How to solve for $\boldsymbol{\theta}$ in $[A]^{\theta}x = b$ if $\boldsymbol{[A]}$, $\boldsymbol{x}$, and $\boldsymbol{b}$ are known?

Question

I am trying to find $\boldsymbol{\theta}$ when the output $\boldsymbol{b}$, input $\boldsymbol{x}$, and the matrix $\boldsymbol{[A]}$ are given. Here, $\boldsymbol{[A]}$ is a diagonal square matrix with complex entries and $\boldsymbol{x}$ and $\boldsymbol{b}$ are vectors (dimensionally compatible of course). The matrix $\boldsymbol{[A]}$ is a product of matrices $[a_{1}][a_{2}][a_{3}]....[a_{n}]$: $$[a_{1}][a_{2}][a_{3}]....[a_{n}] = [A]$$

What I noticed is that,

$$[a_{1}]^{\theta}[a_{2}]^{\theta}[a_{3}]^{\theta}....[a_{n}]^{\theta} \ne ([a_{1}][a_{2}][a_{3}]....[a_{n}])^{\theta}$$

when the matrices have complex entries. $\theta$ is always a real number here. In my case, $\boldsymbol{[a_{i_{}}]}$ are always diagonal and consequentially so is $\boldsymbol{[A]}$.

So, to summarize, I want to know how to solve for $\boldsymbol{\theta}$ in the following 2 equations:

1. $$([a_{1}]^{\theta}[a_{2}]^{\theta}[a_{3}]^{\theta}....[a_{n}]^{\theta})x = b$$

And

2. $$([a_{1}][a_{2}][a_{3}]....[a_{n}])^{\theta}x = b$$

Given that all $\boldsymbol{[a_{i}]}$ are square, diagonal, and have complex entries, $\boldsymbol{x}$ and $\boldsymbol{b}$ vectors are also known and have complex entries as well, and $\boldsymbol{\theta}$ is a real number.

And maybe if someone has time after this, I wouldn't mind knowing a general solution where $\boldsymbol{[a_{i}]}$ are not diagonal (out of curiosity). I am also open to numerical solutions.

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I tried the following approach for the second equation (which failed): $$([a_{1}][a_{2}][a_{3}]....[a_{n}])^{\theta}x = [A]^{\theta}x$$ $$[A]^{\theta}x = b$$ So, given that $\boldsymbol{[A]}$ is a diagonal square matrix with complex entries, $$([A]_{i,i})^{\theta}x_{i} = b_{i}$$ Which gives us: $$\theta_{i} = \frac{\log{(\frac{b_{i}}{x_{i}})}}{\log{([A]_{i,i})}}$$ However, I get different $\boldsymbol{\theta_{i}}$ for different values of $\boldsymbol{i}$ whereas there should be only one $\boldsymbol{\theta}$ which solves the equation.

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As some comments state that if all $\boldsymbol{[a_{i}]}$ are diagonal with complex entries then:

$$\prod_{i}{[a_i]^\theta} = \left(\prod_{i}{[a_i]}\right)^\theta$$

But this is in fact not true. Even more simply put, let $\boldsymbol{c_{1}} \in \boldsymbol{\mathbb{C}}$, $\boldsymbol{r_{1}} \in \boldsymbol{\mathbb{R}}$, and $\boldsymbol{n_{1}} \in \boldsymbol{\mathbb{N}}$ then in general,

$$\prod_{i=1}^{i=n_{1}}{c_{1}^{r_{1}}} \neq \left(\prod_{i=1}^{i=n_{1}}{c_{1}}\right)^{r_{1}}$$

I urge that you verify this to convince yourself. Try with: $c_{1} = 0.707 + 0.9i$, $r_{1} = 0.664$, and $n_{1} = 4$.

Answer 1

Actually, even if $[a_i]$ are complex matrices, the identity holds: $$\prod{[a_i]^\theta} = \left(\prod{[a_i]}\right)^\theta$$ But as you have identified, it does not really help much since you do have the matrix $[A]$ at hand. As one of the comments mentions, you want all of the $\theta$ to be equal and only then you have a solution. However, notice that $log$ is not a single valued function if the domain is of complex numbers. This is a consequence of the fact that exponentiation in complex numbers is not injective. In fact, the logarithm is defined as follows for a complex number: $$log(z) = ln|z| + i(arg z)$$ And $arg z$ takes infinite values (adding multiples of $2\pi$).

So, when you take the $log$ of $[A]_{i,i}$ you might want to consider all possible solutions.