Solving linear equations under constraints

Given a matrix equation, \(AX=Y\)

[a11a12a1na21a22a2nam1am2amn]×[x1x2xn]=[y1y2ym] \begin{bmatrix} a_{11}&a_{12}&\dots&a_{1n} \\ a_{21}&a_{22}&\dots&a_{2n} \\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\dots&a_{mn} \\ \end{bmatrix}\times \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n\\ \end{bmatrix}= \begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_m\\ \end{bmatrix}

with known AA and YY, and mnm\ne n, give a general method to solve for XX, such that 0<x1,x2,,xn<10<x_1,x_2,\dots,x_n<1. Given that for the given equation(s) there is at least one solution under the given constraints. For multiple solutions any one will work (or maybe the one with least square after dealing with the constraints).

Additional constraints: Also consider the case of m=nm=n, and that AA can be singular. Also assume, if necessary, m<nm<n and/or m=1m=1 (ie. a single linear equation with multiple variables).

My approach: I already know about pseudoinverse and how to use them, but the problem with them is that they returns values outside the range. Consider with A=[123]A=\begin{bmatrix} 1&2&3 \end{bmatrix} and X=[]X=\begin{bmatrix} 0.8\\ 0.9\\ 0.8 \end{bmatrix}

What would be a good approach?

Note by Sherlock Doyle
3 years, 3 months ago

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It's more difficult than tutoriage. Maybe I could do this few years ago, but now forget about it. So, I also would like to know the answer.

Diane Watson - 3 years, 1 month ago

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