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Say that the rank of a set of functions $\{f_1,\ldots,f_n\}$ is the size of the largest linearly independent subset.

If you have two sets of linearly independent functions $\{f_1,\ldots,f_n\}$ and $\{g_1,\ldots,g_k\}$, is it possible for the set of all pairwise products $\{f_i\cdot g_j \}$ to have rank less than $n$?

So far, I have attempted to make use of two characteristics of linear independence, namely:

  1. $f_1,\ldots,f_n$ are linearly independent if and only if $(\exists c_1,\ldots,c_n : \sum_i c_i f_i = 0 \Longrightarrow c_1 = \ldots = c_n=0)$
  2. $f_1,\ldots,f_n$ are linearly independent if and only if $\exists x_1,\ldots,x_n$: $\det([f_i(x_j)]) \neq 0$.

I am particularly interested in the case where the $g$ are everywhere positive. I think it might help because you can divide a set of not-all-zero constants $c_1,\ldots,c_n$ by the value of an everywhere nonzero function, and the result will still be a set of not-all-zero constants.

user326210
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    If even one of the $g_i$s is nonzero everywhere -- suppose it is $g_1$ -- then $f_1g_1, f_2g_1, \ldots, f_ng_1$ are already linearly independent before you start considering the other $g$s. – Troposphere Jun 28 '21 at 22:09

1 Answers1

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I assume that we are considering functions $f_1,\dots,f_n,g_1,\dots,g_k:\Bbb R \to \Bbb R$. If you have some other idea of the domain/codomain of these functions, please clarify this in your original question.

If $\{f_1,\dots,f_n\}$ and $\{g_1,\dots,g_k\}$ are linearly dependent and if there exists a function $g_m$ for some $1 \leq m \leq k$ that is positive for all inputs, then the set $\{f_i \cdot g_j : 1 \leq i \leq n, 1 \leq j \leq k\}$ must have rank at least $n$. Indeed: if $\{f_1,\dots,f_n\}$ is linearly independent, then so is $\{f_1 \cdot g_m, \dots, f_n \cdot g_m\}$.

In general, the set of pairwise products could have rank $0$. For instance, consider the functions given by $$ f_i(x) = \begin{cases} x^i & 0 \leq x \leq 1\\ 0 & \text{otherwise}, \end{cases} \quad g_j(x) = \begin{cases} x^j & 2 \leq x \leq 3\\ 0 & \text{otherwise.} \end{cases} $$ Although each set is linearly independent, the set of pairwise products has rank $0$.

Ben Grossmann
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  • Thanks! Don't know if my clarification qs have a quick answer---if not, I can write a separate q: Does this mean you can in general, reduce the rank to any level between 0 and n? In your ex., the functions were clipped outside an interval --- could you still lower the rank if the functions were smooth? – user326210 Jun 28 '21 at 23:41
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    Yes, the idea is essentially the same. Make it so that the $f_i$ are each non-zero only over $[0,2]$ and the $g_j$ are non-zero over $[1,3]$. Make it so that the portion of the $g_j$ restricted to $[1,2]$ is a set with rank $1$ and that the portion of the $f_i$ restricted to $[1,2]$ make a set of the desired rank (between $1$ and $n$). – Ben Grossmann Jun 29 '21 at 01:38