2
$\begingroup$

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$?

$\endgroup$
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
9
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.