Diagonalizing a matrix with complex numbers hope everyone is doing well.
I'm a bit stuck on the following problem:
$T: C^2 \rightarrow C^2$, where $T$ is defined by $T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y$
So I perform the usual calculations to diagonalize the matrix, right up until I find that the roots of the characteristic polynomial are $\lambda_1 = -1, \lambda_2 = 8$
From here I am having a hard time performing the row operations to get my eigenvectors. I realize that the calculations are fairly trivial but they are confusing me.
If someone could help me with, for example, the matrix associated to $\lambda_1$ I would be grateful:
$$
    \begin{matrix}
    3 & 3+3i \\
    3-3i & 6 \\
    \end{matrix}
$$
 A: We may write
$T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y \tag 1$
as
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}, \tag 2$
and then he characterisic polynomial of $T$ is
$\det (T - \lambda I) = \det \left ( \begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} - \lambda I \right ) = \det \left ( \begin{bmatrix} 2 - \lambda & 3 - 3i \\ 3 + 3i & 5 - \lambda \end{bmatrix} \right) = \lambda^2 - 7 \lambda - 8; \tag 3$
it is easy to see that the roots of
$\lambda^2 - 7 \lambda -  8 = 0 \tag 4$
are
$\lambda = -1, \; 8; \tag 5$
since I'm not too swift at row operations myself, I'll solve for the eigenvectors directly from 
$\begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}; \tag 6$
writing out the first row of this equation yields
$2x + (3 - 3i)y = \lambda x, \tag 7$
or
$(2 - \lambda)x + (3 - 3i) y = 0, \tag 8$
or
$y = -\dfrac{(2 - \lambda)x}{3 - 3i} = \dfrac{(\lambda - 2)x}{3 - 3i}; \tag 9$
since eigenvectors are are scalable, we may take $x = 1$; then the two eigenvectors are characteized by the two $y$-values
$y =  \dfrac{(-1 - 2)x}{3 - 3i} = \dfrac{-3}{3 - 3i} = \dfrac{-1}{1 - i} = \dfrac{-1 - i}{2}, \tag{10}$
$y =  \dfrac{(8 - 2)x}{3 - 3i} = \dfrac{6}{3 - 3i} = \dfrac{2}{1 - i} = \dfrac{2 + 2i}{2} = 1 + i; \tag{11}$
we may thus take the two eigenvectors to be
$v_{-1} = \begin{pmatrix} 1 \\ \dfrac{-1 - i}{2} \end{pmatrix}, \; v_8 = \begin{pmatrix} 1 \\ 1 + i \end{pmatrix}; \tag{12}$
we note that $v_{-1}$ and $v_8$ are in fact orthogonal with respect to the Hermitian inner product on $\Bbb C^2$ defined by
$\langle w, z \rangle = \bar w_1 z_1 + \bar w_2 z_2, \tag{13}$
where $w, z \in \Bbb C^2$:
$w = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix}, \; z = \begin{pmatrix} z_1 \\ z_2 \end{pmatrix}; \tag{14}$
indeed,
$\langle v_{-1}, v_8 \rangle = 1 \cdot 1 + \dfrac{-1 + i}{2} \cdot (1 + i) = 1 - \dfrac{2}{2} = 0; \tag{15}$
the orthogonality of $v_{-1}$, $v_8$ should not surprise us since they belong to distinct eigenvalues of the matrix $T$, which is Hermitian:  $T^\dagger = (T^\ast)^T = T$.  $v_{-1}$ and $v_8$ form the columns of $V$, a diagonalizing matrix for $T$:
$V = \begin{bmatrix} v_{-1} & v_8 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ \dfrac{-1 - i}{2} & 1 + i \end{bmatrix}, \tag{16}$
the determinant of which is
$\det(V) = 1 + i + \dfrac{1 + i}{2} = \dfrac{3}{2}(1 + i); \tag{17}$
the inverse of $V$ is thus
$V^{-1} = \dfrac{2}{3(1 + i)}\begin{bmatrix} 1 + i & - 1 \\ \dfrac{1 + i}{2} & 1 \end{bmatrix} = \begin{bmatrix} \dfrac{2}{3} & -\dfrac{2}{3(1 + i)} \\ \dfrac{1}{3} & \dfrac{2}{3(1 + i)} \end{bmatrix}; \tag{18}$
finally, since
$TV = T\begin{bmatrix} v_{-1} & v_8 \end{bmatrix} = \begin{bmatrix} Tv_{-1} & Tv_8 \end{bmatrix} = \begin{bmatrix} -v_{-1} & 8v_8 \end{bmatrix}, \tag{19}$
we may write
$V^{-1}TV = V^{-1} \begin{bmatrix} -v_{-1} & 8v_8 \end{bmatrix} =  \begin{bmatrix} -V^{-1} v_{-1} & 8V^{-1} v_8 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}. \tag{20}$
