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Can you please help me solve the following nonlinear equation to determine the value of the vector $z$ : $$ \boldsymbol{a}=\boldsymbol{z}^{2} \odot \boldsymbol{K}*\boldsymbol{z}^{-1}$$ Where:

  • $\boldsymbol{a}$ is a given vector with all positive entries.
  • $K$ is an element-wise positive, symmetric, and positive semi-definite matrix.
  • $\odot$ represents the element-wise product.
  • the exponents represent elementwise exponentiation
  • $*$ represents matrix multiplication.

Is there an approach or method to solve this equation?

The equation above can be written as follows for each element of vectors $\boldsymbol{a}$, $\boldsymbol{z}$, and matrix $\boldsymbol{K}$ $a_i=\sum_j z_i^2 K_{i j} z_j^{-1}$

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    $\begingroup$ When you ask, "Is there another approach...", that indicates that you already know (at least) one approach. Please tell us what approach(es) you already know, so we don't waste time and effort telling you things you already know. $\endgroup$
    – Gerry Myerson
    Commented Sep 19, 2023 at 2:25
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    $\begingroup$ I know some folks avoid indices at all costs, but this often makes it more difficult to understand what is intended. Do you mean: $a_i=\sum_{j,k} z_iz_jK_{jk}z_k^{-1}$? If not, please clarify. $\endgroup$
    – Michael Renardy
    Commented Sep 19, 2023 at 2:57
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    $\begingroup$ @JohnOmielan, I removed the extra parameters. $\endgroup$
    – Iman Nodozi
    Commented Sep 19, 2023 at 3:19
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    $\begingroup$ @ImanNodozi I suggest editing the question to use the equation in my comment and dispense with the (to me) novel notation for the operations. Everyone here would be comfortable with index notation. $\endgroup$
    – David Roberts ♦
    Commented Sep 19, 2023 at 6:09
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    $\begingroup$ And one may also absorb the coefficients $a_i$, letting $y_i:=a_i^{1/3}x_i$ and $K'_{ij}:=a_i^{-1/3}K_{ij}a_j^{-1/3}$, still a symmetric semi-definite matrix with non-negative coefficients. The equation then writes $y_i^2=\sum_j K'_{ij}y_j$. $\endgroup$
    – Pietro Majer
    Commented Sep 19, 2023 at 20:22

1 Answer 1

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Here is an existence argument on the lines of the proof of the Perron-Frobenius theorem via the Brouwer fixed point theorem. Note that from the equation, since by assumption $a_i>0$ and $K_{ji}\ge0$, it follows that a necessary condition for existence of a solution $z\in\mathbb R^n$ with $z_i>0$ is (1): $\sum_jK_{ij}>0$ for all $i$ .

To show this is also sufficient, note that may also assume w.l.o.g that (2): $K$ is not a direct sum of smaller blocks (up to reordering of the indices). Otherwise, the system would break in a direct sum of smaller systems of the same kind. Equivalently: some power of $K$ has all coefficients non-zero.

Let $\Delta:=\{u\in\mathbb R^n:u_i\ge0,\sum_j u_j=1\}$. For any $u\in\Delta$ put $v_i:=a_i^{-1}\sum_jK_{ij}u_j^{1/2}$. Note that for any $u\in\Delta$ one has $u_j>0$ for least one index $j$, so that it follows by (1) that $\lambda:=\sum_iv_i\ge (\max_i a_i)^{-1}(\sum_i K_{ij_{0}})u_{j_{0}}^{1/2}>0$ (recall $K_{ij}=K_{ji}$). Consider the map $\Delta\ni u\mapsto \frac{v}{\lambda}\in\Delta.$ By the Brouwer fixed point theorem this map has a fixed point $u\in\Delta$, thus verifying for some $\lambda>0$ $$\lambda u_i=a_i^{-1}\sum_jK_{ij}u_j^{1/2}.$$ By assumption (2) this implies that $u_j>0$ for all $j$. Therefore $z_i:= \lambda^{-2} u_i^{-1}>0$ solves $$ a_i=z_i^2\sum_j K_{ij}z_j^{-1}.$$

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