Here is a simple problem that has stumped me for some time; sharing with the community, as I suspect it has been solved somewhere, or is immediately implied by the correct theorem.

Let $\textbf{diag}: \mathbb{R}^n_+ \to \mathbb{R}^{n \times n}$ represent the operation of taking a vector $v$ and constructing a matrix with $v$ on the diagonal. We'll assume we only operate on vectors with positive elements (this is the meaning of the notation $\mathbb{R}^n_+$), so that the resulting matrix is in particular positive definite.

Let $\textbf{diagpart}: \mathbb{R}^{n \times n} \to \mathbb{R}^n$ represent the operation of extracting the diagonal elements of a matrix as a vector. Let $A$ be some symmetric positive-definite matrix--from the application which serves as the source of this problem, we may take $A$ to have all elements at least 1 if desired, but I think this is likely unnecessary.

Define $ \phi: \mathbb{R}^n_+ \to \mathbb{R}^n_+$ by

$$ \phi(v) = \textbf{diagpart}\left(\sqrt{\textbf{diag}(v)^{1/2} A \textbf{diag}(v)^{1/2}}\right) $$

Conjecture: $\phi$ has a unique fixed point on $\mathbb{R}^n_+$, and iterates of $\phi$ converge to this fixed point from any initial vector $v_0$ (having all positive elements of course).

Numerically, this convergence seems to hold independently of both initial point and $A$, and be extremely rapid--I'll post some code presently, but I suspect this audience may not be particularly interested in empirics.

Is a result like this (or implying this) known? I'll put a little commentary with sketches of solution attempts below.

Commentary and attempted solutions:

First, notice that in the one-dimensional case this essentially reduces to iterating the map $x \to \alpha\sqrt{x}$ for $\alpha > 0$. In fact, if $A$ is diagonal and positive definite, the iterations above reduce to iterating such a mapping elementwise. These parallel iterations can be shown to converge to their unique fixed point by a number of techniques, each of those I've tried becoming somewhat problematic in higher dimensions:

1. Considering $\widetilde{\phi} = \gamma \phi$ restricted away from 0 and using the Banach fixed-point theorem. In 1d, this comes down to balancing the values of the derivative against the values of the function itself (to ensure that $\widetilde{\phi}$ maps the appropriate restricted space to itself). This balancing can be done, but relies heavily the 2 in the denominator of $\frac{d}{dx} \sqrt{x}$. I've taken this to mean any matrix-calculus approach going through this strategy must be extremely precise--losing even a factor of 2 in the analysis would ruin the approach. It may be possible to pursue this approach to the end and prove the conjecture, though the calculations are a little hairy.

2. Using monotonicity. One of the simplest ways of showing convergence in 1D is just noting the monotonicity of the mapping--go up if you're less than 1 and down if you're bigger. In conjunction with boundedness, this monotonicity at least gives us convergence. However, I have yet to determine a basis in which the action described by $\phi$ is monotonic in any particular sense (besides seeming to always make progress towards the fixed point).

3. Using concavity. Given existence of some high-powered convergence results for iterates of concave mappings, and looking at $\phi$, might lead one to believe that surely this mapping must be concave. Alas, this does not seem to be the case in any natural sense (concave as a mapping to the matrix Loewner order once after ditching the $\textbf{diagpart}$ in $\phi$, or elementwise concave as a mapping from vectors to vectors). This conclusion is numerical (wanted to verify the concavity numerically before attempting to prove), and though I have no reason to be suspicious of it, it still strikes me as quite strange and possibly misleading.

None of these three approaches have yielded me any fruit. I suspect Brouwer's fixed point theorem may be used to establish the existence of a fixed point, in conjunction with some direct analysis to identify a compact set which is fixed by the mapping $\phi$, but does not immediately imply uniqueness or convergence of the iterates. It is also worth noting that the matrix square root does indeed converge to the identity when iterated--but given the interplay of the $\textbf{diagpart}$ and the square root above, the proofs I have come up with of this statement (e.g. via the logarithm) fail to be immediately applicable.

Any pointers would be appreciated!!

Edit: partial answer

This fixed-point formulation came from an optimization problem, to which we can show existence of a unique solution. Lemma 3.4 here provides the explicit mapping, and Corollary 3.2 shows that $\phi$ does indeed have a unique fixed point.

However, we have still not shown that iterating $\phi$ converges to this fixed point. An explicit formula for the fixed point of $\phi$ in terms of $A$ would be ideal, but may not be possible. There is a little discussion in the 'future work' section of some partial progress on fixed-points of $\phi$ that might be interesting. my co-authors and I would definitely still be interested in any fruitful directions.