Conditions that a complex matrix has real and identical eigenvalues Given a matrix $H \in \mathbb{C}^{N \times N}$, is there some condition on its elements such that all the $N$ eigenvalues are real and have the same value? Obviously, the trivial case of the identity matrix is not acceptable. 
Thanks in advance.
 A: If we know that $H$ is diagonalizable (i.e. that $H$ has $N$ linearly independent eigenvectors), then the eigenvalues of $H$ are all equal to $\lambda \in \Bbb C$ if and only if $H = \lambda I$.
If $H$ is not necessarily diagonzalizable, we can say that $H$ will have $\lambda$ as its only eigenvalue if and only if $(H - \lambda I)^N = 0$, or equivalently, $(H - \lambda I)^k = 0$ for some $k \leq N$.
A: @Riccardo.Alestra , you give to Omnomnomnom the green chevron and, yet, he does not answer your question. Of course, Omnomnomnom is a very good mathematician; perhaps you are impressed because he has $10^5$ points. Would you give him the chevron if he wrote that $2 + 2 = 5$ ?
In fact, a solution to your question is about that follows:
Let $H=[h_{i,j}]$ and $\chi_H(x)=\det(xI_n-H)$.
The NS condition is $trace(H)\in\mathbb{R}$ and $\chi_H(x)=(x-\dfrac{trace(H)}{n})^n$.
Clearly, this condition links the $(h_{i,j})$.
EDIT. Answer to  @Omnomnomnom . I hope that you are not upset. In fact, we often tell students they do not answer the question. If we do the same, then we are not credible. Here, the OP's question is clear. The required eigenvalue is explicitly $trace(A)/n$; moreover, it is real; moreover the fact that $A$ is diagonalizable or not, is interesting but has nothing to do here; in particular, the calculation of eigenvectors of $A$ can only be done if one knows the eigenvalues!
A: Elaborating on Hagen von Eitzen's comment:
If an $N\times N$ matrix has $N$ eigenvalues equal to $\lambda$, we usually speak of a single eigenvalue $\lambda$ with geometric multiplicity $N$. The latter means that there exist $N$ linearly independent eigenvectors with eigenvalue $\lambda$. Since the space if $N$-dimensional, these eigenvectors already span the whole space. Now, vectors with eigenvalue $\lambda$ always form a subspace. All this means that the the whole space is the eigenspace corresponding to $\lambda$.
By definition, the matrix $A$ acts as $Ax = \lambda x$ on an eigenvector $x$. In our case, any vector is an eigenvector, hence $\forall x : Ax = \lambda x$, which is the same as $A = \lambda I$.
