I was just wondering: is it necessarily the case that if $A$ is a $(0, 1)$ matrix, and $SAS^{-1}=B$, where $B$ is also a $(0,1)$ matrix, then do $A$ and $B$ have the same number of $1$s?
I have the gut feeling that this may not be true. However, let us consider this special case which arises in graph theory (coherent configurations): suppose $\{A_1, A_2,..., A_r\}$ is a set of $(0,1)$ matrices such that $\sum_iA_i=J$, with $J$ being the all $1$s matrix, and some subset of it sums to the identity matrix. Let $\{B_1, B_2,...,B_r\}$ be another set satisfying the same conditions. If $SA_iS^{-1}=B_i$, then do $A_i$ and $B_i$ have the same number of $1$s for all $i$?
Thanks!
For a counterexample to the first question, take $A$ to be any upper triangular matrix with more than $n-1$ off-diagonal ones, and $B$ its Jordan normal form. These have the same number of ones on the diagonal, but a different number of off-diagonal ones.
The second statement is true. If $SJS^{-1}=J$ and $SAS^{-1}=B$ then $bJ=JBJ=SJAJS^{-1}=aJ$ where $a$ and $b$ are the totals of the entries of $A$ and $B$ respectively.