91Ó°ÊÓ

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 \(C = {P^{ - 1}}AP\).

Given that\(A = \left( {\begin{aligned}{}5&{}&{ - 2}\\1&{}&3\end{aligned}} \right)\).

Now from Exercise 4, we get that the eigenvalues of\(A\)are\(4 - i,4 + i\)and

\({v_1} = \left( {\begin{aligned}{}{1 - i}\\1\end{aligned}} \right)\)is an eigenvector corresponding to the eigenvalue\(4 - i\).

Then by using Theorem 9 we have,

\(\begin{aligned}{}P &= \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} {{\bf{v}}_1}}&{}&{{\mathop{\rm Im}\nolimits} {{\bf{v}}_1}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}1&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\end{aligned}\)

After that, we can find the matrix by this\(C\)such that,

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\ &= \left( {\begin{aligned}{}0&{}&1\\{ - 1}&{}&1\end{aligned}} \right)\left( {\begin{aligned}{}5&{}&{ - 2}\\1&{}&3\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}4&{}&{ - 1}\\1&{}&4\end{aligned}} \right)\end{aligned}\)

Thus the required matrices are\(P = \left( {\begin{aligned}{}1&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\;{\rm{and}}\;C = \left( {\begin{aligned}{}4&{}&{ - 1}\\1&{}&4\end{aligned}} \right)\).