Let $\mathbb F$ be a finite field and let $T: \mathbb F^n \rightarrow \mathbb F^m$ be a linear transformation.
Suppose $n < m$, and that $T$ is injective (i.e. $T$ has a left inverse).
I am curious if I have counted the left inverses of $T$ correctly.
Attempt
It is clear that if $(e_1,...,e_n)$ is a basis of $\mathbb F^n$, then $(Te_1,...,Te_n)$ is a basis of Im $T$.
Extending this to a basis of $\mathbb F^m$ gives $(Te_1,...,Te_n, w_{n+1},...,w_m)$,
whence we can define $T^{-1}_L$ (the left inverse of $T$) by
$$T^{-1}_L (T(e_i)) = e_i$$ $$T^{-1}_L (w_j) = v_j$$
where $v_j \in \mathbb F^n$ is arbitrary.
This gives $(| \mathbb F |^n)^{m-n}$ choices for the left inverse of $T$.
$\square$
Next, suppose $n > m$, and that $T$ is surjective (i.e. $T$ has a right inverse).
Again, I attempt to count the right inverses of $T$.
Attempt
By the dimension formula, dim Ker $T = n - m$.
Let $(w_1,...,w_m)$ be a basis for $\mathbb F^m$. Since $T$ is surjective, we can find $(v_1,...,v_m)$ s.t. $T(v_i) = w_i$.
We define $T^{-1}_R$ (the right inverse of $T$) by
$$T^{-1}_R (T(v_i)) = v_i + a_i$$
where $a_i \in $ Ker $T$ is arbitrary.
This gives $|$Ker $T|^m = (| \mathbb F |^{n-m})^m$ choices for the right inverse of $T$.
$\square$
Background: This is a followup to a similar question of mine
