The matrix of the quadratic form is
A = [ 1 1 3 1 5 1 3 1 1 ] A=\begin{bmatrix}
1 & 1 & 3 \\
1 & 5 & 1 \\
3 & 1 & 1 \\
\end{bmatrix} A = ⎣ ⎡ 1 1 3 1 5 1 3 1 1 ⎦ ⎤ Find the eigenvalues and eigenvectors of A
A − λ I = [ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ] A-\lambda I=\begin{bmatrix}
1-\lambda & 1 & 3 \\
1 & 5-\lambda & 1 \\
3 & 1 & 1-\lambda \\
\end{bmatrix} A − λ I = ⎣ ⎡ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ⎦ ⎤
∣ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ∣ = ( 1 − λ ) ∣ 5 − λ 1 1 1 − λ ∣ \begin{vmatrix}
1-\lambda & 1 & 3 \\
1 & 5-\lambda & 1 \\
3 & 1 & 1-\lambda \\
\end{vmatrix}=(1-\lambda)\begin{vmatrix}
5-\lambda & 1 \\
1 & 1-\lambda
\end{vmatrix} ∣ ∣ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ∣ ∣ = ( 1 − λ ) ∣ ∣ 5 − λ 1 1 1 − λ ∣ ∣
− ( 1 ) ∣ 1 1 3 1 − λ ∣ + ( 3 ) ∣ 1 5 − λ 3 1 ∣ -(1)\begin{vmatrix}
1 & 1 \\
3 & 1-\lambda
\end{vmatrix}+(3)\begin{vmatrix}
1 & 5-\lambda \\
3 & 1
\end{vmatrix} − ( 1 ) ∣ ∣ 1 3 1 1 − λ ∣ ∣ + ( 3 ) ∣ ∣ 1 3 5 − λ 1 ∣ ∣
= ( 1 − λ ) 2 ( 5 − λ ) − 1 + λ − 1 + λ + 3 + 3 − 45 + 9 λ =(1-\lambda)^2(5-\lambda)-1+\lambda-1+\lambda+3+3-45+9\lambda = ( 1 − λ ) 2 ( 5 − λ ) − 1 + λ − 1 + λ + 3 + 3 − 45 + 9 λ
= − λ 3 + 7 λ 2 − 36 =-\lambda^3+7\lambda^2-36 = − λ 3 + 7 λ 2 − 36 This is a characteristic polynomial.
Solve the equation
− λ 3 + 7 λ 2 − 36 = 0 -\lambda^3+7\lambda^2-36=0 − λ 3 + 7 λ 2 − 36 = 0
− λ 2 ( λ − 6 ) + ( λ − 6 ) ( λ + 6 ) = 0 -\lambda^2(\lambda-6)+(\lambda-6)(\lambda+6)=0 − λ 2 ( λ − 6 ) + ( λ − 6 ) ( λ + 6 ) = 0
− ( λ − 6 ) ( λ − 3 ) ( λ + 2 ) = 0 -(\lambda-6)(\lambda-3)(\lambda+2)=0 − ( λ − 6 ) ( λ − 3 ) ( λ + 2 ) = 0 The roots are: λ 1 = 6 , λ 2 = 3 , λ 3 = − 2. \lambda_1=6,\lambda_2=3,\lambda_3=-2. λ 1 = 6 , λ 2 = 3 , λ 3 = − 2.
These are the eigenvalues.
Find the eigenvectors.
