Answer to Question #223176 in Math for Masande

Question

$Question1 Given A= [■(2&2&1@1&3&1@1&2&2)] 1.1 find the eigenvalues of A 1.2. Determine an eigenvector for the eigenvalue = 5 Question2 A periodic function f(x) with period 2 π is defined by : f(x)= [ (x+π)/2 -π˂ X˂ 0, (x-π)/2 0˂ X˂ π] Determine the Fourier series for the periodic function$

Answer 1

1.

$$A=\begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{vmatrix}=0$$

$$(2-\lambda)^2(3-\lambda)+2+2-3+\lambda-4+2\lambda-4+2\lambda=0$$

$$(4-4\lambda+\lambda^2)(3-\lambda)+5\lambda-7=0$$

$$12-4\lambda-12\lambda+4\lambda^2+3\lambda^2-\lambda^3+5\lambda-7=0$$

$$-\lambda^3+7\lambda^2-11\lambda+5=0$$

$$-\lambda^2(\lambda-1)+6\lambda(\lambda-1)-5(\lambda-1)=0$$

$$-(\lambda-1)(\lambda^2-6\lambda+5)=0$$

$$-(\lambda-1)^2(\lambda-5)=0$$

The roots are $\lambda_1=1, \lambda_2=1, \lambda_3=5.$

These are the eigenvalues.

$\lambda=5$

$$\begin{pmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{pmatrix}=\begin{pmatrix} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{pmatrix}$$

$R_1=R_1/(-3)$

$$\begin{pmatrix} 1 & -2/3 & -1/3 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & -4/3 & 4/3 \\ 1 & 2 & -3 \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & -4/3 & 4/3 \\ 0 & 8/3 & -8/3 \end{pmatrix}$$

$R_2=-3R_2/4$

$$\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & 1 & -1 \\ 0 & 8/3 & -8/3 \end{pmatrix}$$

$R_1=R_1+2R_2/3$

$$\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 8/3 & -8/3 \end{pmatrix}$$

$R_3=R_3-8R_2/3$

$$\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}$$

If we take $x_3=t,$ then $x_1=t, x_2=t.$

$$\vec x=\begin{pmatrix} t\\ t\\ t \end{pmatrix}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}t$$ $$\vec v=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}$$

This is the eigenvector.

$\lambda=1$

$$\begin{pmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{pmatrix}=\begin{pmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 1 & 2 & 1 \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 2 & 1\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}$$

If we take $x_2=t, x_3=s,$ then $x_1=-2t-s.$

$$\vec x=\begin{pmatrix} -2t-s\\ t\\ s \end{pmatrix}=\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}t+\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}s$$

$$\vec u=\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}, \vec w=\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$$

These are the eigenvectors.

2.

$$a_0=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx$$

$$=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}dx+\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}dx$$

$$=\dfrac{2}{\pi}[\dfrac{x^2}{4}+\dfrac{\pi x}{2}]\begin{matrix} 0 \\ -\pi & d \end{matrix}+\dfrac{2}{\pi}[\dfrac{x^2}{4}-\dfrac{\pi x}{2}]\begin{matrix} \pi \\ 0 \end{matrix}$$

$$=\dfrac{\pi}{2}-\dfrac{\pi}{2}=0$$

$$a_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\cos{nx}dx$$

$$=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\cos{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\cos{nx}dx$$

$$=-\dfrac{\cos{\pi n}-1}{2\pi n^2}+\dfrac{\cos{\pi n}-1}{2\pi n^2}=0$$

$$b_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\sin{nx}dx$$

$$=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\sin{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\sin{nx}dx$$

$$=-\dfrac{1}{2n}-\dfrac{1}{2n}=-\dfrac{1}{n}$$

$$f(x)=-\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n}\sin{nx}$$