Question

Find the orthogonal canonical reduction of
the quadratic form Q = 3x2+ 2y2— 2.5 xy.
Also give its principal axes. Finally, draw a
rough sketch of the orthogonal canonical
reduction of Q = 4.

Answer 1

Solution:

$$\begin{pmatrix} 3 & -1.25 \\ -1.25 & 2 \end{pmatrix}$$

For find the ortogonal canonical reduction we make the following determinant:

$$\begin{vmatrix} 3-\lambda & -1.25 \\ -1.25 & 2-\lambda \end{vmatrix} =0$$

$$\lambda^2-5\lambda+\frac{71}{16}=0$$

This equation has irrational roots.

$$\lambda_1=\frac{10+\sqrt{29}}{4}$$

$$\lambda_2=\frac{10-\sqrt{29}}{4}$$ $$f=\frac {10+\sqrt{29}}{4}y_1^2+\frac{10-\sqrt{29}}{4}y_2^2$$

f is a ortogonal canonical reduction.

To go to the main axes, we solve the following systems of equations.

$$\begin{alignedat}{2} (3-\lambda)&x_1- &1.25&x_2 = 0 \\ -1.25&x_1+&(2-\lambda)&x_2 = 0 \end{alignedat}$$

Next, $\lambda=\lambda_1$ and $\lambda=\lambda_2$ are considered.

This means that the given quadratic form is reduced to the principal axes by an orthogonal linear transformation.

$$y_1=\frac{5}{\sqrt{58-4\sqrt{29}}}x_1+\frac{2-\sqrt{29}}{\sqrt{58-4\sqrt{29}}}x_2$$

$$y_2=\frac{5}{\sqrt{58+4\sqrt{29}}}x_1+\frac{2+\sqrt{29}}{\sqrt{58+4\sqrt{29}}}x_2$$

For rough sketch of the orthogonal canonical reduction of Q = 4 need to sketch an ellipse given by equation:

$$\frac{\tilde{x}^2}{\frac{16}{10-\sqrt{29}}}+\frac{\tilde{y}^2}{\frac{16}{10+\sqrt{29}}}=1$$