Dear Karim, tridiagonal or not - if the matrix Q is non-singular and diagonalizable (has a complete basis of eigenvectors), then is your statement true.

It is based on the fact that inverse matrices have equal eigenspaces.

Farid´s matrix is not diagonalizable.

The attached file contains a 3x3 counter-example. The statement is generally false, even when the eigenvalues of Q are simple. The counter example is one for which the eigenvalues are distinct (hence simple). However, it might be true if Q is Hermitian, tri-diagonal, and with simple eigenvalues.

Dear Karim, tridiagonal or not - if the matrix Q is non-singular and diagonalizable (has a complete basis of eigenvectors), then is your statement true.

It is based on the fact that inverse matrices have equal eigenspaces.

Farid´s matrix is not diagonalizable.

Let Q be an nxn Hermitian matrix with distinct nonzero eigenvalues. Let D=diag(L1,L2,...,Ln) be the nxn diagonal matrix with the eigenvalues of Q along the diagonal. Let Vj be an eigenvector belonging to Lj where ||Vj||=1. Let V be the nxn matrix whose jth column is Vj. Then,

inv(Q) = V * inv(D) * V',

where the prime denotes conjugate transpose.

Note that the rows of V' are left eigenvectors of Q and the columns of V are right eigenvectors of Q. Note also that Q need not be tri-diagonal.

Using the facts that the eigenvalues of a Hermitian matrix are real and that a pair of eigenvectors (of a Hermitian matrix) belonging to distinct eigenvalues are orthogonal, the proof is straight forward. From QV=VD and V'Q=DV', we only need to show [V*inv(D)*V']Q is the nxn identity matrix, which is a straight forward calculation. [V*inv(D)*V']Q = V*V' = nxn identity matrix.