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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 '13 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 '13 at 13:05
@user58986: There's a $k$ in there that you haven't introduced. –  joriki Jan 20 '13 at 13:13
@DonAntonio: I already wrote that there cannot be any bijective linear functions since $n<m$. –  user58986 Jan 20 '13 at 13:48

1 Answer 1

up vote 2 down vote accepted

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 '13 at 13:49

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