The given quadratic form is
3 x 2 + 5 y 2 + 3 z 2 − 2 x y − 2 y z + 2 x z 3x^2+5y^2+3z^2-2xy-2yz+2xz 3 x 2 + 5 y 2 + 3 z 2 − 2 x y − 2 yz + 2 x z
The matrix of the given quadratic form is
A = ( 3 − 1 1 − 1 5 − 1 1 − 1 3 ) A=\begin{pmatrix}
3 & -1 &1 \\
-1& 5 & -1 \\
1 & -1 & 3
\end{pmatrix} A = ⎝ ⎛ 3 − 1 1 − 1 5 − 1 1 − 1 3 ⎠ ⎞
We write , A = I A I A= IAI A = I A I
i,e, ( 3 − 1 1 − 1 5 − 1 1 − 1 3 ) \begin {pmatrix}
3&-1&1 \\
-1&5&-1 \\
1&-1&3
\end{pmatrix} ⎝ ⎛ 3 − 1 1 − 1 5 − 1 1 − 1 3 ⎠ ⎞ = ( 1 0 0 0 1 0 0 0 1 ) =\begin{pmatrix}
1&0&0 \\
0&1&0\\
0&0&1
\end{pmatrix} = ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ A ( 1 0 0 0 1 0 0 0 1 ) A\begin{pmatrix}
1&0&0 \\
0&1&0 \\
0&0&1
\end{pmatrix} A ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞
Now we shall reduce A A A to diagonal form by applying congruence operation on it . Performing R 2 → 3 R 2 + R 1 , R 3 → 3 R 3 − R 1 ; C 2 → C 2 + 1 3 C 1 , R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3-R_1;C_2\rightarrow C_2+\frac{1}{3}C_1, R 2 → 3 R 2 + R 1 , R 3 → 3 R 3 − R 1 ; C 2 → C 2 + 3 1 C 1 ,
C 3 → C 3 − 1 3 C 1 ; R 3 → 7 R 3 + R 2 ; C 3 → C 3 − 1 7 C 2 C_3\rightarrow C_3-\frac{1}{3}C_1;R_3\rightarrow 7R_3+R_2;C_3\rightarrow C_3-\frac{1}{7}C_2 C 3 → C 3 − 3 1 C 1 ; R 3 → 7 R 3 + R 2 ; C 3 → C 3 − 7 1 C 2
We get ,
( 3 0 0 0 14 0 0 0 54 ) \begin{pmatrix}
3&0&0\\
0&14&0\\
0&0&54
\end{pmatrix} ⎝ ⎛ 3 0 0 0 14 0 0 0 54 ⎠ ⎞ = = = ( 1 0 0 1 3 0 − 6 3 21 ) \begin{pmatrix}
1&0&0 \\
1&3&0\\
-6&3&21
\end{pmatrix} ⎝ ⎛ 1 1 − 6 0 3 3 0 0 21 ⎠ ⎞ A ( 1 1 3 − 8 21 0 1 − 1 7 0 0 1 ) A\begin{pmatrix}
1& \frac{1}{3} & \frac{-8}{21} \\
0&1&\frac{-1}{7} \\
0&0&1
\end{pmatrix} A ⎝ ⎛ 1 0 0 3 1 1 0 21 − 8 7 − 1 1 ⎠ ⎞
Performing , R 1 → 1 3 R 1 , C 1 → 1 3 C 1 R_1\rightarrow \frac{1}{\sqrt{3}}R_1,
C_1\rightarrow \frac{1}{\sqrt{3}}C_1 R 1 → 3 1 R 1 , C 1 → 3 1 C 1 ;
R 2 → 1 14 R 1 , C 2 → 1 14 ; R_2\rightarrow \frac{1}{\sqrt{14}}R_1,C_2\rightarrow \frac{1} {\sqrt{14}}; R 2 → 14 1 R 1 , C 2 → 14 1 ;
R 3 → 1 54 R 3 , C 3 → 1 54 C 2 R_3\rightarrow \frac{1}{\sqrt{54}}R_3,C_3\rightarrow \frac{1}{\sqrt{54}}C_2 R 3 → 54 1 R 3 , C 3 → 54 1 C 2
We get ,
( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix}
1&0&0 \\
0&1&0\\
0&0&1
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ = ( a 0 0 b 3 b 0 − 6 c 3 c 21 c ) A =\begin{pmatrix}
a&0&0\\
b&3b&0\\
-6c&3c&21c
\end{pmatrix}A = ⎝ ⎛ a b − 6 c 0 3 b 3 c 0 0 21 c ⎠ ⎞ A ( a b 3 − 8 c 21 0 b − c 7 0 0 c ) \begin{pmatrix}
a&\frac{b}{3}&\frac{-8c}{21} \\
0&b&\frac{-c}{7} \\
0&0&c
\end{pmatrix} ⎝ ⎛ a 0 0 3 b b 0 21 − 8 c 7 − c c ⎠ ⎞
Where a = 1 3 , b = 1 14 , c = 1 54 a=\frac{1}{\sqrt{3}},b=\frac{1}{\sqrt{14}},c=\frac{1}{\sqrt{54}} a = 3 1 , b = 14 1 , c = 54 1
Thus the linear transformation ,
X = P Y X=PY X = P Y where ,
P = ( a b 3 − 8 c 21 0 b − c 7 0 0 c ) P=\begin{pmatrix}
a&\frac{b}{3}&\frac{-8c}{21} \\
0&b&\frac{-c}{7} \\
0&0&c
\end{pmatrix} P = ⎝ ⎛ a 0 0 3 b b 0 21 − 8 c 7 − c c ⎠ ⎞
X = [ x , y , z ] ′ , Y = [ y 1 , y 2 , y 3 ] ′ X=[x,y,z]',Y=[y_1,y_2,y_3]' X = [ x , y , z ] ′ , Y = [ y 1 , y 2 , y 3 ] ′
This transformation give quadratic form to the normal form
y 1 2 + y 2 2 + y 3 2 . . . . . . . . . . . . . ( 1 ) {y_1}^2+{y_2}^2+{y_3}^2.............(1) y 1 2 + y 2 2 + y 3 2 ............. ( 1 )
The rank r r r of the given quadratic form = = = The number of nonzero terms in its normal form ( 1 ) (1) ( 1 )
= = = 3.
The signature of the given quadratic form = = = the excess of the number of positive terms over the number of negative terms in its normal form= 3 − 0 = 3 =3-0=3 = 3 − 0 = 3
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