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If we have vector space $V$ with dimension $n$ then how many subspaces of $V$ with dimension $m<n$ are there?

In my opinion the answer should be the number of ways to choose $m$ linearly independent vectors out of $n$ linearly independent vectors, regardless of the order of picking. This number is the binomial coefficient $\binom{n}{m}$?

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Consider $m=1$ and $n=2$. Note that each subspace of dimension $1$ is a line through the origin in the plane. It is obvious that there are infinitely many such lines, and therefore infinitely many subspaces of dimension $1$.

In a more general setting, if $U \subset V$ is a subspace of $V$, $E$ is a basis of $U$ and $F$ is a basis of $V$, it is not necessarily true that $E \subset F$. Therefore it is impossible to generate all subspaces of $V$ with dimension $m$ by starting with a particular basis for $V$ and pick $m$ vectors in this basis and span the $m$ vectors.

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If the vector space $V$ has finte dimension $n$ over a field of characteristic $0$, then the answer is infinte. But if the vector space has dimension $n$ over a finite field $\mathbb{F}_q$, then the subspaces of $V$ of dimension $m$, with $m<n$, are precisely:

$$\frac{(q^n-1)\cdots(q^n-q^{m-1})}{|GL_m(\mathbb{F}_q)|} $$

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In the (2,1) case over the reals, the question is "how many lines in the plane pass through the origin?".

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  • $\begingroup$ Infinitely many! $\endgroup$ – Pekov Jul 23 '16 at 9:03

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