I am quite confused about this. I know that zero eigenvalue means that null space has non zero dimension. And that the rank of matrix is not the whole space. But is the number of distinct eigenvalues ( thus independent eigenvectos ) is the rank of matrix?

**Answer**

Well, if A is an n×n matrix, the rank of A plus the nullity of A is equal to n; that’s the rank-nullity theorem. The nullity is the dimension of the kernel of the matrix, which is all vectors v of the form:

Av=0=0v.

The kernel of A is precisely the eigenspace corresponding to eigenvalue 0. So, to sum up, the rank is n minus the dimension of the eigenspace corresponding to 0. If 0 is not an eigenvalue, then the kernel is trivial, and so the matrix has full rank n. The rank depends on no other eigenvalues.

**Attribution***Source : Link , Question Author : Shifu , Answer Author : Theo Bendit*