It seems that only if an eigenvalue is repeated that the complex matrix may fail to be diagonalizable ( https://www.jstor.org/stable/52325 ). In another word, as long as no eigenvalues are repeated, the complex matrix is diagonalizable. Then how to know if or not there is repeated eigenvalues for a complex matrix?

Dear Xiaoning Zang,

To test if a given matrix has repeated eigenvalues, first, we determine the characteristic polynomial P(x).

Then calculate the resultant between P(x) and its derivative dP(x)/dx.

If Resultant(P(x) ,dP(x)/dx) =0, then the matrix has repeated eigenvalues.

To compute the resultant, see

https://en.wikipedia.org/wiki/Resultant.

Best regards

C= A+iB, where A and B are real matrices, C is the complex symmetric matrix under consideration.

since CTRANSPOSE =C this implies ATRANSPOSE + iBTRANSPOSE= A+iB, which implies A and B are Symmetric, real matrices, therefore, if A and B are commuting matrices they can be simultaneously diagonalized in an Ortho-normal basis of R^n , the respective eigenvalues of A be a1, a2,...an and B be b1,b2,...bn, then the respective eigenvalues of C=A+iB are a1+ib1,a2+ib2......an+ibn.

Let Z = A+Bi be symmetric. Then since A and B are each symmetric If the real and complex parts commute , A and B are simultaneously diagonalizable hence Z is diagonalizable.

A complex symmetric matrix diagonalizable ,Write this as M=A+iB, where both A,B are real and A is positive definite. It follows that AA is invertible. From Horn and Johnson, in the first edition, define C=A−1B. If there is a real invertible matrix R with R−1CR diagonal, there is then a nonsingular real matrix S with SAST and SBST diagonal. It follows that SMST is diagonal complex.In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, matrix A is symmetric if A=AT.

Because equal matrices have equal dimensions, only square matrices can be symmetric.The entries of a symmetric matrix are symmetric with respect to the main diagonal. So if the entries are written as A = (aij), then aij = aji, for all indices i and j.