Say I have the matrix $\pmatrix{0 & i \cr -i & 0 \cr}$ I want to change to the basis $\pmatrix{1 & 0 \cr 0 & 1 \cr}$.
So I calculate the eigenvalues and get $\lambda = \pm 1$. so my basis based on the eigenvectors ends up being $\pmatrix{1 & i \cr -i & 1 \cr}$ and I can normalize that to have$\pmatrix{\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \cr \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \cr}$. (normalized each eigenvector individually). Now to change it over, I just need to do $BAB^{-1}$, where B is $\pmatrix{\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \cr \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \cr}$ and A is $\pmatrix{0 & i \cr -i & 0 \cr}$
Did I understand this right?
You have a certain linear transformation $A$ whose matrix with respect to the given standard basis is $$\left[\matrix{0&i\cr -i&0\cr}\right]\ .$$ You have computed the eigenvalues of this matrix, and they came out to $\lambda_1=1$, $\lambda_2=-1$. Since $\lambda_1\ne\lambda_2$ the transformation $A$ can be diagonalized: If the eigenvectors $e_1$, $e_2$ of $A$ are used as a basis the matrix of $A$ becomes $$\left[\matrix{1&0\cr 0&-1\cr}\right]\ .$$ No more calculations are needed for that.
It is another thing if you want to express other data present in your problem in terms of the basis $(e_1,e_2)$. In this case you have to set up the transformation matrix $T$ from the old (=standard) basis to the new basis $(e_1,e_2)$. This matrix has the old coordinates of the eigenvectors $e_k$ in its columns.