Question

Find the orthogonal canonical reduction of the quadratic form
−

Answer 1

Find the orthogonal canonical reduction of the quadratic form $-x^2+y^2+z^2-6xy-6xz+2yz$

The matrix of the quadratic form

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

Find the eigenvalues

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

The characteristic equation is $det (A-\lambda I)=0$

$$A-\lambda I=\begin{vmatrix} -1-\lambda & -3 & -3 \\ -3 & 1-\lambda & 1\\ -3 & 1 & 1-\lambda \end{vmatrix}=$$ $$=( -1-\lambda)\begin{vmatrix} 1-\lambda & 1\\ 1 & 1-\lambda \end{vmatrix}-(-3)\begin{vmatrix} -3 & 1\\ -3 & 1-\lambda \end{vmatrix}+$$ $$+(-3)\begin{vmatrix} -3 & 1-\lambda \\ -3 & 1 \end{vmatrix}=(-1-\lambda)((1-\lambda)^2-1)+$$ $$+3(-3(1-\lambda)+3)-3(-3+3(1-\lambda))=0$$

$$2\lambda-\lambda^2+2\lambda^2-\lambda^3+9\lambda+9\lambda=0$$ $$\lambda(-\lambda^2+\lambda+20)=0$$ $$-\lambda(\lambda+4)(\lambda-5)=0$$

$\lambda_1=-4$

$\lambda_2=0$

$\lambda_3=5$

Find the eigenvectors

$\lambda=-4$

$$\begin{pmatrix} -1-\lambda & -3 & -3 \\ -3 & 1-\lambda & 1\\ -3 & 1 & 1-\lambda \end{pmatrix}=\begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 1\\ -3 & 1 & 5 \end{pmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 1\\ -3 & 1 & 5 \end{pmatrix}\sim\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -1\\ 0 & 0 & 0 \end{pmatrix}$$

Solve the matrix equation

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

If we take $v_3=t,$ then $v_1=2t, v_2=t, v_3=t.$

Therefore

$$\textbf{v}=\begin{pmatrix} 2t \\ t \\ t \end{pmatrix}=\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} t$$

The principal axis is

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

$\lambda=0$

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

Perform row operations to obtain the rref of the matrix:

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

Solve the matrix equation

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

If we take $v_3=t,$ then $v_1=0, v_2=-t, v_3=t.$

Therefore

$$\textbf{v}=\begin{pmatrix} 0 \\ -t \\ t \end{pmatrix}=\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} t$$

The principal axis is

$${1 \over \sqrt{2}}\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$$

$\lambda=5$

$$\begin{pmatrix} -1-\lambda & -3 & -3 \\ -3 & 1-\lambda & 1\\ -3 & 1 & 1-\lambda \end{pmatrix}=\begin{pmatrix} -6 & -3 & -3 \\ -3 & -4 & 1\\ -3 & 1 & -4 \end{pmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{pmatrix} -6 & -3 & -3 \\ -3 & -4 & 1\\ -3 & 1 & -4 \end{pmatrix}\sim\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1\\ 0 & 0 & 0 \end{pmatrix}$$

Solve the matrix equation

$$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

If we take $v_3=t,$ then $v_1=-t, v_2=t, v_3=t.$

Therefore

$$\textbf{v}=\begin{pmatrix} -t \\ t \\ t \end{pmatrix}=\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} t$$

The principal axis is

$${1 \over \sqrt{3}}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$

$$P=\begin{pmatrix} 2/\sqrt{6} & 0 & -1/\sqrt{3} \\ 1/\sqrt{6} & -1/\sqrt{2} & 1/\sqrt{3}\\ 1/\sqrt{6} & 1/\sqrt{2} & 1/\sqrt{3} \end{pmatrix}$$

$$P^{-1}=P^T=\begin{pmatrix} 2/\sqrt{6} & 1/\sqrt{6} & 1/\sqrt{6} \\ 0 & -1/\sqrt{2} & 1/\sqrt{2}\\ -1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3} \end{pmatrix}$$

$$D=\begin{pmatrix} -4 & 0 & 0 \\ 0 & 0 &0\\ 0 & 0 & 5 \end{pmatrix}$$

$$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}=P^T\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

$$x_1= (2/\sqrt{6})x+( 1/\sqrt{6})y+( 1/\sqrt{6})z$$ $$y_1= (0)x+( -1/\sqrt{2})y+( 1/\sqrt{2})z$$ $$z_1= (-1/\sqrt{3})x+( 1/\sqrt{3})y+( 1/\sqrt{3})z$$

The orthogonal canonical reduction of the quadratic form is

$$-4x_1^2+5z_1^2=0$$

The principal axes are

$$p_1={1 \over \sqrt{6}}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}, p_2={1 \over \sqrt{2}}\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}, p_3={1 \over \sqrt{3}}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$