The whole "hack" of your solution is that you say "use <Belarusian name> Theorem", which I don't know/remember (and some "<Belarusian name> circles"). What you do in fact in my view is constructing a matrix [$]M[$] with such elements [$]m_{ij} > 0[$] that [$]\sum_{j\neq i} m_{ij} < m_{ii}[$], which implies that its eigenvalues are non-zero, and hence [$]M[$] is invertible. This can be shown as follows: for any [$]\vec{x}[$] s.t. [$]||\vec{x}|| > 0[$],
[$]M \vec x = \lambda \vec x = \sum_{j} m_{ij} x_j [$]
[$](\lambda-m_{ii}) x_i = \sum_{j\neq i} m_{ij} x_j [$]
[$]|\lambda-m_{ii}| |x_i| = |\sum_{j\neq i} m_{ij} x_j| [$]
[$]|\lambda-m_{ii}| |x_i| \leq \sum_{j\neq i} m_{ij} |x_j|[$](from the triangle inequality[$]|\sum_{j\neq i} m_{ij} x_j| \leq \sum_{j\neq i} m_{ij} |x_j| ) [$]
Since you assumed [$]\sum_{j\neq i} m_{ij} < m_{ii}[$], then [$]|\lambda-m_{ii}| |x_i| \le m_{ii} |x_i|[$] where [$]i = \text{argmax}_j |x_j|[$].
Dividing by [$]|x_i|[$], which is strictly positive, you obtain [$]|\lambda-m_{ii}| \le m_{ii}[$], which shows that [$]\lambda \neq 0[$].
Am I right that this is your reasoning? It leads to a class of solutions, which are matrices with diagonal elements larger than the sum of the off-diagonal ones.
Alternatively, I proposed (off the top of my sleepy head - I again admit I made silly errors in the first post, bu they don't invalidate the solution) another class of solutions: Hermitian matrices [$]M = \sum_i \lambda_i P_i[$], where [$]\lambda_i[$] is some strictly positive constant and the projection operator [$]P_i = \vec v_i \vec v_i^T[$], where [$]\{\vec v\}[$] is a set of linearly independent normalised vectors with non-negative elements. It's trivial to prove that [$]M[$] are invertible.
For every [$]\vec x[$] such that [$]||\vec x||>0[$] we have [$]\vec x^T M \vec x = \sum_i \lambda_i \vec x^T \vec v_i \vec v_i^T \vec x = \sum_i \lambda_i (\vec x^T \vec v_i)^2 > 0[$] (strict inequality comes from the assumption that [$]\{\vec v_i\}[$] are linearly independent, i.e. there is such [$]i[$] that [$]\vec x^T \vec v_i \neq 0[$]). At the same time [$]M= \sum_i \lambda_i' \vec v_i' \vec v_i'^T[$] (because it's symmetric by construction), where [$]\vec v_i'^T \vec v_j' = \delta_{ij}[$], and [$]\vec x^T M \vec x = \sum_i \lambda'_i (\vec x^T \vec v_i)^2[$]. I put [$]\vec x=\vec v'_i[$] and I see that [$]\lambda'_i > 0[$] for every [$]i[$]. Thus [$]M[$] is invertible ([$]M^{-1} = \sum_i \lambda_i ^{-1} \vec v_i' \vec v_i'^T[$]).
Every mathematical construction is what you call "circular reasoning" to a certain degree, depending on a number of steps one considers obvious - some people need to write out the vector multiplication, while others just approvingly nod at the Perron-Froebenius theorem (I forgot this one too).
Finally, someone is talking about mathematics, instead of muh feelings. Some people are in need of a safe space. If you are indeed a woman as I boldly surmised, I am in for a surprise and disappointment, since you are much more rational than some of the men here.
Yes, that is my hack and my rationale. The derivation is an extremely simple application of the triangular inequality as you have verified. It should not take long to find that. The Balurassian name only serves as a mnemonic to save me from typing out all that LaTeX. I am lazy and guilty of being terse, as Cuchulainn has charged me with. Very simple and clean idea with a complexity of only double the number of elements, isn't it? The complexity is much lower than the M-matrix method with complexity [$]O(n^3)[$] that Cuchulainn first mentioned and you found out later via google. Do you agree?
The invertibility of your matrix construction is not under contention here, provided that you stipulate the number of vectors to be the same as the dimension of the vector (you did not mention how many vectors you have, but that is not important). Now before we go on examining the details of your construction, can we review the problem requirement? That is:
"Construct an invertible positive n-by-n matrix where positive means all matrix entries are positive."
Hint: please note the strict positivity of the entry in the requirement. The entries of your [$]v_i[$] are nonnegative only, allowing zero's. Do you need a chance to mend your proposition before we go on?
As for your philosophical rumination, shall we leave it for the after-party?
