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Let $V$ be a 3-dimentional vector space over the field $F_3=\Bbb Z/3 \Bbb Z$ of $3$ elements.the number of distinct 1 dimentional subspaces of $V$ is

  1. $13$
  2. $26$
  3. $9$
  4. $15$
kinkar
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1 Answers1

9

Hint:

If $V_n(q)$ be a vector space of dimension $n$ on a field $F=GF(q)$, the number of all one-dimensional subspace of $V$ will be: $$\frac{q^n-1}{q-1}$$ In fact, all non zero scalar products of $\langle e\rangle\subset V$ generate the same subspaces of dimension one.

Mikasa
  • 67,374
  • where can I get the proof of this result? – kinkar May 17 '13 at 05:43
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    @kinkar: Let $\langle e\rangle\subset V$. So it has a form like $\langle e\rangle={\alpha e\mid \alpha\in F}$. We have a total number of $q^n-1$ of such these subspaces. But all scalar products of $e$ make similar subspaces. So, we should eliminate them by that dividing. – Mikasa May 17 '13 at 05:51
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    $+\left(\sin^2 x + \cos^2 x\right);;\ddot\smile$ – amWhy May 18 '13 at 01:21
  • @kinkar Check it out here –  May 19 '13 at 10:25