Is it possible to have such a matrix? I am searching for a square $n\times n$ nonnegative integer matrix whose all diagonal entries are zero and which contains at least one eigenvalue whose geometric multiplicity is less than its algebraic multiplicity. Besides, the matrix should contain at least two nonreal eigenvalues.  Is there such a matrix? I have tried using companion matrix. But that didn't work. 
For my work if the matrix size is less than 6 and the entries are small, then it would be better. 
Any help is appreciated. 
 A: The best you can do, in terms of small size and entries, is a $5\times5$ matrix with four $1$s and the rest $0$s, e.g. $$A=\begin{pmatrix}0&1&0&0&0\\0&0&1&0&0\\1&0&0&0&0\\0&0&0&0&1\\0&0&0&0&0\end{pmatrix}$$ Edit:
There are no $4\times4$ matrices with the desired properties.
Proof: 
If you have a nonnegative $4\times4$ matrix with $0$s on the diagonal, then its characteristic polynomial's constant term is $\leq0.$ If it has a real eigenvalue with algebraic multiplicity two and two nonreal eigenvalues, then its characterisitic polynomial is a product of two monic quadratics, one of which has a constant term $\geq0,$ corresponding to the real eigenvalue with algebraic multiplicity two, and one of which has a positive constant term, corresponding to the two nonreal eigenvalues. Contra-diction.
The minimal number of nonzero entries is four.
Proof:
If you have a nonnegative $5\times5$ matrix  $A$ with $0$s on the diagonal, then its monic characteristic polynomial's cubic term coefficient is $\leq0.$ If the number of nonzero entries in $A$ is less than four, then it has at least two zero rows and columns, hence the nullity of $A$ is at least two, and $0$ has algebraic multiplicity three. But now the remaining two eigenvalues are also real.
There are other configurations. Choose a quadratic term from the characteristic polynomial; the entries in its coefficient are $1$s. Next, choose an entry that's not in the span of the previous three, and not on the diagonal; it's a $1.$ The rest are $0$s.
