Short and simple proof that matrix rank $\geq n-1$

Question

An $n \times n$ matrix has zeros on the main diagonal and the off-diagonal entries are either $1$ or $1980$. Prove that its rank is $\geq n-1$.

Thanks for helping.

Answer 1

Note that $1979$ is a prime number. Now consider the matrix modulo $1979$. Then it is the matrix $J-I$ where $J$ is the matrix of all $1$'s. Now $J$ has one eigenvalue that is $n$ and $n-1$ eigenvalues that are $0$ so $J-I$ has one eigenvalue that is $n-1$ and $n-1$ eigenvalues that are $-1$. This means that the rank of this matrix modulo $1979$ is at least $n-1$ (if $n-1$ is not $0$ modulo $1979$ then the rank is $n$). It follows that the rank of the matrix is $\ge n-1$.