I was reading the following lecture: https://www.stat.cmu.edu/~cshalizi/statcomp/13/lectures/15/markov.pdf, but I had a problem with a certain paragraph which is this:
Since q is a square, $K \times K$ stochastic matrix, it will have $K$ eigenvectors, say $v_1,...v_k$, and $K$ eigenvalues, $\lambda_1,...\lambda_k$. (Not all of the eigenvalues are necessarily different.) The eigenvectors will form a basis for $R^k$, meaning that we can write an arbitrary vector as a linear combination of the eigenvectors.
But if there were eigenvectors with the same eigenvalues wouldn't they just be multiple of each others? So they wouldn't form a basis. Also a stochastic matrix might not have an inverse which would make it impossible for it's rank to equal $K$. Is there something wrong with my reasoning, or have I misunderstood what the writer actually means?
The paragraph is the second one in section 2.1, if anybody wants to take a look at it.
The eigenspace for a given eigenvalue can be bigger than one-dimensional. Consider the $2\times 2$ identity matrix. Every vector is an eigenvector for the eigenvalue $1$. The eigenspace is the entire $\mathbb{R}^2$. Not all vectors in there are multiples of each other.
In general, you can choose a basis for each eigenspace. Stitching all these bases together gives you a linearly independent system consisting of eigenvectors. In your case, the matrix is diagonalizable, which makes this system also a basis.