Question

In Example 2, solve the first equation in (2) for \({x_2}\) in terms of \({x_1}\), and from that produce the eigenvector \(y = \left( {\begin{aligned}{}2\\{ - 1 + 2i}\end{aligned}} \right)\) for the matrix A. Show that this \({\bf{y}}\) is a (complex) multiple of the vector \({v_1}\) used in Example 2.

Short answer

Thus, it is showed that\(y = \frac{{ - 1 + 2i}}{5}{v_1}\).

Step-by-step solution

Finding eigen vector

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 \(Av = \lambda v\).

Prove the matrix

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 \(y = \left( {\begin{aligned}{}2\\{ - 1 + 2i}\end{aligned}} \right)\).

Prove further for the matrix

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 \({v_1}\).