Let $V$ be an $n$-dimensional vector space over a finite field $F$. For $ 0\le m \le n$, the number of $m$-dimensional subspaces of $V$ is same as the number of $(n-m)$-dimensional subspaces.

I tried using the following argument:

Corresponding to every $m$ dimensional subspace, we can find a complementary subspace of dimension $(n-m)$. So, $$ \text{number of subspaces of dimension }m\le \text{number of subspaces of dimension }(n-m) $$

Reversing the argument, we get they are equal.

Is this correct?

  • $\begingroup$ No, a subspace can be complementary to several different subspaces, so the map you are thinking to is not necessarily injective. $\endgroup$
    – egreg
    Sep 22, 2015 at 9:45
  • 1
    $\begingroup$ I think that the easiest method to do this exercise is by using the dual space $V^*$. $\endgroup$
    – Crostul
    Sep 22, 2015 at 9:55
  • $\begingroup$ @Crostul, how do I do it using dual spaces? $\endgroup$ Sep 22, 2015 at 9:58

3 Answers 3


HINT: Denote by $\operatorname{Sub}(V)$ the set of subspaces of the vector space $V$. Use the map $$\begin{matrix} \operatorname{Sub}(V) &\longrightarrow &\operatorname{Sub}(V^*)\\ U & \longmapsto & U^{\bot} \end{matrix}$$ Where $U^{\bot} = \{ \phi \in V^* : \phi (U) = \{ 0 \} \}$. Since $V$ and $V^{**}$ are canonically isomorphic, you can consider an analogous map $$\operatorname{Sub}(V^*) \longrightarrow \operatorname{Sub}(V^{**}) \ "=" \ \operatorname{Sub}(V)$$ And note that these two are inverse each other.

To conclude, you should know that $\dim U^{\bot} = n- \dim U$.

  • $\begingroup$ So, you are taking every subspace to its annihilator? But why is the mapping injective? Can't I just directly say that for every subspace of dimension m, there is a unique annihilator of dimension (n-m) and use the argument as earlier? For that to be valid, one subspace can't be annihilator of two subspaces, which is same as showing the mapping is injective, but how will we show that? $\endgroup$ Sep 22, 2015 at 10:33
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    $\begingroup$ Because the annihilator of the annihilator is the subspace itself. $\endgroup$
    – Crostul
    Sep 22, 2015 at 10:41
  • $\begingroup$ Oh, of course. Thanks. $\endgroup$ Sep 22, 2015 at 10:48
  • $\begingroup$ @crostul, this is a problem from "Finite-dimensional vector spaces". Why it is mentioned that $F$ is a finite field? As I understand - your argument works for any field. $\endgroup$
    – Andreo
    May 26, 2019 at 6:02
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    $\begingroup$ @Andreo If the field $F$ is infinite you still have a bijection, however $\mathrm{Sub}(V)$ becomes an infinite set, thus you cannot say how many elements it has. $\endgroup$
    – Crostul
    May 26, 2019 at 6:35

No this is not correct. Not only there is more than one subspace complementary to a given subspace $U$, but choosing one such complementary subspace $W$, it is also complementary to many other subspaces $U'$ of the same dimension as $U$. So one cannot exclude the possibility that the same subspace $W$ is chosen as the complementary subspace many times, and this means one does not get any inequality between their numbers.

What one would need for this kind of argument is a standardised method to make a choice of complementary subspace that would guarantee that for $U'\neq U$ the complementary subspaces chosen will be different. However there is no easy such method.


Problem: How do you choose the complementary subspace? Maybe you choose the same complement for different subspaces. Then your estimate does not work...

Example: $\mathbb F_2^2$, the subspace $\langle (1,1)\rangle$ is complementary to $\langle (0,1)\rangle$ and $\langle (1,0)\rangle$. Why should you use a different one?

Solution: Using a nondegenerate bilinear form $b$, you can find $b$-orthogonal complements. This way your choice of complement becomes unique.

In the example above with the product $(a,b)\cdot(c,d)=ac+bd$, the space $\langle (0,1)\rangle$ is orthogonal to $\langle (1,0)\rangle$ and $\langle (1,1)\rangle$ to itself. Thus there are as many 1-dimensional subspaces as (2-1)-dimensional subspaces. (ok, the statement is stupid but the correspondence works)

  • $\begingroup$ Your last paragraph is not clear to me. It is ok that with "non-degenerate" bilinear form, we can give a bijection between m dimensional and n−m dimensional subspaces. But, does this happens for dot product over any field? $\endgroup$
    – Groups
    Sep 22, 2015 at 10:55
  • $\begingroup$ By "dot" I denoted just a choice of the product (it appears standard, but this depends on your choice of basis.). $\endgroup$ Sep 22, 2015 at 11:17
  • $\begingroup$ Better look at the answer of crostul. In his context my choice of bilinear form corresponds to an isomorphism of $U$ and $U^*$, sending $v$ to $b(v,\dot)$. $\endgroup$ Sep 22, 2015 at 11:26
  • $\begingroup$ The term "orthogonal complement" is confusing in this context, since as the answer shows this "complement" need not be complementary to the original subspace (and could even be equal to it). $\endgroup$ Sep 22, 2015 at 11:41
  • $\begingroup$ True. But nonetheless the terminology gets used (At least this is the case in symplectic geometry). Maybe when using the orthogonal with respect to $b$, one should call it the $b$-orthogonal complement. $\endgroup$ Sep 22, 2015 at 13:00

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