Let's assume that we have an $n\times n$ matrix $A$ with entries in $\mathbb{R}$. I know that the theorem says that $A$ is triangulable if and only if its minimal polynomial $m_A(x)$ splits over $\mathbb{R}$. Do we actually need for the minimal polynomial to split over $\mathbb{R}$, or can we relax the condition on the statement to simply say that at least one of the eigenvalues of $A$ lives in $\mathbb{R}$? That is, if I start with $A$, which has one eigenvalue in $\mathbb{R}$ and corresponding eigenvector, can I construct a flag of $A$-invariant subspaces of $\mathbb{R}^n$ + basis, which in this basis $A$ becomes triangular?
For example, can one put $A = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ in triangular form? If not, where is the obstruction?
Sorry if this is not a good question, but I've been going through Hoffman and Kunze and I am a bit lost in the details surrounding Theorem 5 (and the proceeding lemma) in Section 6.4.
No, you need all the eigenvalues to be real. After all, when you put $A$ in triangular form, the diagonal entries must be the eigenvalues (in some order). In particular, for your example, the obstruction is that the diagonal entries will have to be $1$, $i$, and $-i$.