Answer to Question #211219 in Functional Analysis for Sani

Question

Find the eigenvalues and eigenvectors of the matrices A =

"

1 2

−8 4#

and B =

"

a b

−b a#

.

Answer 1

Solution.

1.

$$A=\begin{pmatrix} 1 & 2\\ -8& 4 \end{pmatrix}$$

Solve equation for finding eigenvalues:

$|A-\lambda I|=0$

$$\begin{vmatrix} 1- \lambda& 2 \\ -8 & 4-\lambda \end{vmatrix}=\lambda^2-5\lambda+20.$$

From here we have eigenvalues$\lambda_1=\frac{5-\sqrt{55}i}{2}, \lambda_2=\frac{5+\sqrt{55}i}{2}.$

Solve equation for finding the first eigenvector

$(A-\lambda_1 I)v_1=0.$

Hence

$$\begin{pmatrix} \frac{-3+\sqrt{55}i}{2} & 2 \\ -8 & \frac{3+\sqrt{55}i}{2} \end{pmatrix}\to \begin{pmatrix} 1 & \frac{-3-\sqrt{55}i}{16}\\ 0& 0 \end{pmatrix}$$

then $x_2=x_2, x_1=\frac{3+\sqrt{55}i}{16}.$

So, $v_1=\begin{pmatrix} \frac{3+\sqrt{55}i}{16} \\ 1 \end{pmatrix}.$

Solve equation for finding the second eigenvector

$(A-\lambda_2 I)v_2=0.$

Hence

$$\begin{pmatrix} \frac{-3-\sqrt{55}i}{2} & 2 \\ -8 & \frac{3-\sqrt{55}i}{2} \end{pmatrix}\to \begin{pmatrix} 1 & \frac{-3+\sqrt{55}i}{16}\\ 0& 0 \end{pmatrix}$$

then $x_2=x_2, x_1=\frac{3-\sqrt{55}i}{16}.$

So, $v_2=\begin{pmatrix} \frac{3-\sqrt{55}i}{16} \\ 1 \end{pmatrix}.$

2.

$$B=\begin{pmatrix} a& b\\ -b& a \end{pmatrix}$$

Solve equation for finding eigenvalues:

$|B-\lambda I|=0$

$$\begin{vmatrix} a- \lambda& b \\ -b & a-\lambda \end{vmatrix}=a^2-\lambda^2+b^2.$$

From here we have eigenvalues$\lambda_1=\sqrt{a^2+b^2}, \lambda_2=-\sqrt{a^2+b^2}.$

Solve equation for finding the first eigenvector

$(B-\lambda_1 I)v_1=0.$

Hence

$\begin{pmatrix} a- \sqrt{a^2+b^2}& b\\ -b & a-\sqrt{a^2+b^2} \end{pmatrix}\to \begin{pmatrix} b & -a-\sqrt{a^2+b^2}\\ 0& 2b\sqrt{a^2+b^2} \end{pmatrix}$ then $x_2=2b\sqrt{a^2+b^2}, x_1=2a\sqrt{a^2+b^2}+2(a^2+b^2).$

So, $v_1=\begin{pmatrix} 2a\sqrt{a^2+b^2}+2(a^2+b^2)\\ 2b\sqrt{a^2+b^2} \end{pmatrix}.$

Solve equation for finding the second eigenvector

$(B-\lambda_2 I)v_2=0.$

Hence

$\begin{pmatrix} a+\sqrt{a^2+b^2}& b\\ -b & a+\sqrt{a^2+b^2} \end{pmatrix}\to \begin{pmatrix} -b & a+\sqrt{a^2+b^2}\\ 0& -2b\sqrt{a^2+b^2} \end{pmatrix}$ then $x_2=-2b\sqrt{a^2+b^2}, x_1=2a\sqrt{a^2+b^2}-2(a^2+b^2).$

So, $v_2=\begin{pmatrix} 2a\sqrt{a^2+b^2}-2(a^2+b^2)\\ -2b\sqrt{a^2+b^2} \end{pmatrix}.$