Question

Find the orthogonal canonical reduction of the quadratic form
x
2 +y
2 +z
2 −2xy−2xz−2yz. Also, find its principal axes.

Answer 1

Answer on Question #79870 – Math – Linear Algebra

Question

Find the orthogonal canonical reduction of the quadratic form

$$Q = x ^ {2} + y ^ {2} + z ^ {2} - 2 x y - 2 x z - 2 y z.$$

Also, find its principal axes.

Solution

$$\begin{array}{l} A = \left[ \begin{array}{c c c} 1 & - 1 & - 1 \\ - 1 & 1 & - 1 \\ - 1 & - 1 & 1 \end{array} \right] \\ - \det (\lambda I - A) = \det (A - \lambda I) = \left| \left[ \begin{array}{c c c} 1 - \lambda & - 1 & - 1 \\ - 1 & 1 - \lambda & - 1 \\ - 1 & - 1 & 1 - \lambda \end{array} \right] \right| = (1 - \lambda) (- 2 \lambda + \lambda^ {2}) - (- 1) (- 2 + \\ \lambda) + (- 1) (2 - \lambda) = (\lambda - 2) (- \lambda^ {2} + \lambda + 1 + 1) = (\lambda - 2) (2 - \lambda) (\lambda + 1) = 0 \\ \lambda_ {1} = \lambda_ {2} = 2, \lambda_ {3} = - 1 \text{ are eigenvalues}. \\ Q = 2 a ^ {2} + 2 b ^ {2} - c ^ {2} \text{ is the orthogonal canonical reduction.} \\ \end{array}$$ $$\begin{array}{l} \operatorname {K e r} \left[ \begin{array}{c c c} 1 - 2 & - 1 & - 1 \\ - 1 & 1 - 2 & - 1 \\ - 1 & - 1 & 1 - 2 \end{array} \right] = \operatorname {K e r} \left[ \begin{array}{c c c} - 1 & - 1 & - 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] = \operatorname {s p a n} \left\{\left[ \begin{array}{c} 1 \\ - 1 \\ 0 \end{array} \right], \left[ \begin{array}{c} 1 \\ 0 \\ - 1 \end{array} \right] \right\} = \\ = \operatorname {s p a n} \left\{\frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ - 1 \\ 0 \end{array} \right], \frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ 0 \\ - 1 \end{array} \right] \right\}. \\ \end{array}$$ $$\begin{array}{l} \operatorname {K e r} \left[ \begin{array}{c c c} 1 + 1 & - 1 & - 1 \\ - 1 & 1 + 1 & - 1 \\ - 1 & - 1 & 1 + 1 \end{array} \right] = \operatorname {K e r} \left[ \begin{array}{c c c} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{array} \right] = \operatorname {K e r} \left[ \begin{array}{c c c} 2 & - 1 & - 1 \\ - 2 & 4 & - 2 \\ - 2 & - 2 & 4 \end{array} \right] = \\ \operatorname {K e r} \left[ \begin{array}{c c c} 2 & - 1 & - 1 \\ 0 & 3 & - 3 \\ 0 & - 3 & 3 \end{array} \right] = \operatorname {K e r} \left[ \begin{array}{c c c} 2 & 0 & - 2 \\ 0 & 1 & - 1 \\ 0 & 0 & 0 \end{array} \right] = \operatorname {K e r} \left[ \begin{array}{c c c} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 0 \end{array} \right] = \operatorname {s p a n} \left\{\left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] \right\} = \operatorname {s p a n} \left\{\frac {1}{\sqrt {3}} \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] \right\}; \\ p _ {1} = \frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ - 1 \\ 0 \end{array} \right], p _ {2} = \frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ 0 \\ - 1 \end{array} \right] \text{ and } p _ {3} = \frac {1}{\sqrt {3}} \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] \text{ are the principal axes}. \\ \end{array}$$

Answer:

1. $Q = 2a^{2} + 2b^{2} - c^{2}$ is the orthogonal canonical reduction.

2. $p_1 = \frac{1}{\sqrt{2}}\left[ \begin{array}{c}1\\ -1\\ 0 \end{array} \right], p_2 = \frac{1}{\sqrt{2}}\left[ \begin{array}{c}1\\ 0\\ -1 \end{array} \right]$ and $p_3 = \frac{1}{\sqrt{3}}\left[ \begin{array}{c}1\\ 1\\ 1 \end{array} \right]$ are the principal axes.

Answer provided by https://www.AssignmentExpert.com