Question

Find the orthogonal canonical reduction of the quadratic form
−x2+y2+z2+ 2xy− 2xz+ 2yz. Also, find its principal axes

Answer 1

Answer on Question #64855 – Math – Linear Algebra

Question

Find the orthogonal canonical reduction of the quadratic form

$$- x ^ {2} + y ^ {2} + z ^ {2} + 2 x y - 2 x z + 2 y z$$

Also, find its principal axes.

Solution

The matrix of the quadratic form:

$$A = \left( \begin{array}{ccc} - 1 & 1 & - 1\\ 1 & 1 & 1\\ - 1 & 1 & 1 \end{array} \right).$$

The characteristic equation:

$$\left| \begin{array}{ccc} - 1 - \lambda & 1 & - 1\\ 1 & 1 - \lambda & 1\\ - 1 & 1 & 1 - \lambda \end{array} \right| = 0$$ $$(- 1 - \lambda) \left| \begin{array}{cc} 1 - \lambda & 1 \\ 1 & 1 - \lambda \end{array} \right| - \left| \begin{array}{cc} 1 & 1 \\ - 1 & 1 - \lambda \end{array} \right| - \left| \begin{array}{cc} 1 & 1 - \lambda \\ - 1 & 1 \end{array} \right| = 0$$ $$(- 1 - \lambda) ((1 - \lambda) ^ {2} - 1) - (1 - \lambda + 1) - (1 + 1 - \lambda) = 0$$ $$2 \lambda + 2 \lambda^ {2} - \lambda^ {2} - \lambda^ {3} + 2 \lambda - 4 = 0$$ $$\lambda^ {3} - \lambda^ {2} - 4 \lambda + 4 = 0$$ $$\lambda^ {2} (\lambda - 1) - 4 (\lambda - 1) = 0$$ $$(\lambda^ {2} - 4) (\lambda - 1) = 0$$ $$\lambda_ {1} = 1; \lambda_ {2} = - 2; \lambda_ {3} = 2.$$

The orthogonal canonical reduction:

$$Q = \lambda_ {1} x ^ {\prime 2} + \lambda_ {2} y ^ {\prime 2} + \lambda_ {3} z ^ {\prime 2}$$ $$Q = x ^ {\prime 2} - 2 y ^ {\prime 2} + 2 z ^ {\prime 2}.$$

For $\lambda_1 = 1$:

$$\begin{array}{l} \left( \begin{array}{ccc} -1 - 1 & 1 & -1 \\ 1 & 1 - 1 & 1 \\ -1 & 1 & 1 - 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \\ \left( \begin{array}{ccc} -2 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \\ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right). \end{array}$$

The principal axis:

$$\frac{1}{\sqrt{3}} \left( \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right).$$

For $\lambda_2 = -2$:

$$\begin{array}{l} \left( \begin{array}{ccc} -1 + 2 & 1 & -1 \\ 1 & 1 + 2 & 1 \\ -1 & 1 & 1 + 2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \\ \left( \begin{array}{ccc} 1 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \\ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 2 \\ -1 \\ 1 \end{array} \right). \end{array}$$

The principal axis:

$$\frac{1}{\sqrt{6}} \left( \begin{array}{c} 2 \\ -1 \\ 1 \end{array} \right).$$

For $\lambda_3 = 2$:

$$\left( \begin{array}{ccc} -1 - 2 & 1 & -1 \\ 1 & 1 - 2 & 1 \\ -1 & 1 & 1 - 2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)$$ $$\begin{array}{l} \left( \begin{array}{rrr} -3 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \\ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right). \end{array}$$

The principal axis:

$$\frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right).$$

Answer: $x^{\prime 2} - 2y^{\prime 2} + 2z^{\prime 2}; \frac{1}{\sqrt{3}}

\left( \begin{array}{c}

1 \\

1 \\

-1

\end{array} \right), \frac{1}{\sqrt{6}}

\left( \begin{array}{c}

2 \\

-1 \\

1

\end{array} \right), \frac{1}{\sqrt{2}}

\left( \begin{array}{c}

0 \\

1 \\

1

\end{array} \right).$

Answer provided by https://AssignmentExpert.com