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 #64856 – 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

Denote by $f$ this quadratic form. Then $f(R) = R^T A R$, where

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

is a $3 \times 3$ matrix,

$$R = \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$

is a column $(3 \times 1$ matrix).

So, it is required to find a orthogonal matrix $Q$ such that

$$Q ^ {T} Q = E$$

and

$$f (R) = d _ {1} x ^ {\prime 2} + d _ {2} y ^ {\prime 2} + d _ {3} z ^ {\prime 2},$$

where

$$R = Q R ^ {\prime},$$ $$R ^ {\prime} = \left( \begin{array}{c} x ^ {\prime} \\ y ^ {\prime} \\ z ^ {\prime} \end{array} \right),$$

i.e., $f(R) = R^{\prime T} D R^{\prime}$, where $D$ is a diagonal $3 \times 3$ matrix,

$$D = \left( \begin{array}{ccc} d _ {1} & 0 & 0 \\ 0 & d _ {2} & 0 \\ 0 & 0 & d _ {3} \end{array} \right),$$

$E$ is a unit matrix.

Then

$$f (R) = R ^ {\prime T} Q ^ {T} A Q R ^ {\prime}.$$

Therefore $D = Q^T A Q$ and $A Q = Q D$.

Then $\sum_{j}^{3} A_{ij} Q_{jk} = d_k Q_{ik}$ (the first index is the number of a row and the second index is the number of a column). So the columns of $Q$ are eigenvectors of $A$ and $d_1, d_2, d_3$ are eigenvalues of $A$.

Eigenvalues of $A$ are roots of the polynomial

$$\det (A - d \mathrm {E}) = \det \left( \begin{array}{ccc} 1 - d & 1 & -1 \\ 1 & 1 - d & 1 \\ -1 & 1 & 1 - d \end{array} \right) = - d ^ {3} + 3 d ^ {2} - 4 = - (d + 1) (d - 2) ^ {2}.$$

This polynomial has a single root $-1$ and double root $2$.

The coordinates of the eigenvector of $A$ associated with eigenvalue $-1$ are a solution of a linear system $(A + E)U = 0$ with respect to $U$. This system can be written as

$$\left\{ \begin{array}{l} 2 u _ {1} + u _ {2} - u _ {3} = 0 \\ u _ {1} + 2 u _ {2} + u _ {3} = 0 \\ - u _ {1} + u _ {2} + 2 u _ {3} = 0 \end{array} \right.$$

Hence $u_{1} = u_{3}$, $u_{2} = -u_{3}$. $Q$ is orthogonal, it follows that $u_{1}^{2} + u_{2}^{2} + u_{3}^{2} = 1 = 3u_{3}^{2}$. Therefore $u_{1} = \frac{1}{\sqrt{3}}, u_{2} = -\frac{1}{\sqrt{3}}, u_{3} = \frac{1}{\sqrt{3}}$.

The coordinates of the eigenvectors of $A$ with eigenvalues 2 are a solution of a linear system $(A - 2E)U = 0$ with respect to $U$:

$$\left\{ \begin{array}{l} - u _ {1} + u _ {2} - u _ {3} = 0 \\ u _ {1} - u _ {2} + u _ {3} = 0 \\ - u _ {1} + u _ {2} - u _ {3} = 0 \end{array} \right.$$

The rank of this system is equal to 1 and there are two linearly independent solutions of this system. It follows from $u_{1} = -u_{3} / 2, u_{2} = u_{3} / 2, u_{1}^{2} + u_{2}^{2} + u_{3}^{2} = 1 = \frac{3}{2} u_{3}^{2}$ that the first solution is

$$u _ {1} = - \frac {1}{\sqrt {6}}, u _ {2} = \frac {1}{\sqrt {6}}, u _ {3} = \frac {\sqrt {2}}{\sqrt {3}}.$$

