Answer to Question #103741 in Linear Algebra for Rahul

Question

Find the orthogonal canonical reduction of the quadratic form -x²+y²+z²+xy+xz− 6

Answer 1

$-x^2+y^2+z^2-6xy-6xz+2yz$

The matrix of the quadratic form $-x^2+y^2+z^2-6xy-6xz+2yz$

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

The characteristics equation is

$\begin{vmatrix} -1-\lambda & -3 &-3\\ -3 & 1-\lambda&1\\ -3&1&1-\lambda\\ \end{vmatrix}$ $=0$

Hence

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

where

$D_1=trace(A)=-1+1+1=1;$

$D_2=\begin{vmatrix} -1 & -3\\ -3& 1 \end{vmatrix}$ + $\begin{vmatrix} -1 & -3\\ -3& 1 \end{vmatrix}$ +$\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix}$ $=$

$=(-1.1-(-3)(-3))+(-1.1-(-3)(-3))+(1.1-1.1)$

$=(-1-9)+(-1-9)+(1-1)=-10-10=-20$ ;

$D_3=det(A)=$ $\begin{vmatrix} -1 & -3 &-3\\ -3 & 1&1\\ -3&1&1\\ \end{vmatrix}$ $=0$ ,

because the second and the third rows of the matrix  A are equal

Hence,

$\lambda^3-\lambda^2-20\lambda=0$

It is easy to see that $\lambda=0$ is the root of the equation.

$\lambda^3-\lambda^2-20\lambda=\lambda(\lambda^2-\lambda-20)$ =$\lambda(\lambda+4)(\lambda-5)$

Therefore, the orthogonal canonical reduction of the quadratic form

$-x^2+y^2+z^2-6xy-6xz+2yz$ is

$Q=$ $(x',y',z')\begin{pmatrix} -4 & 0 &0\\ 0& 0&0\\ 0&0&5 \\ \end{pmatrix}$ $\begin{pmatrix} x' &\\ y'&\\ z'\\ \end{pmatrix}$ $=-4x'^2+5z'^2$