Page 312 - Advanced engineering mathematics

292 CHAPTER 9 Eigenvalues, Diagonalization, and Special Matrices

Proof Let Q be an orthogonal matrix that diagonalizes A. Then
n n
t
a ij x i x j = X AX
j=1 k=1
t
t
t
= (QY) AQY = (Y Q )AQY
t
t
= Y (Q AQ)Y
t
−1
= Y (Q AQ)Y
⎛ ⎞⎛ ⎞
0 0
λ 1 ··· y 1
0 0
⎜ λ 2 ··· ⎟⎜ ⎟
y 2

⎜ ⎟⎜ ⎟
= y 1 y 2 ··· y n ⎜ . . . . ⎟⎜ . ⎟
⎝ . . . . . . . . ⎠⎝ . ⎠
.
0 0 ··· λ n y n
= λ 1 y + λ 2 y + ··· + λ n y .
2
2
2
1 2 n
The expression n λ j y is called the standard form of X AX.
t
2
j=1 j
EXAMPLE 9.17
Consider the quadratic form
2
2
x − 7x 1 x 2 + x .
1 2
t
This is X AX, where

1 −7/2
.
−7/2 1
In general, the real quadratic form
2
ax + bx 1 x 2 + cx 2
1 2
t
can always be written as X AX, with A the real symmetric matrix

a b/2
A = .
b/2 c
In this example, the eigenvalues of A are −5/2,9/2 with corresponding eigenvectors

1 −1
and .
1 1
Divide each eigenvector by its length to obtain columns of an orthogonal matrix Q that
diagonalizes A:
√ √
1/ 2 −1/ 2
Q = √ √ .
1/ 2 1/ 2
The change of variables X = QY is equivalent to setting
1
x 1 = √ (y 1 − y 2 )
2
1
x 2 = √ (y 1 + y 2 ).
2

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October 14, 2010 14:49 THM/NEIL Page-292 27410_09_ch09_p267-294