Not triangulirazable over $\Bbb R$ implies diagonalizable over $\Bbb C$ [closed]

Question

Let $A$ be a real matrix. Suppose $A$ is not triangulirazable over $\Bbb R$ then $A$ is diagonalizable over $\Bbb C$.

My proof: Since $A$ is not triangularizable over $\Bbb R$ it has a complex eigenvalue. But complex eigenvalues occur in pairs for real matrices. Hence, not all the eigenvalues of $A$ are equal. is this sufficient to conclude that $A$ is diagonalizable over $\Bbb C$?

Thank you.

Answer 1

I was going to sketch a construction for a counterexample in the comments, but on reflection it may be more sensible to write it out in full as an answer.

Let $B=\begin{pmatrix} i & 0 \\ 0 &-i \end{pmatrix}$ and let $T=\begin{pmatrix} 1 & 1 \\0 & 1 \end{pmatrix}$. Then the $4\times 4$ matrix $$ B\otimes T = \begin{pmatrix} iT & 0 \\ 0 & -iT \end{pmatrix} = \begin{pmatrix} i & i & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & -i \\ 0 & 0 & 0 & -i \end{pmatrix} $$ has no real eigenvalue and is not diagonalizable over ${\mathbb C}$. On the other hand, note that for a suitable complex unitary matrix $U\in M_2({\mathbb C})$, $A_0:=UBU^*\in M_2({\mathbb R})$. We could take $A_0 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

Therefore if we take $$ A:= A_0 \otimes T = \begin{pmatrix} 0 & - T \\ T & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & - 1 & -1 \\ 0 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} $$ then $A$ is a real matrix, which is similar in $M_4({\bf C})$ to $B\otimes T$, and hence has no real eigenvalue nor has any diagonalizable over ${\bf C}$.

Answer 2

Consider $$A = \begin{bmatrix}1 & 1 & & \\ 0& 1 & & \\ & & 0& 1 \\ & & -1 & 0\end{bmatrix}$$ $A$ has complex eigenvalues, so it is not triangulizable over $\mathbb R$. $A$ is not diagonalizable over $\mathbb C$ since $\mathrm{rank} (A-I)^2 < \mathrm{rank} (A - I)$.