Let $F_q$ be a finite field with $q$ elements.

$T_n(F_q) := \{ A = (a_{ij}) \in F^{n \times n}$ | $a_{ij} = 0$ for $i < j,$ and $a_{ii} \neq 0$ $\forall i \}$.

Determine the number of elements in $T_n(F_q)$.

My solution is as follows:

Starting with the last row going upwards, there are:

$q-1$ possibilities for the last row;

$(q-1)q$ possibilities for the row before the last;




$(q-1)q^{n-1}$ possibilities for the first row.

Therefore, in total there are $(q-1)^nq^{\sum_{i=1}^{n-1} i} = (q-1)^nq^{\frac{n(n-1)}{2}}$ elements.

Could you, please, check my solution?

  • $\begingroup$ You mean $a_{ii}\neq 0$, not $a_{ij}\neq 0$, in the definition of $T_n$, $\endgroup$ – user10354138 May 26 at 22:34
  • $\begingroup$ Yes, my bad. I edited the question accordingly. $\endgroup$ – user314159 May 26 at 22:37
  • $\begingroup$ Yes. Another way: there are $n$ diagonal entries, each can be chosen from $q-1$, and the $\binom{n}{2}$ entries below can be chosen from $q$. So it is $(q-1)^n q^{n(n-1)/2}$. $\endgroup$ – user10354138 May 26 at 22:41

Yes, your solution is right. An easier way would be to simply count the possible diagonal entries, of which there are $(q-1)^n$ (since there are $n$ entries each with $q-1$ choices), and just multiply this by all possible choices of the entries below the diagonal ($q$ choices for each entry, and there are $n(n-1)/2$ entries). There's not really a need to count these by rows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.