91Ó°ÊÓ

Let \(A\) be square matrix of order \(n\). Then the eigen vector \(v\) corresponding to the eigen value \(\lambda \) can found by the following equation \(Av = \lambda v\).

From the example 2, the system of linear equations are:

\(\begin{aligned}{l}\left( { - .3 + .6i} \right){x_1} - .6{x_2} &= 0\\0.75{x_1} + \left( {.3 + .6i} \right){x_2} &= 0\end{aligned}\)

From the first equation we get

\(\begin{aligned}{}\left( { - .3 + .6i} \right){x_1} - .6{x_2} &= 0\\{x_2} &= \frac{{ - .3 + \cdot 6i}}{{.6}}{x_1}\end{aligned}\)

Hence the eigenvector is,

\(y = \left( {\begin{aligned}{}{{x_1}}\\{{x_2}}\end{aligned}} \right) \Rightarrow y = \left( {\begin{aligned}{}{{x_1}}\\{\frac{{ - .3 + .6i}}{{.6}}{x_1}}\end{aligned}} \right)\)

When \({x_1} = 2\), we get \(y = \left( {\begin{aligned}{}2\\{ - 1 + 2i}\end{aligned}} \right)\).

Also,

\(\begin{aligned}{}y &= \left( {\begin{aligned}{}2\\{ - 1 + 2i}\end{aligned}} \right)\\ &= \frac{{ - 1 + 2i}}{5}\left( {\begin{aligned}{}{ - 2 - 4i}\\5\end{aligned}} \right)\\ &= \frac{{ - 1 + 2i}}{5}{v_1}\end{aligned}\)

This means that \(y\) is a complex multiple of \({v_1}\).