How to prove that if A is a diagonalizable matrix, then the rank of A

Re: Proving

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.

 

Re: Proving

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