Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') operators. The eigenvalues of the operator are the allowed values of the observable. Since Hermitian operators have a real spectrum, all is well.
However, there are non-Hermitian operators with real eigenvalues, too. Consider the real triangular matrix:
$$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 8 & 4 & 0 \\ 5 & 9 & 3 \end{array} \right) $$
Obviously this matrix isn't Hermitian, but it does have real eigenvalues, as can be easily verified.
Why can't this matrix represent an observable in QM? What other properties do Hermitian matrices have, which (for example) triangular matrices lack, that makes them desirable for this purpose?
Answer
One problem with the given $3\times 3$ matrix example is that the eigenspaces are not orthogonal.
Thus it doesn't make sense to say that one has with 100% certainty measured the system to be in some eigenspace but not in the others, because there may be a non-zero overlap to a different eigenspace.
One may prove$^{1}$ that an operator is Hermitian if and only if it is diagonalizable in an orthonormal basis with real eigenvalues. See also this Phys.SE post.
$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.
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