# Completeness relation for non-hermitian matrices

Basic question. Suppose you have a complex matrix $A$ with different (non-zero) eigenvalues associated with their corresponding eigenvectors. Does it exist a completeness relation for these matrices? For hermitian matrices, I think this is always true and the relation is

$$\sum_{n=1}^{dim} |\Psi\rangle_n (|\Psi\rangle_n )^T = I,$$

where $T$ denotes the transpose and $I$ the identity operator. My concern is that in general for $A$ one would have left and right eigenvectors, but consider for simplicity that the left eigenvector $|\Psi\rangle_L$ fulfills:

$$|\Psi\rangle_L = (|\Psi\rangle_R^*)^T$$

where the asterisk denotes complex conjugation.

• user2820579! I am also interested in this kind of question (studying non-hermitian problems in quantum mechanics); I solved one problem for the case of degeneracy and there is a thing called "generalized eigenvectors", which you use to construct the Jordan normal form [Chappers provided the link to wiki], but do these vectors along with normal eigenvectors (in regular sense) create a complete basis or mb it is overcomplete? I do not remember any theorem on that from undergrad linear algebra course, mb I have forgotten something. Have you found the answer? Thank you in advance – Sl0wp0k3 Jun 2 '17 at 9:34
• I don't know the answer, but for the down answer I would say no in the general sense. Still, I think it is highly possible to construct a complete basis when the matrix has a special structure (for example scattering matrices, or transport-related Green's functions). We have to dig in the journals or ask somebody who works with these things. – user2820579 Jun 2 '17 at 17:46

No; the usual example is $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},$$ which has only one eigenvector: $(1,0)$.
• The answer's still no: take $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},$$ for example, with single eigenvector $(1,1)$. – Chappers Oct 30 '15 at 1:04