Answer to Question #103668 in Linear Algebra for Sourav mondal

Question

Find the orthogonal canonical reduction of the quadratic form
-x²+y²+z²+xy+xz-6yz
. Also, find its principal axes.

Answer 1

First, determine the matrix of the quadratic form $Q = -x^2 + y^2 + z^2 + xy + xz - 6yz$.

$A = \begin{bmatrix} -1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & -3 \\ \frac{1}{2} & -3 & 1 \end{bmatrix}$

Find an eigenvalues of matrix $A$. To do it, solve the characteristic equation.

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

$\det(A-\lambda I) = \begin{vmatrix} -1-\lambda & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1-\lambda & -3 \\ \frac{1}{2} & -3 & 1-\lambda \end{vmatrix}$

$= (-1-\lambda)\begin{vmatrix} 1 - \lambda & -3 \\ -3 & 1- \lambda \end{vmatrix} - \frac{1}{2} \begin{vmatrix} \frac{1}{2} & -3 \\ \frac{1}{2} & 1- \lambda \end{vmatrix} + \frac{1}{2} \begin{vmatrix} \frac{1}{2} & 1 - \lambda \\ \frac{1}{2} & -3 \end{vmatrix}$

$= (-1- \lambda)[(1- \lambda)^2 - 9] - \frac{1}{2}[\frac{1}{2}(1- \lambda) + \frac{3}{2}] + \frac{1}{2}[-\frac{3}{2} - \frac{1}{2}(1- \lambda)]$

$= (-1- \lambda)(-2- \lambda)(4- \lambda) - \frac{1}{4}(4- \lambda) - \frac{1}{4}(4 - \lambda)$

$= (4 - \lambda)( \lambda^2 + 3 \lambda +\frac{3}{2})$

Using the formula for discriminant for the second term, obtain

$(4 - \lambda)(\lambda + \frac{3}{2} + \frac{\sqrt{3}}{2})(\lambda + \frac{3}{2} - \frac{\sqrt{3}}{2}) = 0$

Therefore, the eigenvalues, in the decreasing order, are

$\lambda_{1} = 4, \lambda_{2} = -\frac{3}{2} + \frac{\sqrt{3}}{2}, \lambda_{3} = -\frac{3}{2} - \frac{\sqrt{3}}{2}$

Thus, the orthogonal canonical reduction of $Q$ is

$Q = 4x_{1}^{2} + ( -\frac{3}{2} + \frac{\sqrt{3}}{2})y_{1}^{2} + ( -\frac{3}{2} - \frac{\sqrt{3}}{2})z_{1}^{2} \\ \\ \quad=4x_{1}^{2} - (\frac{3}{2} - \frac{\sqrt{3}}{2})y_{1}^{2} - ( \frac{3}{2} + \frac{\sqrt{3}}{2})z_{1}^{2}$

where $x_{1}, y_{1}, z_{1}$ are the new coordinates.

If $\lambda = 4$, then

$\begin{bmatrix} -1-4 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1-4 & -3 \\ \frac{1}{2} & -3 & 1-4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} -5 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -3 & -3 \\ \frac{1}{2} & -3 & -3 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

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

The first principal axis is

$p_{1} = \frac{1}{\sqrt{0^2 + (-1)^{2}+1^2}} \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}$

If $\lambda = -\frac{3}{2} + \frac{\sqrt{3}}{2}$, then

$\begin{bmatrix} -1 +\frac{3}{2} - \frac{\sqrt{3}}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1+\frac{3}{2} - \frac{\sqrt{3}}{2} & -3 \\ \frac{1}{2} & -3 & 1+\frac{3}{2} - \frac{\sqrt{3}}{2} \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2} - \frac{\sqrt{3}}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{5}{2} - \frac{\sqrt{3}}{2} & -3 \\ \frac{1}{2} & -3 & \frac{5}{2} - \frac{\sqrt{3}}{2} \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 1+\sqrt{3} \\ 1 \\ 1 \end{bmatrix}$

The second principal axis is

$p_{2} = \frac{1}{\sqrt{(1+\sqrt{3})^2 + 1^{2}+1^2}} \begin{bmatrix} 1+\sqrt{3} \\ 1 \\ 1 \end{bmatrix} \\ \quad = \frac{1}{\sqrt{6+2\sqrt{3}}} \begin{bmatrix} 1+\sqrt{3} \\ 1 \\ 1 \end{bmatrix}$

If $\lambda = -\frac{3}{2} - \frac{\sqrt{3}}{2}$, then

$\begin{bmatrix} -1 +\frac{3}{2} + \frac{\sqrt{3}}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1+\frac{3}{2} + \frac{\sqrt{3}}{2} & -3 \\ \frac{1}{2} & -3 & 1+\frac{3}{2} + \frac{\sqrt{3}}{2} \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2} -+\frac{\sqrt{3}}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{5}{2} + \frac{\sqrt{3}}{2} & -3 \\ \frac{1}{2} & -3 & \frac{5}{2} + \frac{\sqrt{3}}{2} \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 1-\sqrt{3} \\ 1 \\ 1 \end{bmatrix}$

The third principal axis is

$p_{3} = \frac{1}{\sqrt{(1-\sqrt{3})^2 + 1^{2}+1^2}} \begin{bmatrix} 1-\sqrt{3} \\ 1 \\ 1 \end{bmatrix} \\ \quad= \frac{1}{\sqrt{6-2\sqrt{3}}} \begin{bmatrix} 1-\sqrt{3} \\ 1 \\ 1 \end{bmatrix}$

 The change of basis needed to convert $Q$ into the orthogonal canonical reduction is given by

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 0 & \frac{1 + \sqrt{3}}{\sqrt{6+2\sqrt{3}}} & \frac{1 - \sqrt{3}}{\sqrt{6+-\sqrt{3}}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6+2\sqrt{3}}} & \frac{1}{\sqrt{6-2\sqrt{3}}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6+2\sqrt{3}}} & \frac{1}{\sqrt{6-2\sqrt{3}}} \end{bmatrix} \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix}$