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Let's $\mathbb{F}_{p}^{5}$ is $5-$dimension space over $\mathbb{F}_{p}$, where $p$ is prime. How many ways can be decomposed the space $\mathbb{F}_{p}^{5}$ into a direct sum of two subspaces of dimension $2$ and $3$, i.e. present as $$\mathbb{F}_{p}^{5}=V_1\oplus V_2,$$ $$\dim V_1 = 2,~~~\dim V_2 = 3.$$ Thanks.

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Dear Aspirin: One line of attack would be this one. The group $G_5:=GL(5,\mathbb F_p)$ acts transitively on the decompositions you're interested in. Thus it suffices (?) to compute the order of $G_5$ and of the stabilizer $H$ of a given decomposition. I don't know if it's a good idea... Note that $H$ is $G_2\times G_3$. (I hope my notation is clear.) –  Pierre-Yves Gaillard Feb 9 '12 at 9:44
    
@aspirin are you preparing for the exam on sunday? –  Sergey Filkin Feb 9 '12 at 10:51
    
@Sergey Filkin in Independent University?=) –  Aspirin Feb 9 '12 at 11:20
    
@Aspirin yep... –  Sergey Filkin Feb 9 '12 at 11:25
    
@aspirin give me your contacts please. Maybe we would be helpful to each other. My email is in the profile. –  Sergey Filkin Feb 9 '12 at 13:11

2 Answers 2

up vote 4 down vote accepted

Let's do it in the, perhaps, most straightforward way. Any basis $e_1,\dots,e_5$ gives a desired decomposition: $V=\langle e_1,e_2\rangle\oplus\langle e_3,e_4,e_5\rangle$ -- but each decomposition can be obtained in multiple ways (which correspond to different choises of bases for $V_1$ and $V_2$). Namely, if $N_k$ is the number of bases in $k$-dimensional vector space, the answer is $$ \frac{N_5}{N_2\cdot N_3}= \frac{(q^5-1)(q^5-q)(q^5-q^2)(q^5-q^3)(q^5-q^4)}{(q^2-1)(q^2-q)\cdot (q^3-1)(q^3-q)(q^3-q^2)}= q^6\frac{(q^5-1)(q^4-1)}{(q^2-1)(q-1)} $$ (cf. Pierre-Yves Gaillard's comment; oh, and $q=p$, if you will).

So the answer is almost a q-binomial coefficient, but not quite: it's $q^6\binom 52_q$. This similarity can also be explained. Namely, consider a decomposition $V=V_1\oplus V_2$. A subspace $V_2'$ also gives a decomposition $V_1\oplus V_2'$ iff $V_2'\subset V_1\oplus V_2$ is transversal to $V_1$ -- i.e. $V_2'$ is a graph of some linear map $V_2\to V_1$. So for each 2-dimensional subspace there are exactly $|\operatorname{Mat}_{2\times 3}(F_q)|$ ways to complement it to a decomposition -- hence the answer $q^6\binom 52_q$.

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What I've mean: Lat's $N_1(n, k)$ is the quantity of ordered sets consist of $k$ linear independent vectors (in $\mathbb{F}_{p}^{n}$). And let's $N_2(n, k)$ is the quantity of k-dimension sub-spaces of $\mathbb{F}_{p}^{n}$. It's obvious that $$N_1(n, k) = (p^{n}-p^{0})(p^{n}-p^{1})(p^{n}-p^{2})\ldots(p^{n}-p^{k-1})$$ and $$N_2(n, k)=\frac{N_1(n, k)}{N_1(k, k)}.$$ Because $N_1(k, k)$ is a quantity of different basis of $\mathbb{F}_{p}^{k}$. As I see $$N = N_2(5, 3)\frac{(p^{5}-p^{3})(p^{5}-p^{4})}{N_1(2,2)}.$$

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Can you explain how did you arrive at N? –  user21436 Feb 9 '12 at 9:21
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@KannappanSampath, why do you think that would be a problem? Given any 2-dimensional subspace surely there are several choices for the 3-dimensional complementary subspace? –  Jyrki Lahtonen Feb 9 '12 at 10:50

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