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

1. Mar 15, 2010

### 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.

2. Mar 15, 2010

### Landau

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.

3. Mar 17, 2010

### Zorba

Re: Proving

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.