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I have the following quadratic matrix equation in symmetric positive semidefinite covariance matrix $\mathbf{C}$

$$\mathbf{C}\mathbf{S}\mathbf{C}+\mathbf{C}+\mathbf{A}=\mathbf{0}$$

where $\mathbf{S}$ is a singular matrix and $\mathbf{A} = \mathbf{a} \mathbf{a}^{T}$. I need to iteratively find $\mathbf{C}$.

1) If I try to set up the problem using Bernoulli iteration, in order to find the dominant solvent I would have to go for

$$ \mathbf{S}+\{\mathbf{I}+\mathbf{C}_{k}^{-1}\mathbf{A}\}*\mathbf{C}_{k+1}^{-1} = \mathbf{0} $$

$$ \Longrightarrow \mathbf{C}_{k+1}^{-1} = \{\mathbf{I}+\mathbf{C}_{k}^{-1}\mathbf{A}\}^{-1}*(-\mathbf{S}) $$

Which won't make sense since $\mathbf{S}$ is singular

2) If I try to to find the minimal solvent, I set it up as

$$ (\mathbf{C}_{k}\mathbf{S}+\mathbf{I})*\mathbf{C}_{k+1}+\mathbf{A}=\mathbf{0} $$

$$ \Longrightarrow \mathbf{C}_{k+1}=(\mathbf{C}_{k}\mathbf{S}+\mathbf{I})^{-1}*(-\mathbf{A}) $$

But this doesn't seem to be working either.

3) A last, non-Bernoulli iteration I set up was

$$ \mathbf{C}_{k}\mathbf{S}\mathbf{C}_{k}+\mathbf{C}_{k+1}+\mathbf{A}=\mathbf{0} $$

$$ \Longrightarrow \mathbf{C}_{k+1}=-\mathbf{A}-\mathbf{C}_{k}\mathbf{S}\mathbf{C}_{k} $$

Which seems to be working but I'm not sure which root it converges to. Also the final estimate $\mathbf{C}_{k}$ converges to, somehow appears to be twice the actual value (upon performing a ground truth test).

My queries are

  1. Am I on the right track on dealing with this problem? If so, where am I making mistakes?

  2. Is there any procedure to find the dominant and minimal solvent when the coefficient of the square term is singular?

  3. Is the last, non-Bernoulli iteration valid? If so, why is the result twice the actual one? Which root does it converge to?

Any guidance will be highly appreciated. Thank you.

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    $\begingroup$ is $S$ symmetric? $\endgroup$ Commented Dec 7, 2017 at 19:50
  • $\begingroup$ Yes. Apologies for the lack of clarity. $\mathbf{S}$ is symmetric, it is also the product of a column vector with its transpose $\mathbf{S}=s∗s^{T}$ $\endgroup$ Commented Dec 7, 2017 at 20:14
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    $\begingroup$ In that case your equation can't have positive semidefinite solutions, except in the trivial case $a=0$: $a^T(C S C + C + A) a \ge \|a\|^4 > 0$. $\endgroup$ Commented Dec 7, 2017 at 20:17

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