = r
(iv) negative semidefinite if r<n, s = 0
(v) indefinite in all other cases.
31
NATURE OF QUADRATIC FORM
( using rank, signature)

32. NATURE OF QUADRATIC FORM
(using eigen values)
 POSITIVE DEFINITE: If all the eigen values of
the coefficient matrix are POSITIVE.
 POSITIVE SEMI DEFINITE: If all the eigen
values of the coefficient matrix are POSITIVE and
at least one is zero.
 NEGATIVE DEFINITE: If all the eigen values
of the coefficient matrix are NEGATIVE.
 NEGATIVE SEMI DEFINITE: If all the eigen
values of the coefficient matrix are NEGATIVE
and
at least one is zero.
➢ INDEFINITE: If some of the eigen values are
POSITIVE and some are NEGATIVE.
32

33. EX. Determine the nature ,index and signature of the quadratic form
𝑄(𝑥) = 2𝑥1𝑥2 + 2𝑥1𝑥3 + 2𝑥3𝑥2
Solution. Here
𝐴 =
0 1 1
1 0 1
1 1 0
Now the characteristic equation for A is
𝜆3 − 3𝜆 − 2 = 0
or 𝜆 = 2, −1, −1
Some of the eigen values are positive and
some are negative.
Hence 𝐐(𝐱) is indefinite.
Here the canonical form is 2𝑥1
2
− 𝑥2
2
− 𝑥3
2
.
Index =1 No. of positive term in canonical form
Signature=-1
(difference of positive and negative term in canonical form )
33