Answer to Question #229508 in Linear Algebra for Bhass

Question

Reduce the quadratic form

2 2 2 8 7 3 12 – 8 4 x y z xy yz zx    



to the canonical form

through an orthogonal transformation and hence show that it

is positive Semi-definite.

Answer 1

Answer:-

$8x^2+7y^2+3z^2-8yz+4xz+12xy$

matrix in quadratic form:

$A=\begin{pmatrix} 8 & 6&2 \\ 6 & 7&-4 \\ 2&-4&3 \end{pmatrix}$

$\begin{vmatrix} 8 -\lambda& 6&2 \\ 6 & 7-\lambda&-4 \\ 2&-4&3-\lambda \end{vmatrix}=0$

$(8-\lambda)((7-\lambda)(3-\lambda)-16)-6(18-6\lambda+8)+2(-24-14+2\lambda)=0$

$(8-\lambda)(21-10\lambda+\lambda^2-16)+36\lambda-156+4\lambda-76=0$

$40-80\lambda+8\lambda^2-5\lambda+10\lambda^2-\lambda^3+40\lambda-232=0$

$\lambda^3-18\lambda^2+45\lambda+182=0$

$\lambda_1=-2.1,\lambda_2=6.3,\lambda_3=13.8$

eigenvectors:

for $\lambda_1=-2.1$ :

$10.1x+6y+2z=0$

$6x+9.1y-4z=0$

$2x-4y+5.1z=0$

$26.2x+21.1y=0$

$21.1y-19.3z=0$

$y=-1.2x,z=1.1y$

$x_1=\begin{pmatrix} 1 \\ -1.2 \\ -1.4 \end{pmatrix}$

for $\lambda_2=6.3$ :

$1.7x+6y+2z=0$

$6x+0.7y-4z=0$

$2x-4y-3.3z=0$

$2.6x-11.3y=0$

$12.7y+5.9z=0$

$y=0.2x,z=-2.2y$

$x_2=\begin{pmatrix} 1 \\ 0.2 \\ -0.4 \end{pmatrix}$

for $\lambda_3=13.8$ :

$-5.8x+6y+2z=0$

$6x-6.8y-4z=0$

$2x-4y-10.8z=0$

$17.6x-18.8y=0$

$5.2y+28.4z=0$

$y=0.9x,z=-0.2y$

$x_3=\begin{pmatrix} 1 \\ 0.9 \\ -0.2 \end{pmatrix}$

​normalized matrix:

$N=\begin{pmatrix} 1/|x_1| & 1/|x_2| &1/|x_3|\\ -1.2/|x_1| & 0.2/|x_2|&0.9/|x_3|\\ -1.4/|x_1|&-0.4/|x_2|&-0.2/|x_3| \end{pmatrix}$

$|x_1|=2.1,|x_2|=1.1,|x_3|=1.4$

$N=\begin{pmatrix} 0.5 & 0.9 &0.7\\ -0.6 & 0.2&0.6\\ -0.7&-0.4&-0.1 \end{pmatrix}$

$N^T=\begin{pmatrix} 0.5 & -0.6 &-0.7\\ 0.9 & 0.2&-0.4\\ 0.7&0.6&-0.1 \end{pmatrix}$

$AN=\begin{pmatrix} 8 & 6&2 \\ 6 & 7&-4 \\ 2&-4&3 \end{pmatrix}\begin{pmatrix} 0.5 & 0.9 &0.7\\ -0.6 & 0.2&0.6\\ -0.7&-0.4&-0.1 \end{pmatrix}=\begin{pmatrix} -1 & 7.6&9 \\ 1.6 & 7&8.8 \\ -2.4&-0.2&-1.3 \end{pmatrix}$

$D=N^TAN=\begin{pmatrix} 0.5 & -0.6 &-0.7\\ 0.9 & 0.2&-0.4\\ 0.7&0.6&-0.1 \end{pmatrix}\begin{pmatrix} -1 & 7.6&9 \\ 1.6 & 7&8.8 \\ -2.4&-0.2&-1.3 \end{pmatrix}=$

$=\begin{pmatrix} -8.4 &0 &0\\ 0 & 8.3&0\\ 0&0&11.7 \end{pmatrix}$

Canonical form: $0.2x^2+8.3y^2+11.7z^2$

quadratic form Q(x) = x^T · A · x is positive semidefinite if Q(x) ≥ 0