Question

Find the orthogonal canonical reduction of the quadratic form
−x2 + y2 + z2− 6xy− 6xz+ 2yz . Also, find its principal axes, rank and signature of the
quadratic form.

Answer 1

Answer on Question #61907 - Math - Linear Algebra

Question

Find the orthogonal canonical reduction of the quadratic form $-x2 + y2 + z2 - 6xy - 6xz + 2yz$ . Also, find its principal axes, rank and signature of the quadratic form.

$$- x ^ {2} + y ^ {2} + z ^ {2} - 6 x y - 6 x z + 2 y z$$

Solution

The matrix of the quadratic form $-x^{2} + y^{2} + z^{2} - 6xy - 6xz + 2yz$ is

$$A = \left( \begin{array}{ccc} - 1 & - 3 & - 3 \\ - 3 & 1 & 1 \\ - 3 & 1 & 1 \end{array} \right).$$

The characteristic equation is

$$\left| \begin{array}{ccc} - 1 - \lambda & - 3 & - 3 \\ - 3 & 1 - \lambda & 1 \\ - 3 & 1 & 1 - \lambda \end{array} \right| = 0,$$

hence

$$\lambda^ {3} - D _ {1} \lambda^ {2} + D _ {2} \lambda - D _ {3} = 0,$$

where

$$\begin{array}{l} D _ {1} = \operatorname {t r a c e} (A) = - 1 + 1 + 1 = 1; \\ D _ {2} = \left| \begin{array}{cc} - 1 & - 3 \\ - 3 & 1 \end{array} \right| + \left| \begin{array}{cc} - 1 & - 3 \\ - 3 & 1 \end{array} \right| + \left| \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right| = \\ = (- 1 \cdot 1 - (- 3) \cdot (- 3)) + (- 1 \cdot 1 - (- 3) \cdot (- 3)) + (1 \cdot 1 - 1 \cdot 1) = \\ = (- 1 - 9) + (- 1 - 9) + (1 - 1) = - 10 - 10 = - 20; \\ \end{array}$$ $$D _ {3} = \det (A) = \left| \begin{array}{ccc} - 1 & - 3 & - 3 \\ - 3 & 1 & 1 \\ - 3 & 1 & 1 \end{array} \right| = 0, \text{ because the second and the third rows of the matrix}$$

$A$ are equal.

Hence

$$\lambda^ {3} - \lambda^ {2} - 2 0 \lambda = 0.$$

It is easy to see that $\lambda = 0$ is the root of the equation.

$$\lambda^ {3} - \lambda^ {2} - 2 0 \lambda = \lambda (\lambda^ {2} - \lambda - 2 0) = \lambda (\lambda + 4) (\lambda - 5).$$

Therefore, the orthogonal canonical reduction of the quadratic form

$$- x ^ {2} + y ^ {2} + z ^ {2} - 6 x y - 6 x z + 2 y z \quad \text{is}$$ $$Q = \left( \begin{array}{c c c} x ^ {\prime} & y ^ {\prime} & z ^ {\prime} \end{array} \right) \left( \begin{array}{c c c} - 4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 5 \end{array} \right) \left( \begin{array}{c} x ^ {\prime} \\ y ^ {\prime} \\ z ^ {\prime} \end{array} \right) = - 4 x ^ {\prime 2} + 5 z ^ {\prime 2}.$$

If $\lambda = -4$ , then

$$\left( \begin{array}{c c c} - 1 + 4 & - 3 & - 3 \\ - 3 & 1 + 4 & 1 \\ - 3 & 1 & 1 + 4 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to \left( \begin{array}{c c c} 3 & - 3 & - 3 \\ - 3 & 5 & 1 \\ - 3 & 1 & 5 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right).$$

The principal axis is $\frac{1}{\sqrt{6}}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ .

If $\lambda = 0$ , then

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

The principal axis is $\frac{1}{\sqrt{2}}\left( \begin{array}{c}0\\ 1\\ -1 \end{array} \right)$ .

If $\lambda = 5$ , then

$$\left( \begin{array}{c c c} - 1 - 5 & - 3 & - 3 \\ - 3 & 1 - 5 & 1 \\ - 3 & 1 & 1 - 5 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to \left( \begin{array}{c c c} - 6 & - 3 & - 3 \\ - 3 & - 4 & 1 \\ - 3 & 1 & - 4 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} - 1 \\ 1 \\ 1 \end{array} \right).$$

The principal axis is $\frac{1}{\sqrt{3}}\binom{-1}{1}$ .

The rank is the total number $r$ of square terms (both positive and negative). Using (1) we come to conclusion that the rank is $r = 2$ .

Some authors assume the signature to be $n_{+} - n_{-}$ , where $n_{+}$ is the number of positive coefficients in the canonical representation of the form, $n_{-}$ is the number of negative coefficients in the canonical representation of the form. Using (1) we come to conclusion that the signature is $1 - 1 = 0$ .

Other authors assume the signature to be $(n_{+}, n_{-})$ . Using (1) we come to conclusion that the signature is (1,1) in this case.

Answer: $-4x^{\prime 2} + 5z^{\prime 2};\left\{\frac{1}{\sqrt{6}}\binom{2}{1},\frac{1}{\sqrt{2}}\binom{0}{1}, \frac{1}{\sqrt{3}}\binom{-1}{1}\right\}$ ; $r = 2$ ; either 0 or (1,1) (according to different definitions).

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