Let $F$ be a field. I am trying to compute the dimension of the vector space of all $n$ linear functions $D:F^k \rightarrow F$

I figured this is a pretty standard calculation but the only place so far I have found help is http://en.wikipedia.org/wiki/Exterior_algebra

Question: Is the dimension of vector space of all $n$ linear functions from $D:F^k \rightarrow F$ just $n^k$

If we consider alternating $n$-linear functions instead is the dimension $ n \choose k$?

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    $\begingroup$ Do you mean $k$-linear maps $F^k \to F$? This space is always $1$-dimensional. Or do you want $n$-linear maps $(F^k)^n \to F$? That would have dimension $k^n$. $\endgroup$ – Dylan Moreland Jul 22 '11 at 14:23
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    $\begingroup$ The results you were probably thinking of are that if $V$ is a vector space of dimension $d$, then the dimension of the space of multilinear maps $V^k \to F$ is $d^k$, and the dimension of the space of alternating multilinear maps $V^k \to F$ is ${d \choose k}$. $\endgroup$ – Qiaochu Yuan Jul 22 '11 at 14:35
  • $\begingroup$ @Dylan Thank you for the help, I messed up the notation I should have been asking about maps $(F^k)^n \rightarrow F$. Given your answer is $k^n$ how do we determine the dimension of the space of alternating $n$-linear maps? $\endgroup$ – user7980 Jul 22 '11 at 14:38
  • $\begingroup$ @Qiaochu Thank you that is exactly what I was looking for! $\endgroup$ – user7980 Jul 22 '11 at 14:38
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    $\begingroup$ @user7980: I would guess that it is proven in any reasonably pure book on linear algebra, but as far as resources that are freely available, try sites.google.com/site/winitzki/linalg . $\endgroup$ – Qiaochu Yuan Jul 22 '11 at 14:48

Suppose that $D:F^n\rightarrow F$ is $n$-linear. In other words, $$D(ca_1,\ldots,a_n)=cD(a_1,\ldots,a_n)$$ $$\vdots$$ $$D(a_1,\ldots,ca_n)=cD(a_1,\ldots,a_n)$$ for all $a_1,\ldots,a_n,c\in F$, and $$D(a_1+b,a_2,\ldots,a_n)=D(a_1,a_2,\ldots,a_n)+D(b,a_2,\ldots,a_n)$$ $$\vdots$$ $$D(a_1,\ldots,a_{n-1},a_n+b)=D(a_1,\ldots,a_{n-1},a_n)+D(a_1,\ldots,a_{n-1},b)$$ for all $a_1,\ldots,a_n,b\in F$.

Therefore, for all $a_1,\ldots,a_n\in F$, $$D(a_1,\ldots,a_n)=a_1\cdots a_n D(1,\ldots,1)$$ Thus, the entire function $D$ is determined by where $(1,\ldots,1)\in F^n$ goes. We can choose to send it anywhere in $F$. Therefore, the dimension of the space of $n$-linear functions $D:F^n\rightarrow F$ is 1.

This can also be seen by noting that, for any $F$-vector spaces $V$ and $W$, the space of $n$-linear functions $V^n\rightarrow W$ is naturally isomorphic to the space of linear functions $\underbrace{V\otimes\cdots\otimes V}_{n\text{ times}}\rightarrow W$ (this is by the universal property of tensor products). In our case, we have $V=W=F$. But for any vector space $V$, we also have $$F\otimes V\,\cong\!\!\! V$$ (for the same reason: a bilinear map $f:F\times V\rightarrow V$ satisfies $f(a,v)=af(1,v)$ for all $a\in F$, $v\in V$, and therefore is determined by its action on $V$). Therefore, $$\underbrace{F\otimes\cdots\otimes F}_{n\text{ times}}\,\cong\!\!\! F$$ and therefore the question becomes: what is the dimension of the space of linear maps $F\rightarrow F$, which is clearly 1.

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    $\begingroup$ Well, "bijection" is not what you want: "naturally isomorphic" brings out that you want the correspondence to be linear. $\endgroup$ – Qiaochu Yuan Jul 22 '11 at 14:36
  • $\begingroup$ Excellent point (if $F$ is infinite, a bijection would not have to preserve dimension). I've edited. $\endgroup$ – Zev Chonoles Jul 22 '11 at 14:44

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