λ 1 = 6 \lambda_1=6 λ 1 = 6
[ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ] = [ − 5 1 3 1 − 1 1 3 1 − 5 ] \begin{bmatrix}
1-\lambda & 1 & 3 \\
1 & 5-\lambda & 1 \\
3 & 1 & 1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
-5 & 1 & 3 \\
1 & -1 & 1 \\
3 & 1 & -5 \\
\end{bmatrix} ⎣ ⎡ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ⎦ ⎤ = ⎣ ⎡ − 5 1 3 1 − 1 1 3 1 − 5 ⎦ ⎤ R 2 = R 2 + ( 1 / 5 ) R 1 R_2=R_2+(1/5)R_1 R 2 = R 2 + ( 1/5 ) R 1
[ − 5 1 3 0 − 4 / 5 8 / 5 3 1 − 5 ] \begin{bmatrix}
-5 & 1 & 3 \\
0 & -4/5 & 8/5 \\
3 & 1 & -5 \\
\end{bmatrix} ⎣ ⎡ − 5 0 3 1 − 4/5 1 3 8/5 − 5 ⎦ ⎤ R 3 = R 3 + ( 3 / 5 ) R 1 R_3=R_3+(3/5)R_1 R 3 = R 3 + ( 3/5 ) R 1
[ − 5 1 3 0 − 4 / 5 8 / 5 0 8 / 5 − 16 / 5 ] \begin{bmatrix}
-5 & 1 & 3 \\
0 & -4/5 & 8/5 \\
0 & 8/5 & -16/5 \\
\end{bmatrix} ⎣ ⎡ − 5 0 0 1 − 4/5 8/5 3 8/5 − 16/5 ⎦ ⎤ R 3 = R 3 + ( 2 ) R 2 R_3=R_3+(2)R_2 R 3 = R 3 + ( 2 ) R 2
[ − 5 1 3 0 − 4 / 5 8 / 5 0 0 0 ] \begin{bmatrix}
-5 & 1 & 3 \\
0 & -4/5 & 8/5 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 5 0 0 1 − 4/5 0 3 8/5 0 ⎦ ⎤ R 2 = ( − 5 / 4 ) R 2 R_2=(-5/4)R_2 R 2 = ( − 5/4 ) R 2
[ − 5 1 3 0 1 − 2 0 0 0 ] \begin{bmatrix}
-5 & 1 & 3 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 5 0 0 1 1 0 3 − 2 0 ⎦ ⎤ R 1 = R 1 − R 2 R_1=R_1-R_2 R 1 = R 1 − R 2
[ − 5 0 5 0 1 − 2 0 0 0 ] \begin{bmatrix}
-5 & 0 & 5 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 5 0 0 0 1 0 5 − 2 0 ⎦ ⎤ R 1 = ( − 1 / 5 ) R 1 R_1=(-1/5)R_1 R 1 = ( − 1/5 ) R 1
[ 1 0 − 1 0 1 − 2 0 0 0 ] \begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 1 − 2 0 ⎦ ⎤ Solve the matrix equation
[ 1 0 − 1 0 1 − 2 0 0 0 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] \begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix}\begin{bmatrix}
v _1 \\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 1 − 2 0 ⎦ ⎤ ⎣ ⎡ v 1 v 2 v 3 ⎦ ⎤ = ⎣ ⎡ 0 0 0 ⎦ ⎤
If we take v 3 = t , v_3=t, v 3 = t , then v 1 = t , v 2 = 2 t , v 3 = t . v_1=t,v_2=2t,v_3=t. v 1 = t , v 2 = 2 t , v 3 = t .
v = [ t 2 t t ] = [ 1 2 1 ] t \text{v}=\begin{bmatrix}
t \\
2t \\
t
\end{bmatrix}=\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}t v = ⎣ ⎡ t 2 t t ⎦ ⎤ = ⎣ ⎡ 1 2 1 ⎦ ⎤ t
λ 2 = 3 \lambda_2=3 λ 2 = 3
[ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ] = [ − 2 1 3 1 2 1 3 1 − 2 ] \begin{bmatrix}
1-\lambda & 1 & 3 \\
1 & 5-\lambda & 1 \\
3 & 1 & 1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
-2 & 1 & 3 \\
1 & 2 & 1 \\
3 & 1 & -2 \\
\end{bmatrix} ⎣ ⎡ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ⎦ ⎤ = ⎣ ⎡ − 2 1 3 1 2 1 3 1 − 2 ⎦ ⎤ R 2 = R 2 + ( 1 / 2 ) R 1 R_2=R_2+(1/2)R_1 R 2 = R 2 + ( 1/2 ) R 1
[ − 2 1 3 0 5 / 2 5 / 2 3 1 − 2 ] \begin{bmatrix}
-2 & 1 & 3 \\
0 & 5/2 & 5/2 \\
3 & 1 & -2 \\
\end{bmatrix} ⎣ ⎡ − 2 0 3 1 5/2 1 3 5/2 − 2 ⎦ ⎤ R 3 = R 3 + ( 3 / 2 ) R 1 R_3=R_3+(3/2)R_1 R 3 = R 3 + ( 3/2 ) R 1
[ − 2 1 3 0 5 / 2 5 / 2 0 5 / 2 5 / 2 ] \begin{bmatrix}
-2 & 1 & 3 \\
0 & 5/2 & 5/2 \\
