Proving number of non zero eigen values.

The statement is :- number of non zero eigen values of a matrix A is atmost rank(A).

Can someone help me to prove this?

I read some of the answers to the question similar to this here but I am not able to understand it fully.

• you have read some answers that you are not able to understand, would you like to tell us what are those to avoid people posting those answers? – Siong Thye Goh Jul 28 '17 at 4:20
• I read one one question about number of distinct eigen values of a matrix with all one's and the answer was 1 because the rank of matrix is 1. I am not able to understand that. – Zephyr Jul 28 '17 at 10:08

If a matrix $A$ is diagonalizable, then the algebraic multiplicity of any eigenvalue of $A$ is equal to its geometric multiplicity. In particular, if $A$ is symmetric (Hermitian) or all the eigenvalues of $A$ are distinct, then $A$ is diagonalizable. From rank-nullity theorem, it is known that $$\text{rank}(A)+\text{nullity}(A)=\text{dim} (A).$$ Nullity is the dimension of the kernel space of $A$. Meaning the number of linearly independent eigenvectors x for which $Ax=0\cdot x$. So nullity in this case implies the multiplicity of $0$ as an eigenvalue of $A$ and hence rank implies the number of nonzero eigenvalues of $A$.