Let $A_1,...,A_n,B$ be vectors in the $n$-dimensional Euclidean Space, such that they are never on the same affine $(n-1)$-dimensional subspace. (What? Is that a way to say they span $\Bbb{R}^n$?).
Show that for $u_1,...,u_n\in \Bbb{R}$ close enough to $0$, there exists a vector $C\in\Bbb{R}^n$ such that $|C-A_1|=|B-A_1|+u_1,\\ . \\. \\.\\|C-A_n|=|B-A_n|+u_n$.
This is quite confusing. It is Analysis and I have gone through the notes have barely seen something like that. Could it be about a fixed point? I could use a hint.
Use the inverse function theorem.
For sanity write $V_i := B - A_i$ and $D := C - B$.
The $V_i$ are independent iff the $A_i$ areEdit: Nope! :(, and the conditions become$|V_i + D| = |V_i| + u_i$
or
$\sum_{j=1}^{n} (v_{ij} + d_j)^2 = (|V_i| + u_i)^2$
Define a function from $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ by
$f(\langle d_1,\ldots,d_n\rangle) = \langle \sum_{j=1}^{n} (v_{1j} + d_j)^2, \ldots, \sum_{j=1}^{n} (v_{nj} + d_j)^2\rangle$
The jacobian matrix of partials of $f$ at $D = 0$ is the matrix of coefficients of the $V_i$ multiplied by 2, so the determinant is non-zero by linear independence and we may use the inverse function theorem. There is an open neighborhood $U$ of $f(0) = \langle|V_1|^2, \ldots, |V_n|^2\rangle$ and a function $g:U \rightarrow \mathbb{R}^n$ inverting $f$. For $u_i$ sufficiently small, $\langle(|V_1|+u_1)^2, \ldots, (|V_n|+u_n)^2\rangle$ is in $U$ and maps under g to a vector $D$ satisfying the conditions.