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.

19. \(\left( {\begin{aligned}{}{1.52}&{}&{ - .7}\\{.56}&{}&{.4}\end{aligned}} \right)\)

Short answer

The invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.96}&{}&{ - .28}\\{.28}&{}&{.96}\end{aligned}} \right)\;{\rm{and}}\;P = \left( {\begin{aligned}{}2&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\).

Step-by-step solution

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

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 \(A = \left( {\begin{aligned}{}{1.52}&{}&{ - .7}\\{.56}&{}&{.4}\end{aligned}} \right)\).

Find the Characteristic equation by using \(\det \left( {A - \lambda I} \right) = 0\).

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= 0\\\left( {1.52 - \lambda } \right) \cdot \left( {.4 - \lambda } \right) - \left( { - .7} \right) \cdot \left( {.56} \right) &= 0\\{\lambda ^2} - 1.92\lambda + 1 &= 0\\\left( {\lambda - \left( {.96 - .28i} \right)} \right)\left( {\lambda - \left( {.96 + .28i} \right)} \right) &= 0\end{aligned}\)

Hence the eigenvalues are \(\lambda = .96 \pm .28i\).

Now we have to find an eigenvector corresponding to \(96 - 28i\).

\(\begin{aligned}{}A - \left( {.96 - .28i} \right)I &= 0\\\left( {\begin{aligned}{}{.56 + .28i}&{}&{ - .7}\\{.56}&{}&{ - .56 + .28i}\end{aligned}} \right) &= \left( {\begin{aligned}{}0\\0\end{aligned}} \right)\end{aligned}\)

We can use only one row to obtain an eigenvector.

Find the matrix further 

From the second row:

\(\begin{aligned}{}.56{x_1} + ( - .56 + .28i){x_2} &= 0\\{x_1} &= \frac{{2 - i}}{2}{x_2}\end{aligned}\)

Hence the eigenvector is\(v = \left( {\begin{aligned}{}{2 - i}\\2\end{aligned}} \right)\).

By Theorem 9,

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

Find the matrix by using\(C = {P^{ - 1}}AP\).

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\ &= \frac{1}{2}\left( {\begin{aligned}{}0&{}&1\\{ - 2}&2\end{aligned}} \right) \cdot \left( {\begin{aligned}{}{1.52}&{}&{ - .7}\\{.56}&{}&{.4}\end{aligned}} \right) \cdot \left( {\begin{aligned}{}2&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.96}&{}&{ - .28}\\{.28}&{}&{.96}\end{aligned}} \right)\end{aligned}\)

Thus, the invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.96}&{}&{ - .28}\\{.28}&{}&{.96}\end{aligned}} \right),\;{\rm{and}}\;P = \left( {\begin{aligned}{}2&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\).