I. Quadratic Forms and Canonical Forms
Given a quadratic homogeneou s polynomial with n variable s x1 , x 2 , L , x n .
Def 1：
f ( x1 , x 2 , L , xn ) =
2
a11 x1
+ 2a12 x1 x 2 + 2a13 x1 x3 + L + 2a1n x1 xn
+ a 22 x22 + 2a 23 x2 x3 + L + 2a 2 n x2 xn
+ a33 x32 + L + 2a3n x3 xn
+L +
called n - degree quadratic form, simply, quadratic form.
Definition 2：
If linear
operations
⎧ y1 = c11 x1 + c12 x2 + L + c1n xn
⎪y = c x + c x +L+ c x
⎪ 2
21 1
22 2
2n n
⎨
L
L
L
L
⎪
⎪⎩ y n = cn1 x1 + cn 2 x2 + L + cnn xn
2
a nn xn
Its matrix is ⎛ c11
Cn × n
⎜
⎜ c21
=⎜
M
⎜⎜
⎝ cn1
c12 L c1n ⎞
⎟
c22 L c2 n ⎟
M O M ⎟
⎟⎟
cn 2 L cnn ⎠
If it is invertible, then call it
an invertible linear
operation.
If it is orthogonal, then call it an
orthogonal operation.
Def 3： The degree of all unknowns in the quadratic form is 2, that is
f ( x1 , x 2 , L , xn ) =
2
d1 x1
+
2
d 2 x2
+L+
2
d n xn
then we call it canonical form of quadratic form(or standard form).
II. Matrix of quadratic form：
Let aij = a ji，then
f ( x1 , x 2 , L , xn ) = a11 x12 + a12 x1 x 2 + a13 x1 x3 + L + a1n x1 xn
+ a 21 x 2 x1 + a 22 x22 + a 23 x2 x3 + L + a 2 n x2 xn
+ LL
2
+ a n1 x n x1 + a n 2 x n x 2 + a n 3 x n x3 + L + a nn x n
⎛ a11 x1 + a12 x2 + L + a1n xn ⎞
⎟
= ( x1 , x2 , L , xn ) ⎜
⎜ a 21 x1 + a 22 x2 + L + a 2 n xn ⎟
⎟
⎜
LLLL
⎟⎟
⎜⎜
⎝ a n1 x1 + a n 2 x2 + L + a nn xn ⎠
⎛ a11 a12 L a1n ⎞ ⎛⎜ x1 ⎞⎟
⎜
⎟
a 21 a 22 L a 2 n ⎟ ⎜ x2 ⎟
⎜
= ( x1 , x2 , L , xn )
⎜ M
M O M ⎟⎜ M ⎟
⎜⎜
⎟⎟ ⎜ ⎟
⎝ a n1 an 2 L ann ⎠ ⎜⎝ xn ⎟⎠
Matrix of quadratic form
= X T AX
⎛ a11 a12
⎜
⎜ a 21 a 22
A=⎜
M
M
⎜⎜
⎝ an1 an 2
L a1n ⎞
⎟
L a2n ⎟
O M ⎟
⎟⎟
L a nn ⎠
Matrix of
quadratic form
（obviously, a
real symmetric
matrix）
⎛ x1 ⎞
⎜ ⎟
⎜ x2 ⎟
X =⎜ ⎟
M
⎜⎜ ⎟⎟
⎝ xn ⎠
Def 4： Given quadratic form f ( x1 , x 2 , L , xn ) = X T AX
we say the rank of symmetric matrix A is the rank of quadratic form f.
III. Matrix of quadratic form after invertible operations
=
invertible
operation
X =CY
T
B = C AC
r ( A) = r ( B ).
f ( x1 , x2 , L , xn ) = X T AX
(CY ) T A(CY ) = Y T (C T AC )Y
⇒ B T = B, Y T BY is a quadratic form,
and A, B are congruent matrics,
So:
Theory 1
Given quadratic form f ( x1 , x2 ,L, xn ) = X T AX , by invertible linear transforming of
X = CY , it can be changed into another quadratic form f = Y T BY with new variables
and its matrix is B = C T AC, and r ( A) = r ( B).
IV. Change quadratic to canonical
form by orthogonal transform.
⎛ d1
⎞
⎜
⎟
O
Λ=⎜
⎟
d n ⎟⎠
= ？⎜⎝
Question 1: Matrix of canonical form
Question 2：What’s on earth the problem of changing
quadratic into canonical form?
Find an invertible matrix C , such that C T AC = Λ is a diagonal matrix.
Question 3： Can the quadratic be changed into canonical forms?
Yes! Any a real symmetric matrix is orthogonal
congruent with a diagonal matrix.
