Eigen vector of a non symmetric matrix.

eigenvalues-eigenvectorsmatrices

In my quantum mechanics course, we were finding eigenvalues of a 3 by 3 matrix which was not symmetric. My question is when we find eigenvalues and try to find eigenvectors we have infinite choice of choosing arbitrary values to from a eigenvector of 3 by 1 matrix.

My question is

  1. Are these eigenvectors for different eigenvalues linearly independent to each other always. And if not always is there a possible case where they become linearly independent.

  2. If suppose there are choices of choosing eigenvector, is there a choice of choosing eigenvector for different eigenvalue chosen to be orthogonal, in spite of there exist various choices of choosing eigenvector such that some may produce non orthogonal eigenvector.

Edit. If possible go easy on me if my question is absurd and if possible, can explain by an example of a simple 3 by 3 matrix would be kind of you.

Best Answer

Few useful points:

  1. If matrix $(A)_{n\times n}$ has distinct eigenvalues then there always exist $n$ LI eigenvectors. Though for repeated eigenvalues you have to try your luck as the answer is not always positive unless $A$ is symmetric matrix.

  2. There always exist an orthogonal set of LI eigenvectors (in addition to some non-orthogonal set too) for a symmetric matrix $A$.

Added-

  1. If the eigenvectors are orthogonal then matrix $A$ is necessarily symmetric.The proof is as follows:

Let $P$ be the modal matrix (orthogonal) corresponding to matrix $A$ s.t. $A=PDP^{T}$, where $D$ is diagonal matrix with eigenvalues in its diagonal. Then $A^T=(PDP^T)^T=PDP^T=A$

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