# number of 1-to-1 linear functions on vectorspaces over finite fields

This is not a homework. I just ask this question myself and thought it would be easy to figure out. But I did not get the solution.

Let $\mathbb{F}$ be a finite field with $|\mathbb{F}|=q$. Consider the $\mathbb{F}$-vectorspaces $V_1=\mathbb{F}^n,V_2=\mathbb{F}^m$ with dimensions $n<m$.

How many injective, surjective and bijectiv linear functions $f\colon \mathbb{F}^n \to \mathbb{F}^m$ exists?

My approach is: We have any basis $b_1,\ldots,b_n$ of $V_1$ and $c_1,\ldots,c_m$ of $V_2$. It clear that we only have to treat the function on this basis and there cannot be any bijective linear functions since $n<m$. Counting the functions must be similar to count the possibilities to do a injective map from $b_1,\ldots,b_n$ of $V_1$ to $c_1,\ldots,c_m$ of $V_2$.

How do I get the number of injective functions?

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 You want to count how many ordered $n$-tuples of linearly independent vectors you have in $\Bbb F^m$. – Andrea Mori Jan 20 at 12:54 Is it that: number of n linear independen vectors in $\mathbb{F}^m$ are $\left( \prod _{\ell = 0}^{n-1} \frac{q^n-q^\ell}{q^k-q^\ell}\right)\cdot n!$ – user58986 Jan 20 at 13:05 @user58986: There's a $k$ in there that you haven't introduced. – joriki Jan 20 at 13:13 @DonAntonio: I already wrote that there cannot be any bijective linear functions since $n ## 1 Answer If I understand correctly, you already know that there are no surjective or bijective functions since$|\mathbb F^n|\lt|\mathbb F^m|$. To count the injective functions, choose a non-zero vector in$\mathbb F^m$to map$b_1$to – there are$q^m - 1$choices. Now choose a linearly independent vector to map$b_2$to – there are$q^m-q\$ choices. Continuing like this, you have

$$\prod_{k=1}^n\left(q^m-q^{k-1}\right)$$

possibilities in all.

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 Thanks. I tried to count some mappings twice. – user58986 Jan 20 at 13:49