The solution that is orthogonal to it is a solution of the system

$$\left\{ \begin{array}{c} - u _ {1} + u _ {2} - u _ {3} = 0, \\ - \frac {1}{\sqrt {6}} u _ {1} + \frac {1}{\sqrt {6}} u _ {2} + \frac {\sqrt {2}}{\sqrt {3}} u _ {3} = 0, \end{array} \right.$$

that is,

$$\left\{ \begin{array}{l} - u _ {1} + u _ {2} - u _ {3} = 0, \\ - u _ {1} + u _ {2} + 2 u _ {3} = 0. \end{array} \right.$$

Then

$$u _ {3} = 0, u _ {1} = u _ {2}, u _ {1} ^ {2} + u _ {2} ^ {2} + u _ {3} ^ {2} = 1 = 2 u _ {1} ^ {2}, \text{ hence } u _ {1} = u _ {2} = \frac {1}{\sqrt {2}}, u _ {3} = 0.$$

Therefore

$$Q = \left( \begin{array}{ccc} \frac {1}{\sqrt {3}} & - \frac {1}{\sqrt {6}} & \frac {1}{\sqrt {2}} \\ - \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {6}} & \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {3}} & \frac {\sqrt {2}}{\sqrt {3}} & 0 \end{array} \right),$$ $$x = \frac {1}{\sqrt {3}} x ^ {\prime} - \frac {1}{\sqrt {6}} y ^ {\prime} + \frac {1}{\sqrt {2}} z ^ {\prime}$$ $$y = - \frac {1}{\sqrt {3}} x ^ {\prime} + \frac {1}{\sqrt {6}} y ^ {\prime} + \frac {1}{\sqrt {2}} z ^ {\prime},$$ $$z = \frac {1}{\sqrt {3}} x ^ {\prime} + \frac {\sqrt {2}}{\sqrt {3}} y ^ {\prime},$$ $$x ^ {\prime} = \frac {1}{\sqrt {3}} x - \frac {1}{\sqrt {3}} y + \frac {1}{\sqrt {3}} z,$$ $$y ^ {\prime} = - \frac {1}{\sqrt {6}} x + \frac {1}{\sqrt {6}} y + \frac {\sqrt {2}}{\sqrt {3}} z,$$ $$z ^ {\prime} = \frac {1}{\sqrt {2}} x + \frac {1}{\sqrt {2}} y.$$

Principal axes are such columns $e_1, e_2, e_3$ that are a basis and the coordinates of $R$ in this basis are $x', y', z'$:

$$R = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = x' e_1 + y' e_2 + z' e_3$$

for any $R$.

It means that $R = (e_1, e_2, e_3)R'$, where $(e_1, e_2, e_3)$ is a matrix composed by columns $e_1, e_2, e_3$. Therefore $Q = (e_1, e_2, e_3)$ and the principal axes are the columns of $Q$.

**Answer:**

The orthogonal canonical reduction is

$$\begin{array}{l} x = \frac{1}{\sqrt{3}} x' - \frac{1}{\sqrt{6}} y' + \frac{1}{\sqrt{2}} z' \\ y = -\frac{1}{\sqrt{3}} x' + \frac{1}{\sqrt{6}} y' + \frac{1}{\sqrt{2}} z', \\ z = \frac{1}{\sqrt{3}} x' + \frac{\sqrt{2}}{\sqrt{3}} y' \\ \end{array}$$ $$\begin{array}{l} x' = \frac{1}{\sqrt{3}} x - \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z \\ y' = -\frac{1}{\sqrt{6}} x + \frac{1}{\sqrt{6}} y + \frac{\sqrt{2}}{\sqrt{3}} z. \\ z' = \frac{1}{\sqrt{2}} x + \frac{1}{\sqrt{2}} y \\ \end{array}$$

Then the reduced form is

$$- x'^2 + 2 y'^2 + 2 z'^2.$$

The principal axes are

$$\left( \begin{array}{c} \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \right), \left( \begin{array}{c} -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ \frac{\sqrt{2}}{\sqrt{3}} \end{array} \right), \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right).$$

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