T
Given
real
quadratic
form
f
=
X
AX ,
Theory 2
we have orthogonal transform X = QY , such that
T
T
T
T
T
f = X AX = (QY ) A(QY ) = Y (Q AQ )Y = Y ΛY
= λ1 y12 + λ2 y 22 + L + λn y n2
⎛ λ1
⎞
⎜
⎟
Λ=⎜
O
⎟, λ1 ,L, λn are eigenvallues of f of matrix A.
⎜
⎟
λ
n⎠
⎝
Process of changing quadratic into canonical form.
(i) Write down the matrix of quadratic form: A；
(ii )Find all the unequal eigenvalues of A : λ1 , λ2 ,L , λm ;
(iii )For each multiple eigenvalue λi , find ri linearly independent
m
eigenvectors ξ i1 , ξ i 2 , L , ξ iri；
(i = 1,2, L , m), and we know ∑ ri = n.
i =1
(iv) Use the Gram - Schmidt process to the ri linearly independent
eigenvectors : ξ i1 , ξ i 2 ,L , ξ iri；
(i = 1,2, L , m), which corresponding
to every multiple eigenvalue λi ; first orthogonalize and then unitise,
so we get ηi1 ,ηi 2 ,L ,ηiri；
(i = 1,2, L , m)，they are eigenvectors
which also belong to λi .
(v )Compose an n × n square matrix Q by the above
orthonorma l vectors as column vec tors, then Q is
the orthogonal square matrix we want , and now
Q −1 AQ = Q T AQ = Λ is a diagonal matrix.
(vi) Change by orthogonal transform : X = QY ,
Here we have changed quadratic forms into canonical forms.
f = X T AX = (QY )T A(QY ) = Y T (Q T AQ )Y = Y T ΛY
Changing quadratic into canonical forms
Learn by
by square and inertial theory.
yourself：
e.g.1. : Changing quadratic into canonical forms,
and find the invertible transforming matrix.
f ( x1 , x 2 , x 3 ) = x12 + 2 x 2 2 + 5 x32 + 2 x1 x 2 + 4 x 2 x3
Solution : f ( x1 , x2 , x3 ) = x1 + 2 x2 + 5 x3 + 2 x1 x2 + 4 x2 x3
2
2
2
=（x12 + 2 x1 x 2）+ 2 x 2 2 + 5 x32 + 4 x 2 x3
=（x1 + x 2）2 + x 2 2 + 5 x32 + 4 x 2 x3
=（x1 + x 2）2 +（x 2 + 2 x3）2 + x32 = y12 + y 2 2 + y 32
x1 = y1 − y 2 + 2 y 3
⎧
⎛1 −1 2 ⎞
⎧ y1 = x1 + x 2 ⎪
⎜
⎟
⎪
x2 = y 2 − 2 y3
C = ⎜0 1 − 2⎟
⎨
=
+
y
x
2
x
⎨ 2
2
3 ⎪
⎜0 0
⎟
1
=
x
y
⎪y = x
⎝
⎠
3
3
⎩
3
⎩ 3
e.g.2. Let the rank of quadratic form f ( x1 , x2 , x3 ) = x1 +
2
2
x2 + cx3 + 4 x1 x3 + 4 x2 x3 be 2.
2
1. Find the parameter : c;
2. Find an invertible transform which can be changed into canonical form;
3. What happens, if f ( x1 , x2 , x3 ) = 1?
For the rank of f ( x1 , x2 , x3 ) = x1 + x2 + cx3 + 4 x1 x3 + 4 x2 x3 be 2
2
2
2
⇒ the coefficient matrix of A is 2, ⇒ c = 8
f ( x1 , x 2 , x3 ) = x12 + x 2 2 + 8 x32 + 4 x1 x3 + 4 x 2 x3
2
2 =y 2+y 2
= ( x1 + 2 x3 ) + ( x 2 + 2 x3 )
1
2
⎧ y1 = x1 + 2 x3
⎪
⎨ y 2 = x 2 + 2 x3
⎪y = x
3
⎩ 3
⎧ x1 = y1 − 2 y 3
⎪
⎨ x2 = y 2 − 2 y3
⎪x = y
3
⎩ 3
⎛1 0 − 2⎞
⎜
⎟
C = ⎜0 1 − 2⎟
⎜0 0 1 ⎟
⎝
⎠
For A − λE = 0 ⇒ the eigenvalues of A :
λ1 = 0, λ2 = 1, λ3 = 9。
∴ By orthogonal transform, we can change f = 1 into
y 2 2 + 9 y 32 = 1
It’s an
elliptic cylinder.