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Consider the following matrix with entries being intervals

A={{Interval[{-1.00034, -1.00031}], 
  Interval[{0.00980599, 
    0.00982034}]}, {Interval[{-0.0262465, -0.0262241}], 
  Interval[{-0.207671, -0.207659}]}}

I ask Mathematica to calculate the normalized leading eigenvector which gives:

Normalize[Eigenvectors[A][[1]]]
{Interval[{0.741586, 1.34695}], Interval[{0.0281967, 0.0379947}]}

The error worried me, it seems way too big. So I tried just using the formula for an eigenvector as this is just the $2\times 2$ case.

Normalize[{(A)[[1]][[2]], 
  Eigenvalues[
    A][[1]] - (A)[[1]][[1]]}]
{Interval[{0.997819, 1.00106}], Interval[{0.0281876, 0.038007}]}

This gives much better estimates, what is going on? Mathematica calculates eigenvectors in a weird way (for a general size) and hence loses a lot of precision?

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  • $\begingroup$ What are A and B? $\endgroup$
    – Michael E2
    Oct 29, 2022 at 19:34
  • $\begingroup$ @MichaelE2 Fixed, sorry. $\endgroup$
    – 2132123
    Oct 29, 2022 at 19:36

1 Answer 1

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The problem might be that multiple Intervals in the same expression are treated as independent, even if they should represent the same quantity.

As a simple example, consider

#^2 - #*# &[Interval[{0, 1}]]
(* Interval[{-1, 1}] *)

Of course, the correct value of applying this function is zero

#^2 - #*# &[x]
(* 0 *)
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  • $\begingroup$ I am surprised this causes so much error but makes sense. I assumed Interval Arithmetic was much smarter than that ;/ $\endgroup$
    – 2132123
    Oct 31, 2022 at 0:48

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