Diagonalize a non-hermitian matrix

AI Thread Summary
The discussion centers on the concept of transformation optics and the properties of non-Hermitian matrices, particularly in relation to Pendry's work on photonic band structures. A key point is the confusion surrounding the unitary matrix S, which is defined as the sum of outer products of right and left eigenvectors, and its ability to diagonalize the non-Hermitian transfer matrix T. It is clarified that right and left eigenvectors do not necessarily form an orthogonal basis for all non-Hermitian operators, only for a subclass known as normal operators. The conversation also emphasizes the importance of understanding linear algebra concepts, as demonstrated through examples of simple two-dimensional matrices. Overall, the participants highlight the need for further study in linear algebra to grasp these advanced topics better.
njuclean
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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?
 
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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.
 
DrDu said:
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.
 
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.
 
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.
 
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