Suppose I have distinct real numbers $a\_i \in [-1,1]$, $i \in [k]$. I want to choose real numbers $b\_j, j\in [k]$ such that the matrix $(\arccos(a\_i b\_j))\_{i,j \in [k]}$ is nonsingular. 



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> Is this always possible? Equivalently, does $\arccos$ not satisfy a functional equation of the form $\det(\arccos(a\_i x\_j)) = 0$ for indeterminates $x\_j, j \in [k]$?
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It seems like this should hold for any sufficiently ``non-algebraic'' function in place of $\arccos$. Is there a theory that handles such questions? Does functional analysis deal with these questions? I know very little functional analysis, so some general references may be helpful.



Nonexistence of determinantal functional equation for $\arccos$
A quick counter-example to the question as stated is $a\_0=0$, $a\_1=1$ $a\_2=-1$. Since $2arccos(0)-arccos(b)-arccos(-b)=0$ for all $b$, we have $2M\_1-M\_2-M\_3=0$ where $M\_1, M\_2, M\_3$ are the rows of the matrix.


So it is better to assume that the $a\_i$ are nonnegative. In this case, the answer is yes. More generally, consider this problem for a function $f$. You want to choose $n$ linearly independent vectors from the set
$$
X := \{ (f(a\_1x), f(a\_2x),\dots, f(a\_nx)) \mid x\in\mathbb R \} \subset \mathbb R^n
$$
This is not possible if and only if $X$ lies in an $(n-1)$-dimensional subspace. This means that there are constants $c\_1,\dots,c\_n$ (not all zero) such that $c\_1v\_1+\dots+c\_nv\_n=0$ for all $v\in X$. Or, equivalently,
$$
 c\_1 f(a\_1x) + \dots c\_n f(a\_nx) = 0
$$
for all $x\in\mathbb R$ (such that all $a\_ix$ belong to the domain of $f$). In other words, the functions $x\mapsto f(a\_ix)$ are linearly dependent over $\mathbb R$.


This cannot happen if $f=arccos$. Indeed, assuming that $a\_n$ is the maximum of $a\_i$ such that $c\_i\ne 0$, the above sum is well-defined and analytic on $[0,1/a\_n)$ but its derivative goes to infinity as $x$ approaches $1/a\_n$. Hence it is not constant on $[0,1/a\_n)$, and hence non-constant in any neighborhood of 0.


UPDATE. A similar argument shows that the answer is the same for any non-polynomial function analytic near the origin (assuming $a\_i>0$). Indeed, if $f(x)=\sum\_{j=1}^\infty q\_j x^{k\_j}$ where $q\_j\ne 0$, then Taylor expansion of the identity
$$
 c\_1 f(a\_1x) + \dots c\_n f(a\_nx) = 0
$$
implies that $\sum\_i c\_i a\_i^{k\_j}=0$ for all $j$. This cannot happen because, if $a\_n$ is the maximum of $a\_i$, the term $a\_i^{k\_j}$ grows faster (or decays slower) than all other terms as $k\_j\to\infty$.


