Let $T$ be a subset of vector space $Z_2^n$ and $A$ be an element of $GL(2,n)$ means invertible matrices with entries $\{0,1\}.$ Let $T$ be invariant under A. It means for any $t \in T$, $tA \in T$. I'd like to count or find a bound for number of all $A'$s in $GL(2,n)$ which have this property for arbitrary $T$?
3
-
$\begingroup$ For arbitrary $T$? That's asking, as a special case, that the matrix $A$ fixes every vector, as a one-element subset. $\endgroup$– David Roberts ♦Commented Jan 21, 2019 at 8:46
-
$\begingroup$ @DavidRoberts I suspect you're misinterpreting the quantifiers. Given an arbitrary $T$, the OP wants to count or bound the number of $A \in GL(2,n)$ for which $T$ is invariant, as a function of $T$. $\endgroup$– Robert IsraelCommented Jan 21, 2019 at 16:44
-
$\begingroup$ @Robert thanks, I knew there was a sensible interpretation of the question. $\endgroup$– David Roberts ♦Commented Jan 21, 2019 at 19:13
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Since $A$ is invertible, $A T \subseteq T$ implies $AT = T$. Thus $A$ acts as a permutation on $T$. Suppose $T$ has $|T|$ elements and its linear span has dimension $d$. Then an easy upper bound is $|T|! (2^n - 2^d)^{n-d}$.
answered Jan 21, 2019 at 17:00