Call a matrix boolean $n\times n$ matrix $A$ trivial if it has $|\co{A}|=o(n)$ zeros. Note that every trivial matrix has either $m=n-o(n)$ all-$1$ rows or $m$ all-$1$ columns. So, we can replace all these rows/columns by just one row/colum, and the complexity of the resulting submatrix remains almost the same as that of $A$.
For the XOR complexity of any non-trivial boolean $n\times n$ matrix $A$, we clearly have $\XOR{A}\preccurlyeq |\co{A}|$, since $Ax=(J\oplus \co{A})x=Jx\oplus \co{A}x$ and $\XOR{J}\preccurlyeq n$; here, $J$ is the $n\times n$ all-$1$ matrix. This, however, does not imply similar upper bounds for the SUM or OR complexities: by Theorem 5.8 in the book, we know that there are boolean $n\times n$ matrices $A$ for which the XOR complexity is much smaller: \[ \SUM{A}/\XOR{A}\geq \OR{A}/\XOR{A} \succcurlyeq n/\log^2 n \,. \]
Still, Kulikov, Mikhailin, Mokhov and Podolskii [1] recently proved that an upper bound of about $|\co{A}|$ on $\SUM{A}$ also holds.
Theorem [1]: For every non-trivial boolean $n\times n$ matrix $A$, we have $\SUM{A}\preccurlyeq |\co{A}|$ and, hence, $\SUM{A}\preccurlyeq \min\{|A|,|\co{A}|\}$.This result gives us an opportunity to state an open problem, which we missed to state in the book.
Problem 7.20: How large can a gap $\SUM{\co{A}}/\SUM{A}$ be?The result of [1] refutes a naive hope that this gap could be large for sparse matrices $A$. Namely, we now know that, in order to establish a non-trivial gap (if there is one), we need to consider rather balanced matrices $A$, those with $|\co{A}|=\omega(n)$ zeros and $|A|=\omega(n)$ ones.
References: