Besides being symmetric, when will a matrix have ONLY real eigenvalues?

Another approach is to construct a triangular matrix with pre-determined diagonal entries; they will be the eigenvalues, and the matrix is not symmetric.

There are so-called $\mathcal{PT}$-symmetric matrices which may have purely real eigenvalues. A square matrix $M$ is called $\mathcal{PT}$-symmetric iff it satisfies the property: $$ [\mathcal{PT},M] = 0 \Leftrightarrow \mathcal{P}M = M^*\mathcal{P} $$ where $\mathcal{T}\equiv*$ is the complex conjugation operator and $\mathcal{P}$ is a matrix satisfying $[\mathcal{P},\mathcal{T}]=0$ (implying that $\mathcal{P}$ is a real matrix) and $\mathcal{P}^2=1$, $\mathcal{T}^2=1$ $\Rightarrow (\mathcal{PT})^2=1$. Since $\mathcal{P}$ is an involution, its eigenvalues are $\pm1$ and one may find a basis such that $\mathcal{P}=\mathrm{diag}(1,1,1,1,...,-1,-1,-1)$. A $\mathcal{PT}$-symmetric matrix is said to have 'unbroken' $\mathcal{PT}$ symmetry iff any eigenvector of $M$ is also an eigenvector of $\mathcal{PT}$.

Claim: If $M$ has unbroken $\mathcal{PT}$ symmetry, this implies that $M$ has real eigenvalues.

Proof: First note that the eigenvalues of $\mathcal{PT}$ are non-zero since the combination is an involution: $\mathcal{PT} u = \mu u \Rightarrow \mathcal{PT}^2 u = u = \mu^*\mu u \Rightarrow |\mu|=1$.

Now let $Mv = \lambda v \Rightarrow \mathcal{PT}Mv = \mathcal{PT} \lambda v$. Thus since the combination $\mathcal{PT}$ is an anti-linear operator and $M$ commutes with $\mathcal{PT}$: $\mathcal{PT} M v = \lambda^*\mathcal{PT}v\Rightarrow \lambda \mu = \lambda^*\mu$. Since we've shown that $\mu\neq0$, this implies $\lambda^* = \lambda$. QED

More on $\mathcal{PT}$-symmetry and related concepts in this article: http://arxiv.org/abs/1212.1861 The English in there isn't perfect but the content looks good.