Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$. 
Let $A = \begin{bmatrix} 
1 & 5 \\
-1 & 3 
\end{bmatrix}$ act on $\mathbb{C}^2$.  An eigenvector of $A$ is $\begin{bmatrix} 
1-2i \\
1
\end{bmatrix}$, corresponding to the eigenvalue $2+2i$.  Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to a matrix of the form $\begin{bmatrix} 
a & -b \\
b & a 
\end{bmatrix}$.

I know that Eigenvectors for $\begin{bmatrix} 
a & -b \\
b & a 
\end{bmatrix}$. are  $\begin{bmatrix}1\\-i\end{bmatrix}$ and $\begin{bmatrix}1\\i\end{bmatrix}$. I believe I need to make a augmented matrix and row reduce yet still unsure how to approach it. 
 A: Generalizing, we have
$A u = \mu u; \tag 1$
since $A$ is real the eigenvalues are either real or occur in complex conjugate pairs; thus we may write
$\mu = \mu_R + i\mu_I, \tag 2$
$A$ being a real matrix, if $\mu \in \Bbb C \setminus \Bbb R$, the vector $u \in \Bbb C^2$ may be written in terms of real and imaginary parts
$u = u_R + iu_I, \; u_R, u_I \in \Bbb R^2; \tag 3$
with $u_I \ne 0$; then we have
$Au_R + iAu_I = A(u_R + iu_I)$
$= (\mu_R + i\mu_I)(u_R + iu_I) = \mu_R u_R - \mu_I u_I + i(\mu_I u_R + \mu_R u_I), \tag 4$
which leads to 
$Au_R = \mu_R u_R - \mu_I u_I, \tag 5$
$Au_I = \mu_I u_R + \mu_R u_I; \tag 6$
let
$U = [u_R \; u_I]; \tag 7$
that is, the columns of $U$ are the respective real and imaginary parts $u_R$ and $u_I$ of $u$; then we have
$AU = [Au_R \; Au_I] = [\mu_R u_R - \mu_I u_I \; \mu_I u_R + \mu_R u_I] = \mu_R[u_R \; u_I] + \mu_I[-u_I \; u_R]; \tag 8$
furthermore, in the event that
$\mu_I \ne 0, \tag 9$
the vectors $u_R$ and $u_I$ are linearly independent, for if $a, b \in \Bbb R$ are such that
$au_R + bu_I = 0, \tag{10}$
then
$aAu_R + bAu_I = 0; \tag{11}$
which via (5) and (6) becomes
$a(\mu_R u_R - \mu_I u_I) + b(\mu_I u_R + \mu_R u_I) = 0, \tag{12}$
or
$(a\mu_R + b\mu_I)u_R + (b\mu_R - a \mu_I)u_I = 0; \tag{13}$
if we multiply (10) by $\mu_R$ and subtract the result from this equation we obtain
$b\mu_I u_R - a\mu_I u_I = 0; \tag{14}$
in the event that (9) binds this becomes
$bu_R - au_I = 0; \tag{15}$
multiplying (10) by $b$ and (15) by $a$ and subtracting the resulting two equations yields
$(a^2 + b^2)u_I = 0; \tag{16}$
since $u_I \ne 0$ we find
$a^2 + b^2 = 0 \Longrightarrow a = b = 0, \tag{17}$
which shows that $u_R$ and $u_I$ are linearly independent; therefore the matrix $U$ in (7) is nonsingular and we have
$I = U^{-1}U = U^{-1}[u_R \; u_I] = [U^{-1}u_R \; U^{-1}u_I] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}; \tag{18}$
$U^{-1}AU = [Au_R \; Au_I] = \mu_R[U^{-1}u_R \; U^{-1}u_I] + \mu_I[-U^{-1}u_I \; U^{-1}u_R]$
$= \mu_R\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \mu_I \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} \mu_R & \mu_I \\ -\mu_I & \mu_R \end{bmatrix}, \tag{19}$
which shows that $A$ is similar to a matrix of the form
$\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} \mu_R & \mu_I \\ -\mu_I & \mu_R \end{bmatrix}, \tag{20}$
via the "change-of-coordinates" matrix $P = U$.
In the specific situation at hand, we are given that
$\mu = 2 + 2i; u = \begin{pmatrix} 1 - 2i \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + i \begin{pmatrix} -2 \\ 0 \end{pmatrix}; \tag{21}$
thus,
$u_R = \begin{pmatrix} 1 \\ 1 \end{pmatrix},  \tag{22}$
$u_I = \begin{pmatrix} -2 \\ 0 \end{pmatrix}, \tag{23}$
$P = U = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}, \tag{24}$
$\mu_R = \mu_I = 2, \tag{25}$
and the matrix
$\begin{bmatrix} \mu_R & \mu_I \\ -\mu_I & \mu_R \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -2 & 2 \end{bmatrix}. \tag{26}$
$OE\Delta$.
