Question

Find the linear transformation to reducethe following quadratic forms into cononical form using diagonalation method 3x^2+3y^2+3z^2+2xy+2xz-2yz

Answer 1

Solution;

The matrix representation,A,of the equation is;

$A=\begin{pmatrix} 3 & 1&1\\ 1 & 3&-1\\ 1&-1&3 \end{pmatrix}$

Find the characteristic equation from,

$|A-\lambda I|=0$

i.e;

$\lambda^3-s_1\lambda^2+s_2\lambda-s_3=0$

In which ;

$s_1$ =Sum of the main diagonal elements.

$s_1=3+3+3=9$

$s_2$ = Sum of the minors of the main diagonal elements;

$s_2=\begin{vmatrix} 3 & 1 \\ 1& 3 \end{vmatrix}$ +$\begin{vmatrix} 3 & 1\\ 1 & 3 \end{vmatrix}$ +$\begin{vmatrix} 3& -1 \\ -1 & 3 \end{vmatrix}$

$s_2=24$

$s_3$ =|A|=$\begin{vmatrix} 3& 1&1 \\ 1 & 3&-1\\ 1&-1&3 \end{vmatrix}$ =16

The characteristic equation is ;

$\lambda^3-9\lambda^2+24\lambda-16=0$

Solve for the roots of the equation;

If $\lambda=1$

1-9+24-16=0

Hence ,$\lambda =1$ is a root.

By synthetic Division,other roots are from;

$\lambda ^2-8\lambda+16=0$

$(\lambda-4)(\lambda-4)=0$

$\lambda=4,4$

Hence the eigenvalues are;

4,4,1

Find the eigenvectors;

(A-$\lambda I$ )x=0

$[\begin{pmatrix} 3 &1&1\\ 1& 3&-1\\ 1&-1&3 \end{pmatrix}$ -$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda&0\\ 0&0&\lambda \end{pmatrix}]x=0$

To have;

$\begin{bmatrix} 3-\lambda & 1&1 \\ 1 & 3-\lambda&-1\\ 1&-1&3-\lambda \end{bmatrix}$ $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}=0$

Take;

Case (i),if $\lambda$ =4,the equation becomes;

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

-x+y+z=0

x-y-z=0

x-y-z=0

The equation gives ;

$x_1=\begin{bmatrix} 2\\ 1\\ 1 \end{bmatrix}$

Case(iii),$\lambda =1$ ,the equation becomes;

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

2x+y+z=0

x+2y-z=0

x-y+2y=0

The equations give;

$x_3=\begin{bmatrix} -1\\ 1\\ 1 \end{bmatrix}$

To find the third set of eigenvectors,$x_3$

Since;

$x_1x_2^T=0$

$x_2x_3^T=0$

$x_3x_1^T=0$

We know;

$x_1=\begin{bmatrix} 2 \\ 1\\ 1 \end{bmatrix}$ ,$x_3=\begin{bmatrix} -1\\ 1\\ 1 \end{bmatrix}$

Let ;

$x_2=\begin{bmatrix} a \\ b\\ c \end{bmatrix}$

Hence;

$\begin{bmatrix} 2 & 1&1 \\ -1 & 1&1 \end{bmatrix}$$\begin{bmatrix} a \\ b\\ c \end{bmatrix}=0$

2a+b+c=0

-a+b+c=0

Hence,

$x_2=\begin{bmatrix} 0\\ -1\\ 1\\ \end{bmatrix}$

The model matrix,M from the eigenvectors is;

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

The normalised model matrix ,N,is as follows;

Length of eigenvectors;

$x_1=\sqrt{4+1+1}=\sqrt6$

$x_2=\sqrt{0+1+1}=\sqrt2$

$x_3=\sqrt{1+1+1}=\sqrt 3$

$N=\begin{bmatrix} \frac{2}{\sqrt6}&0&\frac{-1}{\sqrt3} \\ \frac{1}{\sqrt6} & \frac{-1}{\sqrt2}&\frac{1}{\sqrt3}\\ \frac{1}{\sqrt6}&\frac{1}{\sqrt2}&\frac{1}{\sqrt3} \end{bmatrix}$

Transponse normalised model matrix ,$N^T$ ;

$N^T=\begin{bmatrix} \frac{2}{\sqrt6}&\frac{1}{\sqrt6}&\frac{1}{\sqrt6} \\ 0 & \frac{-1}{\sqrt2}&\frac{1}{\sqrt2}\\ \frac{-1}{\sqrt3}&\frac{1}{\sqrt3}&\frac{1}{\sqrt3} \end{bmatrix}$

Find the diagonalized matrix,$D=N^TAN$

$D=\begin{bmatrix} 4 & 0&0 \\ 0 & 4&0\\ 0&0&1 \end{bmatrix}$

Hence D is the matrix of linear transformation of the quadratic form into canonical form.

Thus the linear transformation,X is;

X=Dy

X=$\begin{pmatrix} 4& 0&0 \\ 0 & 4&0\\ 0&0&1 \end{pmatrix}$ $(y_1,y_2,y_3)$