Given Information
We are given with a matrix A\[A = \left[ \begin{array} { c c } { 0.5 } & { - 0.6 } \\ { 0.75 } & { 1.1 } \end{array} \right]\] We have to find eigenvector by solving the first equation for $x_2$ in terms of $x_1$. And show that the vector y is a complex multiple of the vector $v_1$

Step-1:
Write the characteristic polynomial:\[\begin{array}{l} det (A - \lambda I) = 0\\\left| {\begin{array}{*{20}{c}}{0.5 - \lambda }&{ - 0.6}\\{0.75}&{1.1 - \lambda }\end{array}} \right| = 0\\(0.5 - \lambda )(1.1 - \lambda ) - ( - 0.6)(0.75) = 0\\{\lambda ^2} - 1.6\lambda + 1 = 0\end{array}\]Solve for the characteristic polynomial:\[\begin{aligned} \lambda & = \dfrac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a } \\ \lambda & = \dfrac { - ( - 1.6 ) \pm \sqrt { ( - 1.6 ) ^ { 2 } - 4 ( 1 ) ( 1 ) } } { 2 ( 1 ) } \\ & = \dfrac { 1.6 \pm \sqrt { - 1.44 } } { 2 } \\ & = \dfrac { 1.6 \pm 1.2 i } { 2 } \\ & = 0.8 \pm 0.6 i \end{aligned}\]Thus, the eigenvalues are:

\[\lambda = 0.8 + 0.6 i , 0.8 - 0.6 i\]



Step-2:
Eigenvector for $\lambda = 0.8 - 0.6 i$\[\begin{array} { r } { ( A - \lambda I ) \mathbf { x } = \mathbf { 0 } } \\ { \left[ \begin{array} { c } { 0.5 - ( 0.8 - 0.6 i ) } \\ { 0.75 } \end{array} \begin{array} { c } { - 0.6 } \\ { 1.1 - ( 0.8 - 0.6 i ) } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 1 } } \end{array} \right] = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] } \\ { \left[ \begin{array} { c c } { - 0.3 + 0.6 i } & { - 0.6 } \\ { 0.75 } & { 0.3 + 0.6 i } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] } \end{array}\]So, the system of equations are:\[\begin{array} { l } { ( - 0.3 + 0.6 i ) x _ { 1 } - 0.6 x _ { 2 } = 0 } \\ { 0.75 x _ { 1 } + ( 0.3 + 0.6 i ) x _ { 2 } = 0 } \end{array}\]Let, $x_2=k$ and write $x_1$ in terms of k\[\begin{aligned} 0.75 x _ { 1 } + ( 0.3 + 0.6 i ) k & = 0 \\ 0.75 x _ { 1 } & = - ( 0.3 + 0.6 i ) k \\ x _ { 1 } & = - \left( \dfrac { 0.3 + 0.6 i } { 0.75 } \right) k \\ x _ { 1 } & = ( - 0.4 - 0.8 i ) k \\ x _ { 1 } & = \left( \dfrac { - 2 - 4 i } { 5 } \right) k \end{aligned}\]Therefore, the general solution is \[\begin{aligned} \mathbf { x } & = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] \\ &\left. = \left[ \begin{array} { c } { \left( \dfrac { - 2 - 4 i } { 5 } \right) k } \\ { k } \end{array} \right] k \right] \\ & = \left[ \begin{array} { c } { - 2 - 4 i } \\ { 5 } \end{array} \right] \dfrac { k } { 5 } \end{aligned}\]Therefore, the eigenvector is:

\[\mathbf { v } _ { 1 } = \left[ \begin{array} { c } { - 2 - 4 i } \\ { 5 } \end{array} \right]\]



Step-3:
Let, $x_1=k$ and write $x_2$ in terms of k\[\begin{aligned} ( - 0.3 + 0.6 i ) k - 0.6 x _ { 2 } & = 0 \\ 0.6 x _ { 2 } & = ( - 0.3 + 0.6 i ) k \\ x _ { 2 } & = \left( \dfrac { - 0.3 + 0.6 i } { 0.6 } \right) k \\ x _ { 2 } & = \left( - \dfrac { 1 } { 2 } + i \right) k \\ x _ { 2 } & = \left( \dfrac { - 1 + 2 i } { 2 } \right) k \end{aligned}\]Therefore, the general solution is \[\begin{aligned} \mathbf { x } & = \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] \\ & = \left[ \begin{array} { c } { k } \\ { \left( \dfrac { - 1 + 2 i } { 2 } \right) k } \end{array} \right] \\ & = \left[ \begin{array} { c } { 2 } \\ { 2 } \\ { - 1 + 2 i } \end{array} \right] \dfrac { k } { 2 } \end{aligned}\]Therefore, the eigenvector is:

\[\mathbf { y } = \left[ \begin{array} { c } { 2 } \\ { - 1 + 2 i } \end{array} \right]\]



Step-4:
Write y as:\[\begin{aligned} \mathbf { y } & = \left[ \begin{array} { c } { 2 } \\ { - 1 + 2 i } \end{array} \right] \\ & = \dfrac { - 1 + 2 i } { 5 } \left[ \begin{array} { c } { - 2 - 4 i } \\ { 5 } \end{array} \right] \\ & = \dfrac { - 1 + 2 i } { 5 } \mathbf { v } _ { 1 } \end{aligned}\]So,

The vector $\mathbf { y }$ is a complex multiple of the vector $\mathbf { v } _ { \mathbf { 1 } }$