Science:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (a)/Solution 1

To compute the eigenvalues, first form the characteristic equation, then solve for its zeros.

$$\left|{\begin{array}{cc}3-\lambda &1\\-2&1-\lambda \end{array}}\right|=(3-\lambda )(1-\lambda )+2=\lambda ^{2}-4\lambda +5=0$$

To solve for the zeros, use

$$\lambda ={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(5)}}}{2(1)}}={\frac {4\pm {\sqrt {-4}}}{2}}=2\pm i,$$

thus the eigenvalues are ${\textstyle \lambda _{1}=2+i}$ and ${\textstyle \lambda _{2}=2-i}$. Since the eigenvalues are complex conjugates of each other, we know that the eigenvectors are also complex conjugates of each other, meaning we only need to compute one eigenvector to know both.

In question 17 we found the eigenvectors of a matrix $A$ by solving ${\textstyle A{\vec {v}}=\lambda {\vec {v}}}$ for the components or ${\textstyle {\vec {v}}}$ using basic algebra. For this question we will solve that same system using row operations. Solving for ${\textstyle {\vec {v}}_{1}}$, we write the system as ${\textstyle (A-\lambda _{1}I){\vec {v}}_{1}={\vec {0}}}$, row operations on the augmented system of equations gives

$$\left[{\begin{array}{cc|c}1-i&1&0\\-2&-1+i&0\end{array}}\right]\rightarrow \left[{\begin{array}{cc|c}1-i&1&0\\0&-2i&0\end{array}}\right]\rightarrow \left[{\begin{array}{cc|c}1-i&1&0\\0&0&0\end{array}}\right]$$

If ${\textstyle {\vec {v}}_{1}=[v_{1}\ \ v_{2}]^{T}}$, then we have ${\textstyle (1-i)v_{1}+v_{2}=0}$. Choosing (arbitrarily) ${\textstyle v_{1}=1}$, we get ${\textstyle v_{2}=-1+i}$, and the two eigenvectors are

$${\vec {v}}_{1}=\left[{\begin{array}{c}1\\-1+i\end{array}}\right],\ \ {\vec {v}}_{2}=\left[{\begin{array}{c}1\\-1-i\end{array}}\right].$$