Eigen vector of a non symmetric matrix. 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

*

*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.


*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.
 A: Yes, when dealing with finite dimensional vector spaces the set of vectors formed by choosing one eigenvector for each eigenvalue is
always linearly independent. It is one of the first results you will find while studying those topics, and the proof can be found in almost any Linear Algebra book (I can recommend you looking for Foundations of Linear Algebra by J. Golan, p. 130).
I suppose the entries of your matrices are real numbers; if your $3 \times 3$ matrix $A$ is diagonalizable, i.e. if you can find a basis of the vector space $R^3$ whose elements are all eigenvectors of $A$ (i.e. if you can find a basis $D$ of $R^3$ for which the matrix representation of $L_A$ is diagonal), then by the Gram-Schmidt process you can indeed
obtain from this basis an orthonormal basis of the space. The bad news is that this new basis is not necessarily formed by eigenvectors
of the matrix anymore.
Indeed, there is a theorem that states that the fact of finding an orthogonal basis $D$ of $V$ (where $V$ is a finite-dimensional
real inner product space) consisting of eigenvectors of a linear operator $T:V \rightarrow V$ is equivalent to the fact
of $T$ being self adjoint.
A: Few useful points:

*

*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.


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


*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 $A^T=(PDP^T)^T=PDP^T=A$
