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 #81486 – Math – Linear Algebra

Question

Find the orthogonal canonical reduction of the quadratic form

$$x^{2} + y^{2} + z^{2} - 2xy - 2xz - 2yz$$

Also, find its principal axes.

Solution

$$q = x^{2} + y^{2} + z^{2} - 2xy - 2xz - 2yz$$

Matrix of quadratic form

$$A = \begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}$$ $$q = \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ $$\begin{pmatrix} 1 - \lambda & -1 & -1 \\ -1 & 1 - \lambda & -1 \\ -1 & -1 & 1 - \lambda \end{pmatrix}$$

Characteristic equation

$$\begin{vmatrix} 1 - \lambda & -1 & -1 \\ -1 & 1 - \lambda & -1 \\ -1 & -1 & 1 - \lambda \end{vmatrix} = 0$$ $$(1 - \lambda) \begin{vmatrix} 1 - \lambda & -1 & -1 \\ -1 & 1 - \lambda \end{vmatrix} - (-1) \begin{vmatrix} -1 & -1 & 1 \\ -1 & 1 - \lambda \end{vmatrix} + (-1) \begin{vmatrix} -1 & 1 - \lambda \\ -1 & -1 \end{vmatrix} = 0$$ $$(1 - \lambda)(1 - 2\lambda + \lambda^{2} - 1) + (-1 + \lambda - 1) - (1 + 1 - \lambda) = 0$$ $$-2\lambda + \lambda^{2} + 2\lambda^{2} - \lambda^{3} - 2 + \lambda - 2 + \lambda = 0$$ $$- \lambda^{3} + 3 \lambda^{2} - 4 = 0$$ $$-(1 + \lambda) \lambda^{2} + 4 (\lambda^{2} - 1) = 0$$ $$-(1 + \lambda) (\lambda^{2} - 4\lambda + 4) = 0$$ $$-(1 + \lambda) (\lambda - 2)^{2} = 0$$ $$\lambda_{1} = 2$$ $$\lambda_{2} = 2$$ $$\lambda_{3} = -1$$

These are eigenvalues.

Find the eigenvectors

$$\lambda = 2$$ $$\left( \begin{array}{cccc} 1 - \lambda & -1 & -1 \\ -1 & 1 - \lambda & -1 \\ -1 & -1 & 1 - \lambda \end{array} \right) = \left( \begin{array}{cccc} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{array} \right)$$ $$\left( \begin{array}{cccc} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{array} \right) \xrightarrow{R_2 - R_1} \left( \begin{array}{cccc} -1 & -1 & -1 \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{array} \right)$$ $$\left( \begin{array}{cccc} -1 & -1 & -1 \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{array} \right) \xrightarrow{R_3 - R_1} \left( \begin{array}{cccc} -1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$ $$\left( \begin{array}{cccc} -1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \xrightarrow{(-1)R_1} \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

Solve the matrix equation

$$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)$$

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

Therefore,

$$\mathbf{v} = \left( \begin{array}{c} -t - s \\ t \\ s \end{array} \right) = \left( \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \right) t + \left( \begin{array}{c} -1 \\ 0 \\ 1 \end{array} \right) s$$ $$\lambda = -1$$ $$\left( \begin{array}{cccc} 1 - \lambda & -1 & -1 \\ -1 & 1 - \lambda & -1 \\ -1 & -1 & 1 - \lambda \end{array} \right) = \left( \begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right)$$ $$\left( \begin{array}{cccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right) \xrightarrow{R_2 + \left( \frac{1}{2} \right) R_1} \left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ -1 & -1 & 2 \end{array} \right)$$ $$\left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ -1 & -1 & 2 \end{array} \right) \xrightarrow{R_3 + \left( \frac{1}{2} \right) R_1} \left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ 0 & -3/2 & 3/2 \end{array} \right)$$ $$\left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ 0 & -3/2 & 3/2 \end{array} \right) \xrightarrow{R_3 + R_2} \left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ 0 & 0 & 0 \end{array} \right)$$ $$\left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 3/2 & -3/2 \\ 0 & 0 & 0 \end{array} \right) \xrightarrow{\left( \begin{array}{c} 2 \\ 3 \end{array} \right) R_2} \left( \begin{array}{cccc} 2 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right)$$ $$\begin{pmatrix} 2 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \xrightarrow{R_1 + R_2} \begin{pmatrix} 2 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$ $$\begin{pmatrix} 2 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \xrightarrow{R_1/2} \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$

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 $v_1 = t, v_2 = t$

Therefore,

$$\mathbf{v} = \begin{pmatrix} t \\ t \\ t \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} t$$

Eigenvalue:2, eigenvectors: $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$, $\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$

Eigenvalue: $-1$, eigenvector: $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

Thus, the orthogonal canonical reduction is

$$q = \begin{pmatrix} x' & y' & z' \\ \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = 2(x')^2 + 2(y')^2 - (z')^2$$

The normed vector are principal axes

$$\frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \text{ and } \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$$

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