diagonalize a non-hermitian matrix

[njuclean]

I'm learning "the transformation optics" and the first document about this method is "Photonic band structures" ( Pendry, J. B. 1993). In this document, the transfer matrix T is non-hermitian, Ri and Li are the right and left eigenvectors respectively.

Pendry defined a unitary matrix S=\sumRiLi, here RiLi is the outer product between the right eigenvector and the left eigenvector. Then T can be diagonalized through STS-1.

I remember that the sum of the outer product between the right eigenvector and the left eigenvector is the unit matrix, so i cannot understand how can the unitary matrix S diagonalize T. Who can tell me what's wrong in my understanding?

[DrDu]

In general, S won't be the unit matrix. Consider it's operation on one of the left eigenvectors L_i. It will get transformed in the corresponding right eigenvector R_i. If the left and right eigenvectors do not conincide, S cannot be the unit matrix.

[njuclean]

In general, S won't be the unit matrix. Consider it's operation on one of the left eigenvectors L_i. It will get transformed in the corresponding right eigenvector R_i. If the left and right eigenvectors do not conincide, S cannot be the unit matrix.

Thanks for your reply.

Are the right vectors of a non-hermitian matrix in general not the orthogonal and complete bases? I'm lack of the knowledge of linear algebra, so my question maybe too simple.

[DrDu]

The left or right eigenvectors form an orthogonal basis not for all possible non-Hermitean operators but only for a sub-class, the so-called "normal" operators. All normal operators can be expressed in the form A+iB where A and B are Hermitean operators.
Maybe you should try to work out the eigenvalues and eigenvectors for some very simple two dimensional matrices e.g. one with i and -i on the diagonal and zero on the outer diagonal.

[njuclean]

I'm glad to see your reply again.

The eigenvalues of the two dimensional matrix (a11=i, a12=0, a21=0, a22=-i) are \lambda1=i and \lambda2=-i. For \lambda1, R1=(1 0)T and L1=(1 0). For \lambda2, R2=(0 1)T and L2=(0 1). Thus \sumRjLj=R1L1+R2L2=I.

I think it's necessary for me to read some books about the linear algebra. Anyway, thanks again for your help.