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Question Statement

Let $K$ be a finite field of $q$ elements. Let $U$, $V$ be vector spaces over $K$ with $\dim(U) = k$, $\dim(V) = l$. How many linear maps $U \rightarrow V$ are there?

I'm struggling to answer this question. Given that a linear map is uniquely determined by its action on a basis we may as well consider the mappings of the standard basis. Even with this restriction, each of the $k$ basis vectors can map to not only $q$ distinct elements but also into $l$ different positions (or none, or into multiple positions).

What easy thing am I missing?

Mhenni Benghorbal
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Asa Cremin
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3 Answers3

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The answer (if I actually got it right), for anyone still wondering:

Let $\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_k$ be a basis of $U$, $\mathbf{f}_1,\ldots,\mathbf{f}_l$ be a basis of $V$ and $T:U\rightarrow V$ be a linear map.

Then $$T(\mathbf{e}_1) = \alpha_{11}\mathbf{f}_1 + \alpha_{12}\mathbf{f}_2 + \cdots + \alpha_{1l}\mathbf{f}_{l}$$ $$T(\mathbf{e}_2) = \alpha_{21}\mathbf{f}_1 + \alpha_{22}\mathbf{f}_2 + \cdots + \alpha_{2l}\mathbf{f}_l$$ $$\vdots $$ $$ T(\mathbf{e}_k) = \alpha_{k1}\mathbf{f}_1 + \alpha_{k2}\mathbf{f}_2 + \cdots + \alpha_{kl}\mathbf{f}_l$$

defines $T$. The number of different linear transformations is then just the number of different combinations of $\alpha_{ij}$'s. There are a finite number of combinations as there are only $q$ choices for each one.

I found two similar methods, each with a slightly different focus, that lead to the answer:

More algebraic

As there are $q$ different choices for each $\alpha_{ij}$ and they are chosen independently, so on each row there are $q^l$ different choices. Now over $k$ independently chosen rows we have $$\underbrace{q^l \cdot q^l \cdot \cdots \cdot q^l}_{k\text{ times}}=(q^l)^k=q^{kl}$$ different choices

More combinatorial

we may write the linear transformation as a matrix of the coefficients ($\alpha_{ij}$'s) in the following way

$$ \pmatrix{ \alpha_{11} & \cdots & \alpha_{1l}\\ \\ \vdots & \ddots & \vdots\\ \\ \alpha_{k1} & \cdots & \alpha_{kl} } $$

You can now look at the number of ways of choosing combinations of non-zero $\alpha_{ij}$'s, which is every possible matrix and therefore every possible linear transformation (note that choosing no $\alpha_{ij}$ non-zero gives us the zero matrix).

There are

  • $\binom{kl}{0} \cdot (q-1)^0 = 1$ ways of choosing no non-zero coefficients. We are choosing zero of the $k \times l$ coefficients.

  • $\binom{kl}{1} \cdot (q-1)^1 = (q-1)$ ways of choosing one $\alpha_{ij}$ non-zero. 1 of the $k\times l$ coefficients and $q-1$ different choices for it

  • $\binom{kl}{2} \cdot (q-1)^2$ ways of choosing two $\alpha_{ij}$ non-zero. There are $q-1$ non-zero choices for each of the 2 coefficients, so $(q-1)^2$ different combinations

In general there are $\binom{kl}{i} \cdot (q-1)^i$ ways of choosing $i\ \alpha_{ij}$ non-zero. The total number of linear transformations is the number of different matrices, which is the sum of all the numbers above:

$$\sum_{i=0}^{kl} \binom{kl}{i} \cdot (q-1)^i = \sum_{i=0}^{kl} \binom{kl}{i} \cdot (q-1)^i \cdot 1^{kl-i}= (q-1+1)^{kl}=q^{kl}$$

Asa Cremin
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Hint: The space of all linear maps from $U\to V$ is isomorphic to $\text{Mat}_{l\times k}(K).$

Sugata Adhya
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All such linear maps are of the form $c_1y_1+..c_ky_k$ where $y_1,..y_k$ is the image of the basis vectors of $U$, and $c_i$ are scalars. Note that you can have $q$ different choices for each $y_i$

Hope you can finish the problem now.

voldemort
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