|
|
Post by Admin on Nov 22, 2015 13:32:38 GMT
Here is a short document I wrote for myself, which others may find interesting. MobiusMatrices.pdf (34.92 KB) |
|
Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on Aug 18, 2021 23:30:39 GMT
Here is a short document I wrote for myself, which others may find interesting. View AttachmentVasco,
The document looks useful. I read the first part and it's a nice summary of the difference between M and [M] on the one hand, and $[M^{-1}]$ and $[M]^{-1}$ on the other. I had read through the section before rewriting my answer to part (ii) of Ch. 3, Exercise 20, so I was aware of the difference, but I found I was able to get to the right answer by multiplying the matrix result by a constant, which seemed easier than normalizing [M]. That is why I don't necessarily think that the third bullet point on p. 157 is wrong. Needham might have been assuming that one could normalize post hoc.
Gary
|
|
|
|
Post by Admin on Aug 19, 2021 10:38:25 GMT
Here is a short document I wrote for myself, which others may find interesting. View AttachmentVasco,
The document looks useful. I read the first part and it's a nice summary of the difference between M and [M] on the one hand, and $[M^{-1}]$ and $[M]^{-1}$ on the other. I had read through the section before rewriting my answer to part (ii), so I was aware of the difference, but I found I was able to get to the right answer by multiplying the matrix result by a constant, which seemed easier than normalizing [M]. That is why I don't necessarily think that the third bullet point on p. 157 is wrong. Needham might have been assuming that one could normalize post hoc. Gary
Gary But multiplying the coefficients of a Mobius transformation by a constant makes no difference to the transformation whereas it multiplies the corresponding matrix by the constant. Needham is equating two matrices which are not in general equal. Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 19, 2021 15:59:31 GMT
Vasco,
The document looks useful. I read the first part and it's a nice summary of the difference between M and [M] on the one hand, and $[M^{-1}]$ and $[M]^{-1}$ on the other. I had read through the section before rewriting my answer to part (ii), so I was aware of the difference, but I found I was able to get to the right answer by multiplying the matrix result by a constant, which seemed easier than normalizing [M]. That is why I don't necessarily think that the third bullet point on p. 157 is wrong. Needham might have been assuming that one could normalize post hoc. Gary
Gary But multiplying the coefficients of a Mobius transformation by a constant makes no difference to the transformation whereas it multiplies the corresponding matrix by the constant. Needham is equating two matrices which are not in general equal. Vasco Vasco, It's true that they are not equal in general, but one might say they are equal up to multiplication by a constant. Gary
|
|
|
|
Post by Admin on Aug 19, 2021 16:56:45 GMT
Gary But multiplying the coefficients of a Mobius transformation by a constant makes no difference to the transformation whereas it multiplies the corresponding matrix by the constant. Needham is equating two matrices which are not in general equal. Vasco Vasco, It's true that they are not equal in general, but one might say they are equal up to multiplication by a constant. Gary
Gary In other words they are, in general, not equal! So this is an error in the book. They are equal if you think of them as transformations but they are not if thought of as matrices.. Maybe Needham is thinking of them as transformations but I don't think he is. Vasco |
|
