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19-Tanzanite
Aug 14, 2015 8:48 AM UTC

It's just that the M11 eigenvectors have been normalised to 1, whereas the M15 and Prime values haven't.  If you take each column of the M15 and form the square root of the sum of squares of each of its numbers, then divide each of the numbers in that column by that value, you get the M11 result.

Alan

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19-Tanzanite
Aug 14, 2015 8:48 AM UTC

It's just that the M11 eigenvectors have been normalised to 1, whereas the M15 and Prime values haven't.  If you take each column of the M15 and form the square root of the sum of squares of each of its numbers, then divide each of the numbers in that column by that value, you get the M11 result.

Alan

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VladimirN24-Ruby IVAuthor
24-Ruby IV
Aug 17, 2015 12:16 PM UTC

Alan,

Thank you for for an explanation.

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