 Solution

On an eigenvector of $A$ with eigenvalue $\lambda$, the linear operator $B=I-A$ has eigenvalue $1-\lambda$. ThereforeThe multiplicities are exchanged: if the matrix rank of $A$ is $r$, then $B$ has $n-r$ copies of $1$ and $r$ copies of $0$. Again, a value with zero multiplicity is absent.