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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!

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    $\begingroup$ Do you want similarity over $\Bbb F_2$ or over $\Bbb R$? $\endgroup$
    – Angina Seng
    Commented Jan 1, 2018 at 17:18
  • $\begingroup$ Over $\mathbb{C}$. Thanks! $\endgroup$
    – Emil Sinclair
    Commented Jan 2, 2018 at 18:26
  • $\begingroup$ Though if you'd like to answer over $\mathbb{F}_2$ feel free! $\endgroup$
    – Emil Sinclair
    Commented Jan 3, 2018 at 19:46
  • $\begingroup$ In particular you have $SJS^{-1}=S\sum_iA_iS^{-1}= \sum_iB_i=J$, so $SJ=JS$, that is $\sum_{k} s_{ik}=\sum_k s_{kj}$ for each $i$ an $j$, where $S=\|s_{ij}\|$, that is the sums of all rows and columns of the matrix $S$ are equal. $\endgroup$
    – Alex Ravsky
    Commented Jan 4, 2018 at 5:58
  • $\begingroup$ Indeed, $J$ is the only boolean matrix with $n$ as an eigenvalue. $\endgroup$
    – Emil Sinclair
    Commented Jan 4, 2018 at 8:47

1 Answer 1

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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.

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