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.



Not triangulirazable over $\Bbb R$ implies diagonalizable over $\Bbb C$
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 ${\bf C}$.


