3

let E be $Z/pZ$-vector space of dimension n. calculate number of linear maps from $E\to E$ that satisfy $f^2=0$ there is a hint that suggest that to consider the null space and subspace in direct sum with it.

I know that the number subspaces of a given dimension is given a previous post: How to count number of bases and subspaces of a given dimension in a vector space over a finite field? but I don't know where to start, any help please!

Community
  • 1
agre
  • 135
  • 7
  • 1
    $f^2=0$ is equivalent to $Im(f)\subseteq Ker(f)$. So pick a subspace $Ker(f)$, a subspace $Im(f)$ of that $Ker(f)$ and then the only choice left is how your function acts on the complement of the kernel, so it's the same thing as an automorphism of $Im(f)$ (because the complement of the kernel has the same dimension as it). You'll get an ugly sum that you may be able to simplify. – xavierm02 Feb 01 '17 at 21:18
  • Thank you for answering, can you please clarify? – agre Feb 01 '17 at 21:31
  • @xavierm02 Note, however, that the dimensions of the subspaces must sum to $n$. – Ben Grossmann Feb 01 '17 at 21:33
  • @xavierm02 also, I assume you're talking about the orthogonal complement to the kernel (we must choose a distinguished complement for counting purposes). Does the "orthogonal complement" still make sense in finite fields? – Ben Grossmann Feb 01 '17 at 21:36
  • @ Omnomnomnom:is there a simpler way to to solve this problem without using orthogonal complement? – agre Feb 01 '17 at 21:45
  • @Omnomnomnom No because I'm counting functions at this point. And while different complements will make the function look different, each complement yields a bijection between the functions I want to count and automorphisms of the image. – xavierm02 Feb 01 '17 at 21:47
  • @xavierm02 I'm fairly sure that, for a single fixed kernel, one may choose two different complements, but end up defining the same overall $f$ on $E$. Are you accounting for this? – Ben Grossmann Feb 01 '17 at 21:53
  • @Omnomnomnom Yes. But a single complement is sufficient (and necessary) to define all $f$s. – xavierm02 Feb 01 '17 at 21:57
  • @xavierm02 you're right... the existence of a satisfactory choice is sufficient, after all. – Ben Grossmann Feb 01 '17 at 22:05

2 Answers2

2

Let $K:=\operatorname{Ker}(f)$ and $I:=\operatorname{Im}(f)$.

$f^2=0$ is equivalent to $I\subseteq K$.

So you first pick a subspace $K$ of $E$ and then pick a subspace $I$ of $K$. The sets of functions you'll attain for different $K$ and $I$ are clearly disjoint so if you manage to count for each $I$ and $K$, you'll then just have to sum over all $I$ and $K$.

Now, suppose $I$ and $K$ fixed. Take some $H$ so that $E=H\oplus K$. Then it is equivalent to give the action of some $f$ on $E$ or to give it on both $H$ and $K$. Since the action on $K$ is trivial, giving the action of $f$ on $E$ is equivalent to giving its action on $H$, i.e. a function $H\to I$ that has image $I$. Since those functions are surjective and both spaces are or equal dimension, those functions will all be automorphisms. Now take $\varphi$ any (fixed) automorphism between $I$ and $H$. The application $( g : H \to I)\mapsto (\varphi \circ g : H \to H)$ is a bijection. So to count the morphisms when $I$ and $K$ are fixed, you just have to count automorphisms of $I$.

xavierm02
  • 7,495
1

Based on the other answer's explanation, we may compute the total as follows: let $m = \lceil n/2\rceil$. Then in the notation of the post that you linked, the total will be $$ \sum_{j=m}^n \binom nj_p \binom{j}{n-j}_p \prod_{k=0}^{j-1}(p^j - p^k) $$

Community
  • 1
Ben Grossmann
  • 225,327