线性代数II-chapter5(1-2)

that is, f =
( 2 ) Find eigenvalues λ1 , λ2 , , λn of A; ( 3) Find corresponding eigenvectors β1 , β 2 , , βi of
r
i , j =1
∑
n
aij xi x j = xT Ax, where AT = A
a11 a21 = ( x1 , x2 , , xn ) an1
where A = A is a symmetric matrix.
T
Example 1
Express the following quadratic form into a matrix form
2 2 f ( x1 , x2 , x3 ) = 3 x12 + x2 x3 + 2 x1 x2 x1 x3 + 4 x2 x3
Chapter 5 A quadratic form
§5.1-5.2 A quadratic form and its standard form §5.3 A positive definite quadratic form
New Words
A quadratic form 二次型 A standard form 标准型 Congruence 合同, 全等 合同,
λ 2
then λE A = 0 0
0 0 2 λ 3 1 = (λ 2)(λ 4) 1 λ 3
So the eigenvalues of A are λ1 = 2, λ2 = λ3 = 4.
To find an eigenvector corresponding to eigenvalue
2. The congruence of matrices
Definition 2 For given n × n matrices A and B, if there is an invertible
matrix P, such that PT AP = B then A is congruence to B, denoted by A B.
T
For λ2 = λ3 = 4, we solve the homogeneous linear system of equations that is
( 4 E A) x = 0,
0 0 0 x1 0 1 1 x2 = 0, 0 1 1 x 3
we obtain a set of vectors α 2 = (1, 0, 0 ) , α 3 = ( 0,1,1)
The proof of the above theorem is straightforward and is provided here for the interested students.
Procedures to change a quadratic form into the standard form by an orthogonal transformation (1) Rewrite the quadratic form into the matrix form;
Sቤተ መጻሕፍቲ ባይዱlution
Since 1 f ( x1 , x2 , x3 ) = 3 x + x1 x2 x1 x3 2 2 +x2 x1 + x2 + 2 x2 x3
2 1
1 2 x3 x1 + 2 x3 x2 x3 2
thus, 3 f ( x1 , x2 , x3 ) = ( x1 , x2 , x3 ) 1 1 2 1 1 2 x1 1 2 x2 x 2 1 3
2 2 f ( x1 , x2 , , xn ) = a11 x12 + a22 x2 + + ann xn
+ 2a12 x1 x2 + 2a13 x1 x3 + 2a1n x1 xn + 2a23 x2 x3 + + 2an 1,n xn 1 xn is called a quadratic form.
T
T
is a maximal set of linearly independent eigenvectors corresponding to eigenvalue λ2 = λ3 = 4, then we take nuit eignvectors α2 T p2 = = (1, 0, 0 ) , α2
∑
n
aij xi x j = xT Ax,
there is an orthogonal transformation x = Py ( P is an orthogonal matrix) such that
2 2 2 f = xT Ax = y T PT AP y = λ1 y1 + λ2 y2 + + λn yn
α3 2 2 p3 = = 0, , , α3 2 2
thus, under the orthogonal transformation
T
0 x1 2 x2 = 2 x 3 2 2 we have
1 0 0
0 y1 2 y2 , 2 y3 2 2 f = 2y + 4y + 4y
( 4)
every eigenvalue λi ;
Orthogonalize eigenvectors β1 , β 2 , , βir using
the Gram-Schmidt Method;
( 5) Obtain a orthonormal eigenvectors ξ1 , ξ 2 , , ξ n
Notations
(1) If aij = a ji (i, j = 1,2,, n ), then 2 f (x1 , x2 , , xn ) = a11 x1 + a12 x1 x2 + + a1n x1 xn
2 + a21 x2 x1 + a22 x2 + + a2 n x2 xn
+
2 + an1 xn x1 + an 2 xn x2 + + ann xn n
where λ1 , λ2 , , λn are eignvalues of A,
(
)
if let
P = ( p1 , p2 , , pn ),
then p1 , p2 , , pn are eignvectors of A corresponding to λ1 , λ2 , , λn , respectively.
T
there is an invertible matrix C such that B = C T AC , that is A B
3. The standard form of a quadratic form
Definition 3
2 2 The quadratic form f = d1 y12 + d 2 y2 + + d n yn is
Example 2
Transform a quadratic form
2 2 f = 4 x12 + 3 x2 + 3 x3 + 2 x2 x3
into the standard form.
4 0 0 The matrix of f is A = 0 3 1 , 0 1 3
Solution
=
i , j =1
∑a x x
ij i
j
(2) We can write the quadratic form into a matrix form,
f (x1 , x2 , , xn ) = x1 (a11 x1 + a12 x2 + + a1n xn ) that is + x2 (a21 x1 + a22 x2 + + a2 n xn ) + + xn (an1 x1 + an 2 x2 + + ann xn ) a11 x1 + a12 x2 + + a1n xn a21 x1 + a22 x2 + + a2 n xn = (x1 , x2 , , xn ) a x + a x ++ a x n2 2 nn n n1 1
Notations
(1) Obviously, the following properties are satisfied
(I) A A; (II) If A B, then B A; (III) If A B, B C , then A C.
2 ) The quadratic form f = xT Ax can be changed into ( f = y By, under the linear transformation x = Cy
to get a solution α1 = k1 ( 0,1, 1) , which is all
T
eigenvectors corresponding to eigenvalue λ1 = 2, then we take an unit eigenvector, 2 2 α1 p1 = = 0, 2 , 2 , α1
called the standard form.
Notations
(1) To change the quadratic form
f = xT Ax into the satndard form
2 2 2 f = d1 y1 + d 2 y2 + + d n yn