Science:Math Exam Resources/Courses/MATH152/April 2012/Question 08 (b)/Solution 1

We first calculate the eigenvector $\displaystyle x$ of the eigenvalue $\displaystyle -1-2i$,

$0={\begin{pmatrix}-1-(-1-2i)&-2\\2&-1-(-1-2i)\end{pmatrix}}x={\begin{pmatrix}2i&-2\\2&2i\end{pmatrix}}x={\begin{pmatrix}2ix_{1}-2x_{2}\\2x_{1}+2ix_{2}\end{pmatrix}}=0$

Considering the first line, we obtain

${\begin{aligned}\displaystyle 2ix_{1}-2x_{2}&=0\\x_{2}&=ix_{1}\end{aligned}}$

Notice that there is a free variable which is common when finding eigenvectors. Therefore, we can choose $\displaystyle x_{1}=1$, then $\displaystyle x_{2}=i$ and have the eigenvector $\displaystyle x={\begin{pmatrix}1\\i\end{pmatrix}}$.

For the second eigenvector $\displaystyle y$ of the eigenvalue $\displaystyle -1+2i$, we calculate

$0={\begin{pmatrix}-2i&-2\\2&-2i\end{pmatrix}}y={\begin{pmatrix}-2iy_{1}-2y_{2}\\2y_{1}-2iy_{2}\end{pmatrix}}=0$

Considering the first line, we obtain

${\begin{aligned}\displaystyle -2iy_{1}-2y_{2}&=0\\y_{2}&=-iy_{1}\end{aligned}}$

There is a free choice like before and we choose $\displaystyle y_{1}=1$, then $\displaystyle y_{2}=-i$ and have the eigenvector $\displaystyle y={\begin{pmatrix}1\\-i\end{pmatrix}}$.