Answer to Question #234680 in Calculus for Anuj

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

Assuming typo error, and modifying the condition of the question into

"Reduce the quadratic form $8x^2+7y^2+3z^2-12xy-8yz+4zx$

to the canonical form through orthogonal transformation and hence show it is positive semi-definite"

Solution

Converting to matrix form

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

Characteristic equation

det $||A-\lambda\Iota||=0$

det $\begin{Vmatrix} 8-\lambda&-6& 2\\ -6&7-\lambda & -4\\ 2&-4&3-\lambda \end{Vmatrix}$ =0...(i)

Characteristic polynomial

$\lambda^3-D_1\lambda^2+D_2\lambda-D_3=0$

Where $D_1=$ sum of main diagonal element

$=8+7+3=18$

$D_2=$ sum of minors of the main diagonal element

$D_2=\begin{vmatrix} 7& -4\\ -4& 3 \end{vmatrix}$ + $\begin{vmatrix} 8 & 2 \\ 2 & 3 \end{vmatrix}$ +

$\begin{vmatrix} 8 & -6\\ -6& 7 \end{vmatrix}$ $=5+20+20=45$

$D_3=$ Det $|A|$ = $8(21-16)-$

-$6(-18+8)+2(56-36)$ =

Characteristic equation is

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

Solving for $\lambda;$

$\lambda=0, \>\lambda=3,\>\lambda=15$

Finding Eigenvector

Case 1 $\lambda=0$

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

$\begin{pmatrix} 1&0& \frac{1}{2} \\ 0&1 & -1\\ 0&0&0 \end{pmatrix}$

$\begin{pmatrix} 1&0 & \frac{-1}{2}\\ 0&1 & -1\\ 0&0&0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ =

$\implies\>x=\frac{1}{2}z$

$y=z$

Vector =$\begin{pmatrix} \frac{1}{2} \\ 1 \\ 1 \end{pmatrix}$ =$\begin{pmatrix} 1 \\ 2\\ 2 \end{pmatrix}$

Case 2 $\lambda =3,$ substituting $\lambda=3$ in.....(i)

$\begin{pmatrix} 5&-6 & 2 \\ -6&4 & -4\\ 2&-4&0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ =0

Using rref of the matrix

$\begin{pmatrix} 1&0& 1\\ 0&1&\frac{1}{2} \\ 0&0&0 \end{pmatrix}$ $\begin{pmatrix} x\\ y \\ z \end{pmatrix}$ =0

$\implies\>x=-z$

$y=\>\frac{-1}{2}z$ let $z=1$

Vector is $\begin{pmatrix} -1 \\ \frac{-1}{2} \\ 1 \end{pmatrix}$ =$\begin{pmatrix} -2 \\ -1\\ 2 \end{pmatrix}$

Case 3, $\lambda=15,$ substituting $\lambda=15$ in .....(i)

$\begin{pmatrix} -7&-6 & 2 \\ -6&-8& -4\\ 2&-4&-12 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ =0,using rref form

$\begin{pmatrix} 1&0& -2 \\ 0&1&2 \\ 0&0&0 \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix}$ =0

$\implies \>x=2z$

$y= -2z$ let $z=1$

Vector is $\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$

Orthonormal matrix from unit eigenvectors

$Q=\frac{1}{3}$ $\begin{pmatrix} 1&-2& 2 \\ 2&-1 & -2\\ 2&2&1 \end{pmatrix}$

Diagonalizing matrix

$D=Q^TAQ=$

$\frac{1}{9}$ $\begin{pmatrix} 1&2 & 2 \\ -2&-1 & 2\\ 2&-2&1 \end{pmatrix}\begin{pmatrix} 8&-6 & 2 \\ -6&7 & -4\\ 2&-4&3 \end{pmatrix}\begin{pmatrix} 1&-2 & 2 \\ 2&-1 & -2\\ 2&2&1 \end{pmatrix}$

=$\begin{pmatrix} 0&0 & 0\\ 0&3 & 0\\ 0&0&15 \end{pmatrix}$

Orthogonal transformation reduce the quadratic form to canonical form

$3y_2^2+15y_3^2$

Nature of quadratic form

Contain a zero, eigenvalue and 2 eigenvalue $>1$

Therefore it is positive semi-definite.