How to prove that $S=AA^T$ , the eigenvalues of S are non-negative integers when A is a binary matrix Assume a binary matrix $A\in\{0,1\}^{N\times k}$, where $N\geq k$, and only one element for each row of $A$ is $1$, $\forall u, A(u,:)\overrightarrow{1}=1$, where $\overrightarrow{1}$ denotes all-ones vector.
Noting $S=AA^T$, and I found that the eigenvalues of $S$ are all non-negative integers, is there a prove of it?

From the SVD of the matrix $\,\,A = UDV^T$ we can calculate the EVD of the symmetric square matrices
$$\eqalign{
A^TA &= VD^2V^T \in {\mathbb R}^{k\times k} \cr
AA^T &= UD^2U^T \in {\mathbb R}^{N\times N} \cr
}$$
The $D$ matrix is diagonal, so eigenvalues of each product pair are the diagonal elements of $D^2$, which are the same.
Because of the way the $A$ matrix is defined, the matrix $(A^TA)$ is diagonal.
Recall that there are only $k$ possible row vectors.
The diagonal elements of $(A^TA)$ are non-negative integers which count how many times a given row vector is duplicated within the $A$ matrix. 
