In my statistics classes it was stated that if an $n\times n$ matrix $A$ has $k$ zero eigenvalues, we have $rank(A) = n-k$. Is there any straightforward proof of this? Are there any limitations on the matrix needing to be symmetric?
From the relations $det(A) = \Pi_i \lambda_i$ and $ rank(A) < n \leftrightarrow det(A) = 0 $, I understand that one zero eigenvalue must imply that the tank of $A$ is deficient. But in case of several zeros, why must the number of them equal the number of ranks lost?