Is an irreducible, asymmetric adjacency matrix diagonalizable? I am wondering whether the following statement is true or not. My computer has so far not been able to provide me with any counterexamples, nor have I been able to find much in the way of arguments to support it. I might just be missing something obvious, but I'm hoping that someone is able to help me. The statement is:

Let $D$ be a simple, directed graph with adjacency matrix $A\in\{0,1\}^{n \times n}$. If $D$ is strongly connected, then $A$ is diagonalizable.

Kind regards,
Pepijn

Edit: somehow managed to forget the essential requirement in the body of the question.
Take $$ A = 
\begin{pmatrix} 
0 &amp; 1 &amp; 0 \\ 
1 &amp; 1 &amp; 1 \\
1 &amp; 0 &amp; 1
\end{pmatrix} %A={{0,1,0},{1,1,1},{1,0,1}}$$
The directed graph of this matrix is strongly connected,which implies $A$ is irreducible.
But this matrix is not diagonalizable, since $0$ is an eigenvalue with multiplicity $2$, and $\operatorname{rank} A=2$.
Take $$ A = 
\begin{pmatrix} 
0 &amp; 1 &amp; 0 \\ 
1 &amp; 1 &amp; 1 \\
1 &amp; 0 &amp; 1
\end{pmatrix} %A={{0,1,0},{1,1,1},{1,0,1}}$$
The directed graph of this matrix is strongly connected,which implies $A$ is irreducible.
But this matrix is not diagonalizable, since $0$ is an eigenvalue with multiplicity $2$, and $\operatorname{rank} A=2$.