I'm calculating eigenvectors of a high dimension complex symmetric matrix and I know this is the case about the Hermitian matrices but are eigenvectors of a complex symmetric matrix orthogonal too?

Because my matrix always has degenerate eigenvalues and the eigenvectors corresponding to those eigenvectors are not orthogonal. (Done with Matlab)

If not, can they form an orthogonal eigenbasis by orthogonalization?

What are the methods to do such thing? Will Gram-Schmidt be useful here?

EDIT: This matrix is a normal matrix.

  • 2
    $\begingroup$ If your matrix is normal, yes, it has an orthogonal eigenbasis and you can "orthogonalise" the eigenbasis via Gram-Schmidt process. $\endgroup$
    – user1551
    Aug 7 '17 at 7:59
  • $\begingroup$ @MarcvanLeeuwen I think he means a matrix of the form $A=QDQ^T$ where $Q$ is real orthogonal and $D$ is complex diagonal. Then $A$ is normal and complex symmetric but not Hermitian. $\endgroup$
    – user1551
    Aug 7 '17 at 8:01
  • 1
    $\begingroup$ @user1551 I think $AA^*$ means $AA^*$. The comment at the answer said this, though the answer has now been deleted. $\endgroup$ Aug 7 '17 at 8:03
  • $\begingroup$ OP, can you please do full disclosure and tell us what you know about your matrix? $\endgroup$ Aug 7 '17 at 8:04
  • $\begingroup$ @MarcvanLeeuwen it is a symmetric complex matrix (i.e A(i,j)=A(j,i)) so it's not Hermitian. But it is a normal matrix (i.e AA*=A*A) $\endgroup$
    – Alireza
    Aug 7 '17 at 8:08

Eigenvectors of a complex symmetric matrix are orthogonal in a tranpsose sense and not in the usual conjugate transpose sense. That is, if $\lambda_{1}$ (eigenvector $v_1$) and $\lambda_{2}$ (eigenvector $v_2$) are 2 distinct eigenvalues of a complex symmetric matrix A, then $v_1^T v_2=0$, but $v_1^\dagger v_2\neq 0$. eig() function in MATLAB will yield in such orthogonal eigenvectors in the transpose sense. You can check this for yourself in MATLAB. Hope I answered your question.


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