<s>[INST] Diagonalize the matrix A (complex numbers) Diagonalize the matrix A
$A=\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}$
So, i began the problem by finding the characteristic polynomial which was
$λ^3-7λ^2-15λ-27$
using long division i got $(λ-9)(λ^2+2λ+3)$
so i used the quadratic formula and got 
$λ_1=-1+i\sqrt{2}$ and $λ_2=-1-i\sqrt{2}$ and $λ_3=9$
I decided to start with $λ_1$
$A-\left(-1+i\sqrt{2}\right)I=\begin{pmatrix}2-i\sqrt{2} & 2 & 4 \\3 & 6-i\sqrt{2} & 2 \\2 & 6 & 2-i\sqrt{2}\end{pmatrix}$
Now i understand how to diagonalize when i have all numbers but once i get these $i$'s in the equation it's like my brain doesn't comprehend the steps i need to take the get the diagonal form..
I would like to think i'm supposed to start by getting the inverse of this new matrix but fail to see how to do that with $i$'s involved.
 [/INST] A strange way to find the eigenvectors is to use the Cayley-Hamilton Theorem,
$$
          (A-9I)(A^{2}+2A+3I) = 0.
$$
Because of this, the columns of $A^{2}+2A+3I$ are eigenvectors of $A$ with eigenvalue $9$. For example,
$$
\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}
\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+
2\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+
3\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{pmatrix} \\
=\begin{pmatrix}15 & 36 & 12 \\ 22 & 43 & 24 \\22 & 40 & 21  \end{pmatrix}
 + \begin{pmatrix}2 & 4 & 8 \\ 6 & 10 & 4 \\ 4 & 12 & 2\end{pmatrix}
 + \begin{pmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{pmatrix} \\
= \begin{pmatrix}20 & 40 & 20 \\ 28 & 56 & 28 \\ 26 & 52 & 26\end{pmatrix}
$$
Similarly, the columns of the following are eigenvectors with eigenvalue $-1-\sqrt{2}i$
$$
         (A-9I)(A-(-1+\sqrt{2}i)I)=A^{2}+(-8-\sqrt{2}i)A+(-9+9\sqrt{2}i)I
$$
You only need the first column:
$$
         \begin{pmatrix} 15 \\ 22 \\ 22\end{pmatrix}+(-8-2\sqrt{2}i)\begin{pmatrix} 1 \\ 3 \\ 2\end{pmatrix} +
(-9+9\sqrt{2}i)\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \\
          =\begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ 6-4\sqrt{2}i\end{pmatrix} 
$$
And the conjugate of this vector is going to have the conjugate of its eigenvalue. So a basis of eigenvectors is
$$
     \begin{pmatrix}10 \\ 14 \\ 13\end{pmatrix},
    \begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ +6-4\sqrt{2}i\end{pmatrix},
    \begin{pmatrix} -2-7\sqrt{2}i \\ -2+6\sqrt{2}i \\ +6+4\sqrt{2}i\end{pmatrix} 
$$
</s>