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.

17. \(\left( {\begin{aligned}{}1&{}&{ - .8}\\4&{}&{ - 2.2}\end{aligned}} \right)\)

Short answer

The invertible matrix \(P\)and matrix \(C\) are \(C = \left( {\begin{aligned}{}{ - .6}&{}&{ - .8}\\{.8}&{}&{ - .6}\end{aligned}} \right),P = \left( {\begin{aligned}{}2&{}&{ - 1}\\5&{}&0\end{aligned}} \right)\).

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 \(A = \left( {\begin{aligned}{}1&{}&{ - .8}\\4&{}&{ - 2.2}\end{aligned}} \right)\)

The characteristic equation is \({\lambda ^2} + 1.2\lambda + 1 = 0\).

Solving this we get \(\lambda = - 0.6 \pm 8i\).

To find an eigenvector corresponding to \( - .6 - .8i\) we compute,

\(A - \left( { - .6 - .8i} \right)I = \left( {\begin{aligned}{}{1.6 + .8i}&{}&{ - .8}\\4&{}&{ - 1.6 + .9i}\end{aligned}} \right)\)

The equation \(\left( {A - \left( { - .6 - .8i} \right)I} \right)v = 0\) gives the system of linear equations,

\(\begin{aligned}{}\left( {1.6 + 0.8i} \right)x - 0.8y &= 0\\4x + \left( { - 1.6 + .8i} \right)y &= 0\end{aligned}\)

Now from the 2^nd equation, we get,

\(x = \left( {\left( {2 - i} \right)/5} \right)y\) with \(y\) is a free variable.

The eigenvector corresponds to \( - .6 - .8i\) is \(v = \left( {\begin{aligned}{}{2 - i}\\5\end{aligned}} \right)\).

By Theorem 9,

\(P = \left( {\begin{aligned}{}{{\rm{ Re v}}}&{}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right) \Rightarrow P = \left( {\begin{aligned}{}2&{}&{ - 1}\\5&{}&0\end{aligned}} \right)\)

Find the matrix  

After that, we can find the matrix by using\(C = {P^{ - 1}}AP\).

\(\begin{aligned}{}C {}&= {P^{ - 1}}AP\\ {}&= \frac{1}{5}\left( {\begin{aligned}{}0&{}&1\\{ - 5}&{}&1\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&{ - .8}\\4&{ - 2.2}\end{aligned}} \right)\left( {\begin{aligned}{}2&{}&{ - 1}\\5&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - .6}&{}&{ - .8}\\{.8}&{}&{ - .6}\end{aligned}} \right)\end{aligned}\)

Therefore, the matrixes are \(C = \left( {\begin{aligned}{}{ - .6}&{}&{ - .8}\\{.8}&{}&{ - .6}\end{aligned}} \right)\) and \(P = \left( {\begin{aligned}{}2&{}&{ - 1}\\5&{}&0\end{aligned}} \right)\).