1

Let $V_{n}(F)$ be a vector space over the field $F=\mathbb Z_{p}$ with $\dim( V_{n}) = n$, i.e., $ \lvert V_{n}(\mathbb Z_{p}) \rvert = p^{n}$. Assume $ b \in {V_{n}} $ and Hamming weight $b$ is $w$, i.e., $w_H(b)={w}$. If $S \prec V $ and $ \dim(S)=k$ and $\{b\mid w_{H}(b)=w \} $. How can I count number of subspaces $S$ such that $S\cap \{b\mid w_{H}(b)=w \}\neq\varnothing$?

In other words, how can I count the number of $k$-dimensional subspaces of $V_n(F)$ that contain a vector of Hamming weight $w$?

egreg
  • 238,574
Amir
  • 445
  • I don't understand what the edit changed (other than replaced $b$ with $a$). As far as I can tell $a$ is some fixed vector of a known weight, and you still just want to count the number of subspaces of a given dimension that contain $a$. After all, $a$ is in that intersection, if and only if $a\in S$. Please clarify. – Jyrki Lahtonen Aug 06 '16 at 19:04
  • @JyrkiLahtonen No, $a$ is not a fixed vector . I want to count all sub-spaces that ${b \ \vert\ w_{H}=w }\cap {S}\neq \varnothing$ – Amir Aug 06 '16 at 19:45
  • Ahh! Ok.${}{}{}$ – Jyrki Lahtonen Aug 06 '16 at 19:52
  • 2
    I think it's clearer to simply say "I want to count the number of $k$-dimensional subspaces of $V_n(F)$ that contain a vector of Hamming weight $w$". – Greg Martin Aug 07 '16 at 07:33
  • @GregMartin thanks a lot! – Amir Aug 07 '16 at 07:44

0 Answers0