0 & 5/2 & 5/2 \\
\end{bmatrix} ⎣ ⎡ − 2 0 0 1 5/2 5/2 3 5/2 5/2 ⎦ ⎤ R 3 = R 3 − R 2 R_3=R_3-R_2 R 3 = R 3 − R 2
[ − 2 1 3 0 5 / 2 5 / 2 0 0 0 ] \begin{bmatrix}
-2 & 1 & 3 \\
0 & 5/2 & 5/2 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 2 0 0 1 5/2 0 3 5/2 0 ⎦ ⎤ R 2 = ( 2 / 5 ) R 2 R_2=(2/5)R_2 R 2 = ( 2/5 ) R 2
[ − 2 1 3 0 1 1 0 0 0 ] \begin{bmatrix}
-2 & 1 & 3 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 2 0 0 1 1 0 3 1 0 ⎦ ⎤ R 1 = R 1 − R 2 R_1=R_1-R_2 R 1 = R 1 − R 2
[ − 2 0 2 0 1 1 0 0 0 ] \begin{bmatrix}
-2 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ − 2 0 0 0 1 0 2 1 0 ⎦ ⎤ R 1 = ( − 1 / 2 ) R 1 R_1=(-1/2)R_1 R 1 = ( − 1/2 ) R 1
[ 1 0 − 1 0 1 1 0 0 0 ] \begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 1 1 0 ⎦ ⎤ Solve the matrix equation
[ 1 0 − 1 0 1 1 0 0 0 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] \begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}\begin{bmatrix}
v _1 \\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 1 1 0 ⎦ ⎤ ⎣ ⎡ v 1 v 2 v 3 ⎦ ⎤ = ⎣ ⎡ 0 0 0 ⎦ ⎤
If we take v 3 = t , v_3=t, v 3 = t , then v 1 = t , v 2 = − t , v 3 = t . v_1=t,v_2=-t,v_3=t. v 1 = t , v 2 = − t , v 3 = t .
v = [ t − t t ] = [ 1 − 1 1 ] t \text{v}=\begin{bmatrix}
t \\
-t \\
t
\end{bmatrix}=\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}t v = ⎣ ⎡ t − t t ⎦ ⎤ = ⎣ ⎡ 1 − 1 1 ⎦ ⎤ t
λ 3 = − 2 \lambda_3=-2 λ 3 = − 2
[ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ] = [ 3 1 3 1 7 1 3 1 3 ] \begin{bmatrix}
1-\lambda & 1 & 3 \\
1 & 5-\lambda & 1 \\
3 & 1 & 1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
3 & 1 & 3 \\
1 & 7 & 1 \\
3 & 1 & 3 \\
\end{bmatrix} ⎣ ⎡ 1 − λ 1 3 1 5 − λ 1 3 1 1 − λ ⎦ ⎤ = ⎣ ⎡ 3 1 3 1 7 1 3 1 3 ⎦ ⎤ R 2 = R 2 − ( 1 / 3 ) R 1 R_2=R_2-(1/3)R_1 R 2 = R 2 − ( 1/3 ) R 1
[ 3 1 3 0 20 / 3 0 3 1 3 ] \begin{bmatrix}
3 & 1 & 3 \\
0 & 20/3 & 0 \\
3 & 1 & 3 \\
\end{bmatrix} ⎣ ⎡ 3 0 3 1 20/3 1 3 0 3 ⎦ ⎤ R 3 = R 3 − R 1 R_3=R_3-R_1 R 3 = R 3 − R 1
[ 3 1 3 0 20 / 3 0 0 0 0 ] \begin{bmatrix}
3 & 1 & 3 \\
0 & 20/3 & 0 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 3 0 0 1 20/3 0 3 0 0 ⎦ ⎤
R 2 = ( 3 / 20 ) R 2 R_2=(3/20)R_2 R 2 = ( 3/20 ) R 2
[ 3 1 3 0 1 0 0 0 0 ] \begin{bmatrix}
3 & 1 & 3 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 3 0 0 1 1 0 3 0 0 ⎦ ⎤ R 1 = R 1 − R 2 R_1=R_1-R_2 R 1 = R 1 − R 2
[ 3 0 3 0 1 0 0 0 0 ] \begin{bmatrix}
3 & 0 & 3 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 3 0 0 0 1 0 3 0 0 ⎦ ⎤ R 1 = ( 1 / 3 ) R 1 R_1=(1/3)R_1 R 1 = ( 1/3 ) R 1
[ 1 0 1 0 1 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 1 0 0 ⎦ ⎤
Solve the matrix equation
[ 1 0 1 0 1 0 0 0 0 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}\begin{bmatrix}
v _1 \\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 1 0 0 ⎦ ⎤ ⎣ ⎡ v 1 v 2 v 3 ⎦ ⎤ = ⎣ ⎡ 0 0 0 ⎦ ⎤
If we take v 3 = t , v_3=t, v 3 = t , then v 1 = − t , v 2 = 0 , v 3 = t . v_1=-t,v_2=0,v_3=t. v 1 = − t , v 2 = 0 , v 3 = t .
