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The following proposition is easy to verify:

Given $n$ functions $f_1,...,f_n$, they are linearly independent in an interval $I$ if there exists a set of $n$ points in $I$, namely $x_1,...,x_n$, such that the matrix \begin{pmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\ f_1(x_n) & \cdots & f_n(x_n) \end{pmatrix}

has full rank.

My question is: can I replace the if by if and only if? That is, if $n$ functions are linearly independent, is it true that there exists a set of $n$ points such that the above matrix has full rank? I am pretty convinced so, but I am not being able to prove it.

learner
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1 Answers1

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Let's try to prove this by contrapositive. Define $$ M(x_1,\dots,x_n) = \begin{pmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\ f_1(x_n) & \cdots & f_n(x_n) \end{pmatrix} $$ We have supposed, then, that for every collection $x_1,\dots,x_n$, the matrix $M$ fails to have full rank. That is, there exists a function $g:\Bbb R^n \to \Bbb R^n \setminus \{0\}$ such that $$ M(x_1,\dots,x_n)g(x_1,\dots,x_n) = 0 $$ We wish to conclude that $g$ may be taken to be a constant function. Suppose for contradiction that it is impossible to do so. That is, there exists a collection $\mathbf x^1,\dots,\mathbf x^m \in \Bbb R^n$ such that $\bigcap_{i=1}^m \ker M(\mathbf x^i) = \{0\}$. Equivalently, the block matrix $$ \tilde M = \pmatrix{M(\mathbf x^1) \\ \vdots \\ M(\mathbf x^n)} $$ has full rank. This means, however, that an $n \times n$ submatrix of $\tilde M$ has full rank. However, this submatrix is necessarily of the form $M(\mathbf x^{i_1}_{k_{i_1}},\dots,\mathbf x^{i_n}_{k_{i_n}})$. Thus, we have contradicted the supposition that $M$ never has full rank.

Thus, there exists a constant $g = (c_1,\dots,c_n)$ such that $Mg = 0$. In other words, $$ \sum_{i=1}^n c_i f_i(x) = 0 $$ for any $x \in \Bbb R$. That is, the functions $f_i$ are linearly dependent.

Ben Grossmann
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  • A subtle step here is the observation that it suffices to consider finite collections $\mathbf x^i$. In order for this to work, it is important that we are considering the intersection of subspaces of a finite dimensional vector space. I'd be happy to fill the details in there, assuming there are no other aspects of my proof that require explanation. – Ben Grossmann Feb 17 '17 at 17:32
  • I am still trying to understand your proof, actually. I do not understand why $g$ has to be constant for the matrix $M$ to be singular. I would say that it suffices that $g$ is not the zero function... Am I wrong? – learner Feb 17 '17 at 17:41
  • I'm saying that once we know $g$ may be constant, we can conclude that the $f_i$ are library dependent, as is done at the end of the proof – Ben Grossmann Feb 17 '17 at 17:49
  • Yes, I do agree that if $g$ is non-zero and constant, then the $f_i$ are linearly dependent, contradicting the initial hypothesis. But how does that prove that there exists $x_1,...,x_n$ such that $M$ is non-singular? I mean, suppose there exists a function $g(x_1,...,x_n)$ that is zero nowhere and such that $Mg=0$ for all $x_1,...,x_n$. Then, $M$ is singular for every possible $x_1,...,x_n$. How do you prove that this implies that the $f_i$ are linearly dependent? – learner Feb 17 '17 at 17:59
  • @D... I have a paragraph that starts "suppose for contradiction that it is impossible (to make $g$ constant)". Try to see where I go from there – Ben Grossmann Feb 17 '17 at 18:09
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    Ok, I think I got it. Please correct me if I am wrong: 1- you argue that if $M(x_1,...,x_n)$ is singular for any $x_1,...,x_n$ then there exists a non-zero function $g(x_1,...,x_n)$ such that $Mg=0$ for all $x_1,...,x_n$. 2- you prove by contradiction that such $g$ would have to be a constant function. 3- you argue that, if $g$ is constant, then $f_1,...,f_n$ are linearly dependent, contradicting the initial hypothesis.

    Is that right?

     – learner Feb 17 '17 at 18:50
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    Correction: I prove that $g$ necessarily can be a constant function. Other than that, sounds right. – Ben Grossmann Feb 17 '17 at 20:31
  • See also this post – Ben Grossmann Dec 02 '20 at 17:36
  • matrix rank, if finite, works in the same way for infinite matrices ( that is, the equality of row rank, column rank, and largest non-zero minor). – orangeskid Mar 28 '24 at 18:27