Answer to Question #240525 in Linear Algebra for mds

Question

Given               B 2 31 2 21 1 32 find the eigenvalues of B and an eigenvector for B corresponding to  1.

Answer 1

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

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

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

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

$$=(2-\lambda)^3-3(2-\lambda)-6(2-\lambda)+3+6-(2-\lambda)$$

$$=8-12\lambda+6\lambda^2-\lambda^3-20+10\lambda+9$$

$$=-\lambda^3+6\lambda^2-2\lambda-3=0$$

$$-\lambda^2(\lambda-1)+5\lambda(\lambda-1)+3(\lambda-1)=0$$

$$\lambda_1=1$$

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

$$D=(5)^2-4(-1)(3)=37$$

$$\lambda_2=\dfrac{-5+\sqrt{37}}{-2}=-\dfrac{-5+\sqrt{37}}{2}$$

$$\lambda_3=\dfrac{-5-\sqrt{37}}{-2}=\dfrac{5+\sqrt{37}}{2}$$

Eigenvalues: $1, -\dfrac{-5+\sqrt{37}}{2}, \dfrac{5+\sqrt{37}}{2}.$

$\lambda=1$

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

$R_2=R_2-2R_1$

$$\begin{pmatrix} 1 & 3 & 1 \\ 0 & -5 & -1 \\ 1 & 3 & 1 \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & 3 & 1 \\ 0 & -5 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$

$R_2=R_2/(-5)$

$$\begin{pmatrix} 1 & 3 & 1 \\ 0 & 1 &1/5 \\ 0 & 0 & 0 \end{pmatrix}$$

$R_1=R_1-3R_2$

$$\begin{pmatrix} 1 & 0 & 2/5 \\ 0 & 1 &1/5 \\ 0 & 0 & 0 \end{pmatrix}$$

If we take $v_3=t,$ then $v_1=-\dfrac{2}{5}t, v_2=-\dfrac{1}{5}t,$

Thus

$$\vec v=\begin{pmatrix} (-2/5)t \\ (-1/5)t \\ t \end{pmatrix}=\begin{pmatrix} -2/5)\\ -1/5 \\ 1 \end{pmatrix}t$$

Eigenvalue: $1,$ eigenvector $\begin{pmatrix} -2/5)\\ -1/5 \\ 1 \end{pmatrix}.$