TITLE: Eigenvalues are unique?
QUESTION [21 upvotes]: I'm studying eigenvector and eigenvalue but there are some confusing things to me. 
(1) Eigenvectors are not unique
(2) If eigenvectors come from distinct eigenvalues, then eigenvectors are unique.
Here is my question.
Then, eigenvalues are unique all the time?
Or there are any restrictions to make eigenvalues to be unique just like the case of eigenvectors?
Also, when we diagonalize matrix A=$S\lambda S^{-1}$, the eigenvector matrix S is not unique. It means if we multiply each column of S by nonzero constant we can make a new $S'$.
Why is that?
If we can diagonalize matrix A, it means that eigenvectors are unique. But the eigenvetor matrix S is not unique? Then with the new eigenvector matrix$S'$,  $S'\lambda S'^{-1}$ makes the same matrix A?
My last question is that. If eigenvalues are not unique, then eigenvalue matrix $\lambda$ is not unique too? But $S\lambda'S^{-1}$ makes same matrix A?

REPLY [3 votes]: 1.Given a matrix, the superset (a set that allows multiple instances of an element) of eigenvalues is unique. It implies that you can not find a different superset of eigenvalues for a matrix. 
2.Eigenvectors corresponding to eigenvalues of single multiplicity are parametrized by a coefficient which is denoted by $c.\mathbf{x}$. The unit eigenvector $\mathbf{x}$ is unique up to sign (it can be multiplied by -1) for this case.
3.For eigenvalues of multiplicity $n$, the dimension of eigenspace $m$ is such that $m \leq n$. For this case, we can find different sets of unit vectors which span the same eigenspace. In other words, set of eigenvectors is not unique for this case. (Here, we assume that $m>1$, i.e if $m=1$, the unit eigenvector is unique up to sign as noted in 2.