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I am trying to estimate precise values of three 2-D vectors, $x_1$, $x_2$, and $x_3$. I have made multiple, imprecise measurements of their sums. For example:

$2x_3 \approx [ 1.157, -73.111]$

$x_2 + x_3 \approx [ 25.184, -55.829]$

$x_2 + 2x_3 \approx [ 26.407, -86.504]$

$2x_1 + x_2 + 3x_3 \approx [ 76.085, -96.201]$

(many, many similar equations)...

Each measurement has some degree of random error. I don't know the distribution of these errors, so assume whatever error distribution you like.

What is a good way to estimate the values of these three vectors?

Currently I'm estimating my vectors by picking out three of these measurements which isolate $x_1$, $x_2$, or $x_3$. This ignores a lot of information, and often leaves me with substantial error in my estimates. I'm hoping there's a better way that uses more information from the additional measurements.

Other information:

I have the precise constraint that the three vectors form a triangle:

$x_1 + x_2 + x_3 = 0$ (I might have to flip the sign of one of the vectors to make this true)

Generally, this triangle is close to equilateral, but not precisely equilateral.

(This question arose during an image processing problem in experimental physics. Apologies if this is off-topic, but it seems more math than physics to me.)

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If the measurement errors were normally distributed, the linear least squares approach would be your best method to use repeated measurements, a result dating back to Gauss.

If you want to exactly impose the triangle equality, then what you have are four scalar unknowns and more than four linear equations (though you've only shown eight equations, counting the vector components, presumably you have many more that could be adduced). When making the same measurement over and over, the shortcut is to average the result and use that. If different measurements have different numbers of repetitions, you can account for this with a weighted least squares approach.

The basic idea is to choose the four unknowns so as to minimize the (weighted?) sum of squares errors in fitting the measurements.

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    Yep, that was it. Exactly what I was looking for.2011-05-22