I am told that $A\in M_{2}(\mathbb{R})$ s.t $1-i$ is an eigenvalue of $A$ with corresponding eigenvector $\begin{pmatrix}3+2i\\ 1+3i \end{pmatrix}$.
I wish to find a matrix $P$ s.t $P^{-1}AP=\begin{pmatrix}1-i & 0\\ 0 & 1+i \end{pmatrix}$. From what I understand since $\lambda_{1}=1-i$ is an eigenvalue then also $\lambda_{2}=\bar{\lambda}_{2}=1+i$ is an eigenvalue and since $\lambda_{1}\neq\lambda_{2}$ and $A$ is a $2\times2$ matrix I conclude that such $P$ does exist.
My efforts for finding $P$ are: I said that if $v_{1}=\begin{pmatrix}3+2i\\ 1+3i \end{pmatrix}$ is an eigenvector corresponding to $\lambda_{1}$ then $v_{2}=\overline{v_{1}}$ is eigenvector corresponding to $\lambda_{2}$ hence $B=\{v_{1,}v_{2}\}$ is a basis of eigenvectors.
From what I remember $P$ is the matrix that transfers from the standard base to $B$ (or the other way around ?), so I tried writing $\begin{pmatrix}1\\ 0 \end{pmatrix}=\alpha v_{1}+\beta v_{2}$ and failed (I got that such $\alpha,\beta$ does not exist).
Can somene please help me understand what is my mistake and how to find the matrix $P$ ?