v = [ − t 0 t ] = [ − 1 0 1 ] t \text{v}=\begin{bmatrix}
-t \\
0 \\
t
\end{bmatrix}=\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}t v = ⎣ ⎡ − t 0 t ⎦ ⎤ = ⎣ ⎡ − 1 0 1 ⎦ ⎤ t
The Normalized matrix
N = [ 1 / 6 1 / 3 − 1 / 2 2 / 6 − 1 / 3 0 1 / 6 1 / 3 1 / 2 ] N=\begin{bmatrix}
1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\
2/\sqrt{6} & -1/\sqrt{3} & 0 \\
1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\
\end{bmatrix} N = ⎣ ⎡ 1/ 6 2/ 6 1/ 6 1/ 3 − 1/ 3 1/ 3 − 1/ 2 0 1/ 2 ⎦ ⎤
N T = [ 1 / 6 2 / 6 1 / 6 1 / 3 − 1 / 3 1 / 3 − 1 / 2 0 1 / 2 ] N^T=\begin{bmatrix}
1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\
1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\
-1/\sqrt{2} & 0 & 1/\sqrt{2} \\
\end{bmatrix} N T = ⎣ ⎡ 1/ 6 1/ 3 − 1/ 2 2/ 6 − 1/ 3 0 1/ 6 1/ 3 1/ 2 ⎦ ⎤
[ 1 / 6 2 / 6 1 / 6 1 / 3 − 1 / 3 1 / 3 − 1 / 2 0 1 / 2 ] A [ 1 / 6 1 / 3 − 1 / 2 2 / 6 − 1 / 3 0 1 / 6 1 / 3 1 / 2 ] = \begin{bmatrix}
1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\
1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\
-1/\sqrt{2} & 0 & 1/\sqrt{2} \\
\end{bmatrix}A\begin{bmatrix}
1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\
2/\sqrt{6} & -1/\sqrt{3} & 0 \\
1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\
\end{bmatrix}= ⎣ ⎡ 1/ 6 1/ 3 − 1/ 2 2/ 6 − 1/ 3 0 1/ 6 1/ 3 1/ 2 ⎦ ⎤ A ⎣ ⎡ 1/ 6 2/ 6 1/ 6 1/ 3 − 1/ 3 1/ 3 − 1/ 2 0 1/ 2 ⎦ ⎤ =
= [ 6 0 0 0 3 0 0 0 − 2 ] =\begin{bmatrix}
6 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -2 \\
\end{bmatrix} = ⎣ ⎡ 6 0 0 0 3 0 0 0 − 2 ⎦ ⎤
[ y 1 y 2 y 3 ] [ 6 0 0 0 3 0 0 0 − 2 ] [ y 1 y 2 y 3 ] = \begin{bmatrix}
y_1 & y_2 & y_3
\end{bmatrix}\begin{bmatrix}
6 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -2 \\
\end{bmatrix}\begin{bmatrix}
y_1 \\
y_2 \\
y_3
\end{bmatrix}= [ y 1 y 2 y 3 ] ⎣ ⎡ 6 0 0 0 3 0 0 0 − 2 ⎦ ⎤ ⎣ ⎡ y 1 y 2 y 3 ⎦ ⎤ =
= 6 y 1 2 + 3 y 2 2 − 2 y 3 2 =6y_1^2+3y_2^2-2y_3^2 = 6 y 1 2 + 3 y 2 2 − 2 y 3 2 which is the required canonical form.
R a n k = 3 Rank=3 R ank = 3 (the number of non-zero eigenvalues).
I n d e x = 2 Index=2 I n d e x = 2 (the number of positive eigenvalues).
S i g n a t u r e = 2 − 1 = 1 Signature=2-1=1 S i g na t u re = 2 − 1 = 1 (the difference between the number of positive eigenvalues and the number of negative eigenvalues ).
Nature: the eigenvalues are 6 > 0 , 3 > 0 , − 2 < 0. 6>0,3>0,-2<0. 6 > 0 , 3 > 0 , − 2 < 0. Therefore the given quadratic form is indefinite in nature.
Comments