Question

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

18. \(\left( {\begin{aligned}{}1&{}&{ - 1}\\{.4}&{}&{.6}\end{aligned}} \right)\)

Short answer

The invertible matrix and matrix are \(P = \left( {\begin{aligned}{}1&{}&{ - 3}\\2&{}&0\end{aligned}} \right)\;{\rm{and}}\;\;C = \left( {\begin{aligned}{}{.8}&{}&{ - .6}\\{.6}&{}&{.8}\end{aligned}} \right)\).

Step-by-step solution

Finding the matrix \(P\) and the matrix \(C\) 

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 \(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right)\).

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

Find the Invertible matrix

Given that \(A = \left( {\begin{aligned}{}1&{}&{ - 1}\\{.4}&{}&{.6}\end{aligned}} \right)\).

Then,

\(\left( {A - \lambda {I_2}} \right) = \left( {\begin{aligned}{}{1 - \lambda }&{}&{ - 1}\\{.4}&{}&{.6 - \lambda }\end{aligned}} \right)\)

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

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

Further solving we get,

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

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\({X_1} = \left( {\begin{aligned}{}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)be an eigenvector corresponding to the eigenvalue\(.8 - .6i\).

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

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

Therefore,

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

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

Find the matrix further

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

Then we get,

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

And then,

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\ &= \frac{1}{6}\left( {\begin{aligned}{}0&{}&3\\{ - 2}&{}&1\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&{ - 1}\\{.4}&{}&{.6}\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&{ - 3}\\2&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.8}&{}&{ - .6}\\{.6}&{}&{.8}\end{aligned}} \right)\end{aligned}\)

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