# Solving multiplicative system of equations

I am interesting in figuring out when systems with multiplication instead of addition have solutions and how to find them, for example this system:

$$\left\{\begin{matrix} a \cdot c &=& \alpha_0 \\ a \cdot d &=& \alpha_1 \\ b \cdot c &=& \alpha_2 \\ b \cdot d &=& \alpha_3 \\ \end{matrix}\right.$$

has solutions if(f?) the following criteria is met $$\alpha_0 \alpha_3 = \alpha_1 \alpha_2 \quad \wedge \quad [(\alpha_2 \neq 0 \wedge \alpha_1 \neq 0) \vee (\alpha_0 \neq 0 \wedge \alpha_3 \neq 0)]$$

This system was generated when I attempted to write in factors the following: $$\alpha_0 \vec{x}\otimes\vec{x} + \alpha_1 \vec{x} \otimes \vec{y} + \alpha_2 \vec{y} \otimes \vec{x} + \alpha_3 \vec{y} \otimes \vec{y} = (a \vec{x} + b \vec{y}) \otimes (c \vec{x} + d \vec{y})$$

My attempt ultimately is to try and find when this vector can be written in terms of factors: $$(\alpha_0 \vec{x} \otimes \vec{x} \otimes \cdots \otimes \vec{x}) + (\alpha_1 \vec{x} \otimes \vec{x} \otimes \cdots \otimes \vec{y}) + \cdots + (\alpha_{2^n-1} \vec{y} \otimes \vec{y} \otimes \cdots \otimes \vec{y}) = \bigotimes_{i=0}^{n} (\theta_i\vec{x} + \iota_i\vec{y})$$

What are some resources that I can learn more to solve this problem? Textbooks, papers and/or websites are greatly appreciated.

Thank you.

• you can solve for their absolute values by taking logarithms.
– abel
Feb 20, 2015 at 12:36

You can express your equations as $$\begin{bmatrix}a\\b\end{bmatrix} \begin{bmatrix}c&d\end{bmatrix} = \begin{bmatrix}\alpha_{0}&\alpha_{1}\\\alpha_{2}&\alpha_{3}\end{bmatrix}.$$

A necessary condition is thus that the matrix $$A = \begin{bmatrix}\alpha_{0}&\alpha_{1}\\\alpha_{2}&\alpha_{3}\end{bmatrix}$$ has rank at most one, the case of rank $0$ being trivial.

When $A$ has rank one, a solution is easily found. If fact in this case the linear space $R$ generated by the rows $$\begin{bmatrix}\alpha_{0}&\alpha_{1}\end{bmatrix}, \begin{bmatrix}\alpha_{2}&\alpha_{3}\end{bmatrix}$$ has dimension one. If $$\begin{bmatrix}u&v\end{bmatrix}$$ is a non-zero vector in $R$, there are thus $\lambda, \mu$ such that $$\begin{cases} \begin{bmatrix}\alpha_{0}&\alpha_{1}\end{bmatrix} = \lambda \begin{bmatrix}u&v\end{bmatrix}\\ \begin{bmatrix}\alpha_{2}&\alpha_{3}\end{bmatrix} = \mu \begin{bmatrix}u&v\end{bmatrix} \end{cases}.$$ Thus $$A = \begin{bmatrix}\lambda\\\mu\end{bmatrix} \begin{bmatrix}u&v\end{bmatrix}.$$

For a general answer, you may want to see the accepted answer to this question.

There are 4 equations, but not independent, through the relationship $\alpha_0\alpha_3=\alpha_1\alpha_2$. So, with 4 unknows, the system is undetermined.

For example : let $a$ be any number $\neq 0$ $$c=\frac{\alpha_0}{a}$$ $$d=\frac{\alpha_1}{a}$$ $$b=\frac{\alpha_2}{c}=\frac{\alpha_2}{\alpha_0}a$$ or : $$b=\frac{\alpha_3}{d}=\frac{\alpha_3}{\alpha_1}a$$ which is the same because $\frac{\alpha_2}{\alpha_0}=\frac{\alpha_3}{\alpha_1}$

By the way, in the case of $a, b, b, d$ positive, change to unknows : $A=\ln(a) \; ;\;B=\ln(b) \; ;\;C=\ln(c) \; ;\;D=\ln(d)$. The system is transformed to four linear equations (still not independent).