2
$\begingroup$

For example, if my big space is $P_3(\mathbb{R})$, what are all the possible subspace of this vector space?

Obviously $P_3(\mathbb{R})$ itself, $P_2(\mathbb{R})$, $P_1(\mathbb{R})$ and $\{0\}$ are subspaces, but then what if I choose something random like:

$\{x^3-2abx + abc : a, b, c \in \mathbb{R}\}$

or

$\{ax^3-2b-c+(2a-b) : a, b, c \in \mathbb {R}\}$

They might or might not be subspaces, so we check. But can we compile a list of all possible forms a subspace might take?

$\endgroup$
4
  • $\begingroup$ Related: math.stackexchange.com/questions/2229306/… $\endgroup$ Apr 12 '17 at 23:00
  • $\begingroup$ The key word you want is Grassmanian en.wikipedia.org/wiki/Grassmannian $\endgroup$
    – Lee Mosher
    Apr 12 '17 at 23:01
  • $\begingroup$ Is it just the fact that the vectors are polynomials that is throwing you off; do you have an easier time wrapping your head around, say, all subspaces of $\Bbb R^4$? $\endgroup$
    – pjs36
    Apr 13 '17 at 3:05
  • 1
    $\begingroup$ I suppose it was, but I realized they're the same. That is, $P_3(\mathbb{R})$ is isomorphic to $\mathbb{R}^4$. $\endgroup$ Apr 13 '17 at 3:07
0
$\begingroup$

For finite dimensional vector spaces, the answer is yes and can be done via the following theorem

Theorem: Let $F$ be a field and let $k$ be a positive integer. Then every $k$-dimensional vector space over $F$ is isomorphic to $F^k$.

It's not particularly hard to prove that $F^k$ has a subspace of dimension $F^\ell$ for all $0\leq\ell \leq k$ by looking at coordinate projections onto a basis. Thus we know that the subspaces of $F^k$ are precisely $\{F^\ell:0\leq \ell\leq k\}$. Counting the number of each subspace is also possible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.