Let $A$ be a $2 \times 2$ matrix with complex eigenvalues, $\lambda_1 = \alpha + i \beta$ and $\lambda_2 = \alpha – i \beta$ ($\alpha, \beta \in \mathbb{R}$). We know that $u + iv$ is an eigenvector associated with the eigenvalue $\lambda_1$, where $u$ and $v$ are vectors with real coefficients.
Show: If we have the matrix $P = \left[ {\begin{array}{cc}
u & v \\
\end{array} } \right]$, then $P^{-1} A P = \left[ {\begin{array}{cc}
\alpha & \beta \\
-\beta & \alpha \\
\end{array} } \right] =: J$, the real Jordan Canonical Form for $A$.
I don't know how to approach the problem. Since I don't know $P^{-1}$, I would consider $AP = PJ$ and then substitute A with $\tilde{P}\tilde{J} \tilde{P}^{-1}$, $\tilde{P} = \left[ {\begin{array}{cc}
u + iv & u – iv \end{array} } \right] $ and $\tilde{J} = \left[ {\begin{array}{cc}
\alpha + i \beta & 0 \\
0 & \alpha – i \beta \\
\end{array} } \right] $. Unfortunately, I couldn't came up with $\tilde{P}^{-1}$. Would appreciate some hints.
Best Answer
Assuming that $A$ is a matrix with real entries, we have\begin{align}A.(u-vi)&=\overline A.\overline{(u+vi)}\\&=\overline{A.(u+vi)}\\&=\overline{(\alpha+\beta i)(u+vi)}\\&=(\alpha-\beta i)(u-vi)\end{align}and therefore $u-vi$ is an eigenvector of $A$ with eigenvalue $\alpha-\beta i$. So\begin{align}A.u&=A.\left(\frac{u+vi}2+\frac{u-vi}2\right)\\&=\frac{(\alpha+\beta i)(u+vi)+(\alpha-\beta i)(u-vi)}2\\&=\alpha u-\beta v\end{align}and a similar computation shows that $A.v=\beta u+\alpha v$. Therefore, the matrix of $w\mapsto A.w$ with respect to the basis $(u,v)$ is $\left[\begin{smallmatrix}\alpha&\beta\\-\beta&\alpha\end{smallmatrix}\right]$.