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With respect to a matrix $A\in\mathbb{R}^{n\times n}$, is there any if and only if conditions that make sure the eigen values of $A$ remain real and distinct? $A$ here is not symmetric but is stable and also there exists a $P$ such that $PAP^{-1}$ is symmetric. For having real eigen values, $A$ needs to symmetric or symetrizable right. But I have no clue to show that $A$ has distinct eigen values. The theory of interlacing inequalities for symmetric/hermitian matrices cannot help as I cannot say about a submatrix of $A$ having distinct eigen values or not.

The $A$ matrix is of the form $\left( \begin{array}{cccc} -(a+b+c+d)&e &f & g \\ b& -e& 0 & 0 \\ c& 0 & -f &0 \\ d& 0& 0 & -g \\ \end{array} \right)$, i.e. $A$ is a Metzler Matrix. For certain values of $a~,b~,c~,d~,e~,f~,g \ge 0$ the matrix $A$ has real and distinct eigen values. I want to know whether any matrix $A$ which is Metzler, diagonally dominant, and with $a~,b~,c~,d~,e~,f~,g \ge 0$ the eigen values will be real and distinct.

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    Are you interested in easily calculable condition or abstract characterization? – 2017-02-20
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    @Blazej Probably the first, otherwise we could say : The eigenvalues are real and distinct if and only if the characteristic polynomial has real and distinct roots. I do not think that this is the intent of the question. – 2017-02-20
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    A necessary condition is that the matrix can be diagonalized. – 2017-02-20
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    @Peter Yes I am looking for a condition that is necessary for $A$ to have real and distinct eigen values. – 2017-02-20
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    @Peter please elaborate or help me to show that if the characteristic polynomial of $A$ has n distinct roots, then $A$ is diagonalizable? As A is stable we can diagonalize $A$ right, such as the one with eigen values as diagonal entries can be obtained. Thank you for your time – 2017-02-20
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    Can we assume that the characteristic polynomial is known ? In this case, there might be easier criterions. To check, whether a matrix is diagonalizable is probably more difficult than checking the characteristic polynomial. – 2017-02-20
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    @Peter I have edited the question, $A$ has a peculiar form, is essentially nonnegative. Also $\sum_{i=1}^{n}A_{ij}\le 0$. I think assuming that characteristic polynomial known makes the result for a particular system right? Or am I missing something? Thanks – 2017-02-20
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    There is a following abstract characterization not using the characteristic polynomial: Hermitian matrix $A$ has a degenerate eigenvalue if and only if the $*$-algebra of all matrices commuting with $A$ contains a $*$-subalgebra isomorphic to $M_2(\mathbb C)$. However, I doubt this condition is computionally tractable for nontrivial matrices $A$. – 2017-02-20
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    @Blazej I am quite unfamiliar with these terms..not a mathematician....Thank you for your comment. So this is still an open problem...I mean there does not exist a theorem that states necessary conditions for the matrix roots to be real and distinct right? – 2017-02-20

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You already got a sufficient condition for realness (matrix being hermetian) but a condition for distint-ness is not that easy. You could use Gerschgorin-circles for a start.

This theorem gives you circles in the complex plane with midpoints on the diagonal entries. It states, that each eigenvalue lies in it's overlapping circles. If two (or more) overlap, the eigenvalues lie in the union of both circles (not necessarly in the intersection!)

So if every circle is distinct, all eigenvalues are unique.

If the circles overlap, the still can be unique!

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As pointed out by Peter, it is sufficient that the characteristic polynomial $\mathrm{det}(A-t I)$ has no double zeros. While this condition might be difficult to check in general, you have rather special form of $A$. You can compute the characteristic polynomial (by hand or using some software like mathematica to make it faster and less painful) and try to verify this property. This might be possible to do even without finding all roots explicitly. For example you can find all roots of the derivative of characteristic polynomial (this is third order rather than fourth so much easier) and then verify whether any of these is a root of the characteristic polynomial of $A$. Another method (probably more useful for larger matrices) is to extract some partial information (using tools such as polynomial discriminant, Descartes rule of signs or Vieta formulas - look up e.g. wikipedia for these). The approach with Greschgorin circles pointed out in one of the answers also looks promising.

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For the eigenvalues to be real matrix has to be Hermitian $v^{\dagger}Aw=w^{\dagger}A^{\dagger}v$ For any $v, w\in\mathbb{C}$. And here $F^{\dagger}=\bar{F}^{T}$. Don't know what to say about them being distinct.

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    This is not right. $\pmatrix{1&1\\ 0&1}$ is not Hermitian, but its eigenvalues are real. – 2017-02-20