Proving

[DavidLiew]

How to prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.
Thanks and regard.

[Landau]

Let's say D is the diagonalized form of A. Then the diagonal elements of D are precisely the eigenvalues of A. The rank of D is the number of linearly independent columns. Obviously this equals the number of non-zero eigenvalues. Since the rank of A and the rank of D are the same, the conclusion follows.

[Zorba]

If A is similar to B then \textrm{rk}(A)=\textrm{rk}(B), then consider the rational canonical form, and it follows as Landau stated above.