1
$\begingroup$

A sequence of real functions $\{\phi_1,\cdots,\phi_n\}$ is called a Chebyshev system on an interval $I\subseteq\mathbb{R}$, if any real linear combination $\sum_{l=1}^n a_l\phi_l$ has at most $n-1$ distinct zeros on $I$. It is equivalent to say that the square matrix $\Big(\phi_l(t_k)\Big)_{k,l=1}^n$ is invertible where $t_1,\cdots, t_n$ are distinct points of $I$.

As a well-known example, $\phi_l(x)=\sin(lx)$ for $l=1,\ldots,n$ is a Chebyshev system on $I=(0,\pi)$, meaning that $\Big(\sin(lt_k)\Big)_{k,l=1}^n$ is invertible where $t_k$ are distinct points in $I$.

Let us consider the $n\times n$ matrix $\big(\beta_{kl}\big)$ where the entries $\beta_{kl}$ are all either $1$ or $-1$.

Q. Is the matrix $\Big(\beta_{kl}\sin(lt_k)\Big)_{k,l=1}^n$ invertible where $t_k$ are distinct in $I$?

$\endgroup$
2
  • $\begingroup$ $(\sin(lt_k))_{k,l=1}^n$ is not invertible if (say) $t_1=0$. $\endgroup$ Commented Oct 22, 2023 at 0:07
  • $\begingroup$ Yes, you are right. It needs a minor change. $\endgroup$
    – ABB
    Commented Oct 22, 2023 at 8:58

1 Answer 1

1
$\begingroup$

$\newcommand\be\beta$The answer is no.

For instance, suppose that $n=2$, $\be_{11}=\be_{12}=\be_{21}=1=-\be_{22}$. Then for $$D(t_1,t_2):=\det\big(\beta_{kl}\sin(lt_k)\big)_{k,l=1}^n$$ we have $D(1,2)=-0.189\ldots<0<0.106\ldots=D(1,3)$, so that $D(t_1,t_2)=0$ for $t_1=1$ and some $t_2\in(2,3)$. $\quad\Box$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .