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Best answer by AlanStevens

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

2 replies

A
AlanStevens19-TanzaniteAnswer
19-Tanzanite
August 14, 2015

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

V
VladimirN24-Ruby IVAuthor
24-Ruby IV
August 17, 2015

Alan,

Thank you for for an explanation.

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