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Once we have one eigenvector vv corresponding to the eigenvalue λλ, any non-zero scalar multiple of vv is also an eigenvector corresponding to λλ. So rather than talking about the number of eigenvectors, we talk about the number of linearly independent eigenvectors ( for example corresponding to a particular eigenvalue), equivalently the dimensions of the eigenspaces (corresponding to a particular eigenvalue).

One useful thing is that for an eigenvalue, its geometric multeplicity cannot exceed its algebraic multeplicity. In general, one is often forced to solve for the eigenspace. Depending on the information you have, you may be able to make some deductions without solving for the eigenspace/s.

Again in regard to your last question, you need to be clear on what you are asking.

For example [11;01][11;01] has one repeated eigenvalue (multiplicity two) λ=1λ=1. Solving for the corresponding eigenspace, we get v=t[1;0]v=t[1;0], t∈Rt∈R. So we can write down only one eigenvector, if we are asked to give the set of linearly independent eigenvectors (any nonzero scaler multiple of [1;0][1;0] will do).

On the other hand, the 2×22×2 identity matrix again has only one eigenvalue (multiplicity two) but its corresponding eigenspace has dimension 22 (note any nonzero vector is an eigenvector). So we have two linearly independent eigenvectors.

Refer the link below for more information. 

http://math.stackexchange.com/questions/581760/repeated-eigenvalues-how-to-check-if-eigenvectors-are-linearly-independent-or-n

I hope the description helps you. 

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