If A is a real matrix, then eigen values of (A + A' )/2 are Real part of eigen -values of A.It means they are same if eigen-values of A are real otherwise they are real part of eigen-values of A.

First of all, I wish to remark that the eigenvalues of an arbitrary Hermitian matrix are real numbers. Matrix $A $ with entries $ a_{m,n } $ is Hermitian if $A = A^{T*}$, where $A^{T*}$ denote a matrix transposed and complex conjugate.As a corollary.it solves your problem.

If you look in my webpage http://vlcek.fd.cvut.cz/publications.htm, there is another application of matrices $(A + A^{-1})/2$ in our old paper Vlček M., Zima V. : On reciprocity, Proceedings ECCTD'85, Vol. 1, pp. 250-252, Prague 1985.

Thanks for the quick replies.

I want to make it clear that I am only talking about real matrices.

Mittal: I don't what you said is true. Take as a counter example: A=[1 3; 2 4]. The eigenvalues of A are -0.372 and 5.372. The corresponding symmetric matrix is (A+A.')/2 = [1 5/2; 5/2 4] with eigenvalues -0.415 and 5,415. I think it is clear that the two are different.

The question I am more interested in is the correctness of the conjecture: if the real eigenvalues of (A+A.')/2 are positive, then the real parts of the eigenvalues of A will be positive.

Thanks in advance.

Agreed Abdelmalek. The possible reason is that A and A' have same set of eigen values but different eigen-vectors.

R.C.Mittal, your statement is only true if A is a normal matrix, i.e. it is diagonalizable by a unitary matrix. The small example by S. Abdelmalek is not normal.

There are known results saying: if A+A' is positive definite, then A is also positive definite.