TITLE: Is there a non-singular matrix with $k$ ones on each row?
QUESTION [5 upvotes]: Let $n$ be a fixed positive integer. I would like to know for what values of $k$ there exists an $n$ by $n$ $0/1$ matrix that is non-singular with exactly $k$ ones per row.
Clearly if $k=1$ then the identity matrix is non-singular. Also if $k=n$ there are no non-singular matrices.  
What can one say about $1 < k  < n$?
If $n$ is large, is it true for almost all $k$ in the range?

REPLY [3 votes]: I edited my answer because this new proof is much simpler.
Thm: For every $k<n$ there is a non singular $(0,1)-$matrix of order $n$ with exactly $k$ ones on each row.
Proof: If the order is two, consider the identity.
Assume the result is true for matrices with order $n-1$. 
Now if $k<n-1$, by induction hypothesis, we can find  $(0,1)-$matrix $A$ with exactly $k$ ones per row of order $n-1$. 
Let $B$ be a matrix of order $n$ defined by  
$B=\left[\begin{array}{cc}A_{n-1\times n-1} & 0_{n-1\times 1} \\ u_{1\times n-1} & 1_{1 \times 1} \end{array}\right]$, where $u$ is any $(0,1)$ vector with $k-1$ ones.  
Notice that $\det(B)=\det(A)\neq 0$.
If $k=n-1$ consider the matrix of order $n$ with diagonal equal to zero and the other entries equal to 1. $\square$