# Number of elements in the set of invertible lower triangular matrices over a finite field

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?

• You mean $a_{ii}\neq 0$, not $a_{ij}\neq 0$, in the definition of $T_n$, – user10354138 May 26 at 22:34
• Yes, my bad. I edited the question accordingly. – user314159 May 26 at 22:37
• 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 $(q-1)^n q^{n(n-1)/2}$. – 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.