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Let $S$ be a vector space of functions from $\mathbb{R}^n$ to $\mathbb{R}$, say $S := \{ f:\mathbb{R}^n \rightarrow \mathbb{R} \}$.

I am looking for some examples in which the dimension of $S$ is known.

For instance, trivial examples are the following.

Linear functions $f(x) := a^\top x$ implies that $\text{dim}(S) = n$.

Quadratic functions $f(x) := x^\top A x$ implies that $\text{dim}(S) = n^2$, or probably just $n(n+1)/2$ because we can take $A$ symmetric.

What is the dimension of the space of:

  • Sinusoidal functions $f(x) := a \sin( b^\top x + c) $? Is it just $n+2$?

  • Other known less-trivial examples?

Then, if $S_1$ has dimension $d_1$ and $S_2$ has dimension $d_2$, what is the dimension of $S:= \{ f := f_1 \circ f_2 : \ f_1 \in S_1, f_2 \in S_2 \}$?

Also, given a $d$-dimensional vector space $S_0$ of functions $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^m$, and a function $g: \mathbb{R}^m \rightarrow \mathbb{R}$, what is the dimension of the space $S := \{ g \circ f : \ f \in S_0 \}$?

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    $\begingroup$ Is there any reason to think that the set of compositions is a vector space? $\endgroup$ – Gerry Myerson May 4 '13 at 10:06
  • $\begingroup$ Suppose it is, otherwise the question is not well posed and there is no point in answering. $\endgroup$ – user693 May 4 '13 at 10:17
  • $\begingroup$ Absurd to suppose the impossible: if $\,f_1,f_2:\Bbb R^n\to\Bbb R^m\,$ and $\,n\neq m\,$ ,then $\,f_1\circ f_2\,$ cannot be defined... $\endgroup$ – DonAntonio May 4 '13 at 10:29
  • $\begingroup$ Clearly and obviously $f_1: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $f_2: \mathbb{R}^m \rightarrow \mathbb{R}$. Where is it written that both $f_1$ and $f_2$ are from $\mathbb{R}^n$ to $\mathbb{R}^m$??? $\endgroup$ – user693 May 4 '13 at 10:38
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    $\begingroup$ Re your question about sinusoidal functions: try thinking about what a basis would be. (The same advice applies to your question in general.) $\endgroup$ – symplectomorphic May 4 '13 at 10:49
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I may be misinterpreting what you're asking but my first thought after reading your post is that you're looking for a tensor product of vector spaces. In which case the dimension becomes the product of the dimensions.

If you examine the last construction in your post you have S is a new space of functions $\mathbb{R}^n \rightarrow \mathbb{R}$ that simply factors through $\mathbb{R}^m$ However since you're fixing the map $g:\mathbb{R}^m \rightarrow \mathbb{R}$ the basis for $S_0$ and what g does to that basis is what completely determines your new vector space. (I'm assuming you are considering the necessary restrictions on g and the functions of $S_0$ to make the composition a vector space)

Also unrelated to answering the post @Adam I have encountered many discussions like the ones in the comments under your post. There are some mathematicians that will dismiss anything you say unless it fits perfectly into whatever they see as the correct frame of mind. I have encountered this throughout my math career it can be both infuriating and helpful. Try to develop 2 things. 1. try to be as clear as possible in what you're saying, developing your own communication will help. 2. recognize that some people will always be that frustrating in this manner and learn to walk away from those that are just not helpful without letting it get under your skin. What you're doing is a good thing actually and you should keep doing it even though some mathematicians will seem to do their best to try to keep you from doing it, whether that is their intention or not.

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  • $\begingroup$ Thanks for the answer. So if $f$ belongs to a vector space of dimension $d$ and $g$ is scalar valued, besides "regular enough", the dimension of $S := \{ g \circ f \}$ is $d$ as well. Am I right? $\endgroup$ – user693 May 4 '13 at 18:32
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    $\begingroup$ I believe it's a little tricky actually. It's somewhat more straightforward if you define it to be the 1 dimensional space spanned by the basis {g}. That resembles a tensor product construction more. In that sense the dimension is also d. However, if you leave it as g is a scalar not 0 then it should keep the dimension the same. But I believe that as originally written g could actually decrease your dimension depending on what you define it to be. Say you have basis $\{f_1,...,f_d\}$ and $g(f_d) = 0$ then in my interpretation your new space should have dim = d - 1. $\endgroup$ – Q11 May 4 '13 at 18:48
  • $\begingroup$ I see. At least, do you agree that the dimension would be at most $d$? $\endgroup$ – user693 May 5 '13 at 11:20
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    $\begingroup$ Yes, you'd have to add more dimensions to the second space to increase the dimension of your space. $\endgroup$ – Q11 May 5 '13 at 18:21

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