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Problem:

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?

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  • $\begingroup$ 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
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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.

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