So I have a 4x4 matrix from which I get 4 complex eigenvectors and I want to create a real basis for the eigenvectors. I know that for a $2 \times 2 $matrix we can use:
$$\Lambda=V^{-1}AV$$
where $\Lambda$ is the real standard form of the $2 \times 2$ matrix, A and $V$ is the change of basis matrix.
I also know that for the $2 \times 2$ case, $V = [Re(v)|Im(v)]$ (by columns) where $v$ is an eigenvector associated with one of the eigenvalues.
My question is how does the formula for $V$ from the $2 \times 2$ matrix extend for a 4x4 matrix? Because if the same formula for $V$ is applied it will not yield a square matrix, thus, it will not have an inverse to compute $\Lambda$ from it.
Thank you,
Alex
Consider a $4 \times 4$ real matrix $A$ with four distinct non-real eigenvalues $\alpha, \overline{\alpha}, \beta, \overline{\beta}$. If $u$ is an eigenvector for $\alpha$, then its complex conjugate $\overline{u}$ is an eigenvector for $\overline{\alpha}$. Similarly if $v$ is an eigenvector for $\beta$, $\overline{v}$ is an eigenvector for $\overline{\beta}$. Then you can consider the basis $$ \text{Re}(u), \text{Im}(u), \text{Re}(v), \text{Im}(v)$$ Note that $$\eqalign{A (\text{Re}(u)) &= \frac{1}{2} A (u + \overline{u}) \cr &= \frac{1}{2} \left(\alpha u + \overline{\alpha} \overline{u}\right)\cr &= \text{Re}(\alpha) \text{Re}(u) - \text{Im}(\alpha) \text{Im}(u)}$$ etc.