how do i explain which eigenvector the software reports as there are theoretically an infinite number?

A= [1 2 -5; 3 5 3;-4 6 1]

i can find the eigen values and eigen vectors, but i don't know how to find which one has an infinite number. is there a specific command on Matlab to get this report?

[Walter Roberson]

117 Sierra: our working policy is that we do not delete questions that have valid Answers except under extraordinary circumstances (e.g. someone trying to cheat on an exam, or a question we cannot legally discuss such as encryption)

Answer 1

[Adam]

doc eig
doc eigs

will both return normalised eigenvectors, depending on your input arguments.

Answer 2

[John D'Errico]

Return to linear algebra 101. Do not pass go, do not collect $200. Really, if you will be using tools like eig, it would be good if you at least have some understanding of the mathematics behind eigenvalues and eigenvectors. Here you surely read a comment somewhere online that eigenvectors are not unique, without understanding anything at all what was meant by that comment.

All eigenvectors define a basis for a vector subspace. In the case of distinct eigenvalues, the corresponding eigenvectors each live in their own subspace. So any multiple of an eigenvector V, thus k*V, is also an equally valid eigenvector for the corresponding eigenvalue. We can see this by recognizing that if

A*V = lambda*V

then it is equally valid that

A*(k*V) = lambda*(k*V)

for any scalar k.

This effectively means that an eigenvector is not unique, due to the application of an arbitrary scale factor. In general, eigenvectors are scaled to have unit norm, but even then, we can always multiply any eigenvector by k=-1 and not change the norm.

All of that applies to the case of distinct eigenvalues. Suppose you have an eigenvalue of multiplicity higher than 1? It gets messier now. Then the eigenvectors for that eigenvalue live in a higher dimensional subspace. We can now choose any set of orthogonal basis vectors that span that subspace as eigenvectors.

But in the case of the matrix A that you show, the eigenvalues are completely distinct.

eig(A)
ans =
    -5.7582
     5.9008
     6.8574