Question

a) Let a quadratic form have the expression x
2
+ y
2
+ 2z
2
+ 2xy + 3xz with respect to the
standard basis B
1 = f(1; 0; 0); (0; 1; 0); (0; 0; 1)g. Find its expression with respect to the
basis B
2 = f(1; 1; 1); (0; 1; 0); (0; 1; 1)g (3)
b) Consider the quadratic form
Q : 2x
2
4xy + y
2
+ 4xz + 3z
2
i) Find a symmetric matrix A such that Q = X
t
AX .
ii) Find the orthogonal canonical reduction of the quadratic form.
iii) Find the principal axes of the form.
iv) Find the rank and signature of the form. (5)

Answer 1

Answer on Question #46094-Engineering-Other

a) Let a quadratic form have the expression $x^{2} + y^{2} + 2z^{2} + 2xy + 3xz$ with respect to the standard basis $B_{1} = \{(1,0,0),(0,1,0),(0,0,1)\}$. Find its expression with respect to the basis $B_{2} = \{(1,1,1),(0,1,0),(0,1,1)\}$.

b) Consider the quadratic form $Q: 2x^{2} + y^{2} + 3z^{2} - 4xy + 4xz$

i) Find a symmetric matrix $A$ such that $Q = X^{t}AX$.

ii) Find the orthogonal canonical reduction of the quadratic form.

iii) Find the principal axes of the form.

iv) Find the rank and signature of the form.

Solution

a) $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = x' \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + y' \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + z' \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} x' \\ x' + y' + z' \\ x' + z' \end{pmatrix}$.

The expression of a quadratic form with respect to the basis $B_{2}$ is

$$\begin{aligned} x'^{2} + (x' + y' + z')^{2} + 2(x' + z')^{2} + 2x'(x' + y' + z') + 3x'(x' + z') &= 9x'^{2} + y'^{2} + 3z'^{2} + 4x'y' + 11x'z' + 2y'z'. \end{aligned}$$

b) i) $A = \begin{pmatrix} 2 & -2 & 2 \\ -2 & 1 & 0 \\ 2 & 0 & 3 \end{pmatrix}$.

ii) The characteristic equation for $Q$ is $\lambda^3 - D_1\lambda^2 + D_2\lambda - D_3 = 0$.

$$\begin{aligned} D_{1} &= 2 + 1 + 3 = 6; \\ D_{2} &= \left| \begin{array}{cc} 2 & -2 \\ -2 & 1 \end{array} \right| + \left| \begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array} \right| + \left| \begin{array}{cc} 2 & 2 \\ 2 & 3 \end{array} \right| = 3. \\ D_{3} &= |A| = -10. \end{aligned}$$

Hence

$$\lambda^{3} - 6\lambda^{2} + 3\lambda + 10 = 0.$$

Easy to see that $\lambda = -1$ is the root of equation.

$$\lambda^{3} - 6\lambda^{2} + 3\lambda + 10 = (\lambda + 1)(\lambda^{2} - 7\lambda + 10) = (\lambda + 1)(\lambda - 2)(\lambda - 5).$$

Therefore the orthogonal canonical reduction of the quadratic form is

$$Q = \begin{pmatrix} x' & y' & z' \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = -x'^{2} + 2y'^{2} + 5z'^{2}.$$

iii) $\lambda = -1$

$$\begin{pmatrix} 2 + 1 & -2 & 2 \\ -2 & 1 + 1 & 0 \\ 2 & 0 & 3 + 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \to \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}.$$

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

$$\lambda = 2$$ $$\begin{pmatrix} 2 - 2 & -2 & 2 \\ -2 & 1 - 2 & 0 \\ 2 & 0 & 3 - 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \to \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}.$$

The principal axis is $\frac{1}{3}\begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}$ .

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

The principal axis is $\frac{1}{3}\begin{pmatrix} -2 \\ 1 \\ -2 \end{pmatrix}$ .

v) The rank is 3 and signature is $-1 + 1 + 1 = 1$ .

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