In Exercises 13-20, find an invertible matrix \(P\) and a matrix \(C\) of the form \(\left( {\begin{array}{*{20}{r}}a&{ - b}\\b&a\end{array}} \right)\) such that the given matrix has the form\(A = PC{P^{ - 1}}\). For Exercises 13-16, use information from Exercises 1-4.

18. \(\left( {\begin{array}{*{20}{r}}1&{ - 1}\\{.4}&{.6}\end{array}} \right)\)

Let \(A\) be a \(2 \times 2\) with eigenvalues of the form \(a \pm bi\) , and let \(v\) be the eigenvector corresponding to the eigenvalue \(a - bi\), then the matrix \(P\) will be \(P = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm Re}\nolimits} v}&{{\mathop{\rm Im}\nolimits} v}\end{array}} \right)\).

The matrix \(C\) can be obtained by \(C = {P^{ - 1}}AP\).

Given that \(A = \left( {\begin{array}{*{20}{r}}1&{ - 1}\\{.4}&{.6}\end{array}} \right)\).

Then,

\(\left( {A - \lambda {I_2}} \right) = \left( {\begin{array}{*{20}{c}}{1 - \lambda }&{ - 1}\\{.4}&{.6 - \lambda }\end{array}} \right)\)

Let characteristic equation of \(A\) will be,

\(\begin{array}{c}\det \left( {\lambda {I_2} - A} \right) = 0\\(1 - \lambda )(.6 - \lambda ) + .4 = 0\\{\lambda ^2} - 1.6\lambda + .6 + .4 = 0\\{\lambda ^2} - 1.6\lambda + 1 = 0\end{array}\)

Further solving we get,

\(\begin{array}{c}{\lambda ^2} - 1.6\lambda + 1 = 0\\\left( {\lambda - \left( {.8 + .6i} \right)} \right)\left( {\lambda - \left( {.8 - .6i} \right)} \right) = 0\end{array}\).

This implies that the roots of\({\lambda ^2} - 1.6\lambda + 1 = 0\)are \(.8 \pm .6i\).

Hence the eigenvalues of\(A\)are \(.8 \pm .6i\).

Now let\({X_1} = \left( {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}} \right)\)be an eigenvector corresponding to the eigenvalue\(.8 - .6i\).

Therefor\(A{X_1} = \left( {.8 - .6i} \right){X_1}\) then we get,

\(\begin{array}{c}\left( {A - \left( {.8 - .6i} \right)I} \right){X_1} = 0\\\left( {\begin{array}{*{20}{c}}{.2 + .6i}&{ - 1}\\{.4}&{ - .2 + .6i}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\end{array}\)

Therefore,

\(\begin{array}{c}.4{x_1} + \left( { - .2 + .6i} \right){x_2} = 0\\{\rm{ (}}{\rm{.4) }}{x_1} = \left( {.2 - .6i} \right){x_2}\\{x_1} = \frac{{1 - 3i}}{2}{x_2}\end{array}\)

This \({x_2}\) is a free variable.

Therefore\(\left( {\begin{array}{*{20}{c}}{1 - 3i}\\2\end{array}} \right)\)is an eigenvector corresponding to the eigenvalue\(.8 - .6i\).

Then we get,

\(P = \left( {\begin{array}{*{20}{l}}{{\mathop{\rm Re}\nolimits} {X_1}}&{{\mathop{\rm Im}\nolimits} {X_1}}\end{array}} \right) \Rightarrow P = \left( {\begin{array}{*{20}{c}}1&{ - 3}\\2&0\end{array}} \right)\)

And then,

\(\begin{array}{c}C = {P^{ - 1}}AP\\ = \frac{1}{6}\left( {\begin{array}{*{20}{r}}0&3\\{ - 2}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{.4}&{.6}\end{array}} \right)\left( {\begin{array}{*{20}{l}}1&{ - 3}\\2&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{r}}{.8}&{ - .6}\\{.6}&{.8}\end{array}} \right)\end{array}\)

Thus, the required matrices are \(P = \left( {\begin{array}{*{20}{c}}1&{ - 3}\\2&0\end{array}} \right)\quad {\rm{and}}\;C = \left( {\begin{array}{*{20}{r}}{.8}&{ - .6}\\{.6}&{.8}\end{array}} \right)\).