Question

Find the orthogonal canonical reduction of the quadratic form -x^2+y^2+z^2-4xy-4xz.
Also, find it's principal axes.

Answer 1

$$f(x, y, z)=-x^2+y^2+z^2-4xy-4xz$$

The matrix of the quadratic form is

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

The characteristic equation is

$$\begin{vmatrix} -1-\lambda & -2 & -2 \\ -2 & 1-\lambda & 0 \\ -2 & 0 & 1-\lambda \end{vmatrix}=0$$

Hence

$$(1-\lambda)({\lambda}^2-9)=0$$ $$\lambda_1=1, \lambda_2=3, \lambda_3=-3$$

Thus, the orthogonal canonical reduction is

$$Q=\begin{pmatrix} x' & y' & z' \\ \end{pmatrix}*\begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 &-3 \end{pmatrix}*\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}=(x')^2+3(y')^2-3(z')^2$$

Find eigenvectors:

$$\lambda_1=1$$ $$\begin{pmatrix} -2 & -2 & -2 \\ -2 & 0 & 0 \\ -2 & 0 &0 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 &0 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$v=\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$$

The principal axis is

$${1 \over \sqrt{2}}\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$$ $$\lambda_2=3$$ $$\begin{pmatrix} -4 & -2 & -2 \\ -2 & -2 & 0 \\ -2 & 0 &-2 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$v=\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$

The principal axis is

$${1 \over \sqrt{3}}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$ $$\lambda_3=-3$$ $$\begin{pmatrix} 2 & -2 & -2 \\ -2 & 4 & 0 \\ -2 & 0 &4 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}*\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$v=\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$$

The principal axis is

$${1 \over \sqrt{6}}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$$