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.

14. \(\left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&1\end{aligned}} \right)\)

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

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

\(A = \left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&1\end{aligned}} \right)\).

From Exercise 2, the eigenvalues of A is\(\lambda = 3 \pm i\), and the eigenvector corresponds to\(\lambda = 3 - i\)is \(v = \left( {\begin{aligned}{}{2 - i}\\1\end{aligned}} \right)\).

By Theorem 9,\(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right) \Rightarrow P = \left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\).

This implies that,\({P^{ - 1}} = \left( {\begin{aligned}{}0&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\).

Substitute the value of\({P^{ - 1}}\), \(P\) and \(A\) into \(C = {P^{ - 1}}AP\) to obtain matrix \(C\).

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\C &= \left( {\begin{aligned}{}0&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\left( {\begin{aligned}{}5&{}&{ - 5}\\1&1\end{aligned}} \right)\left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\\C &= \left( {\begin{aligned}{}3&{}&{ - 1}\\1&{}&3\end{aligned}} \right)\end{aligned}\)

Hence, the\(P\)and\(C\)are \(P = \left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\)and \(C = \left( {\begin{aligned}{}3&{}&{ - 1}\\1&{}&3\end{aligned}} \right)\) respectively.