Unit vectors with same Geometric Mean

Question

How many N-dimensional unit vectors (norm equals do 1) are there such that the geometric mean of its coordinates (restricted to be non-zero) is the same?

Answer 1

Something like this:

$x = [x_1, \ldots,x_N]$ such that $\|x\|_2 = 1$ and for some $i\in\{1,\ldots, N\}$, you have $x_i=0$. Then, the geometric mean of the values in such unit vectors is the same (equal to zero).

HINT: Now that you changed the question, (restricting geometric means to be non-zero): Then, we will have for all $i$'s $x_i\neq 0$.

Let $a = \sqrt[N]{\prod_{i=1}^Nx_i}$, then we have satisfied the conditions if a pair of elements $x_j$ and $x_k$ have the following properties: $x_kx_j = \frac{a^N}{\prod_{i\not\in\{j,k\}}^Nx_i}(=b)$ and $x_j^2+x_k^2 = 1- \sum_{i\not\in\{j,k\}}x_i^2(=c)$

Obviously, we can switch the values of $x_j$ and $x_k$ without changing the value of $b$ and $c$, therefore are $2!*{{N}\choose{2}} = \frac{N!}{(N-2)!}$ of such combinations for each $a\neq 0$.

Now the question is what happens if we consider three elements, or more? (I.e., $x_kx_jx_l = \frac{a^N}{\prod_{i\not\in\{j,k,l\}}^Nx_i}(=b)$ and $x_j^2+x_k^2+x_l^2 = 1- \sum_{i\not\in\{j,k,l\}}x_i^2(=c)$)

With this logic, we can conclude that for every value of $0<a<\frac{1}{N^\frac{N}{2}}$, there are $N!$ solutions. And for $a=\frac{1}{N^\frac{N}{2}}$, there is one solution. (Assuming $x_i \in \mathbb {R} \quad \forall i $)

Answer 2

Consider instead the problem of maximizing the geometric mean $\mu$ of the coordinates of vectors in the open first orthant of $\Bbb R^n$ (i.e., with $x_1, \ldots, x_n > 0$) subject to the constraint $$g(x_1, \ldots, x_n) = x_1^2 + \cdots + x_n^2 = 1 .$$ This is equivalent to maximizing the product $$f(x_1, \ldots, x_n) = \mu^n = x_1 \cdots x_n ,$$ which is a straightforward application of Lagrange multipliers.

Substituting in $\nabla f = \lambda \nabla g$ and comparing the $i$th entries gives $$2 x_i = \lambda x_1 \cdots x_{i - 1} x_{i + 1} \cdots x_n ,$$ and taking the ratios of the $i$th and $j$th entries and cross-multiplying gives $x_i^2 = x_j^2$, and since all components are positive, $x_i = x_j$. As $i, j$ are arbitrary, we get $x_1 = \cdots = x_n$.

It follows from this that $\frac{1}{\sqrt n} (1, \ldots, 1)$ is the only vector with geometric mean $\frac{1}{\sqrt n}$.