Talk:Similarity (geometry)
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[edit] Matrix congruency
"Two real matrices A and B are called congruent if there is a regular real matrix P such that PtAP = B. Two real symmetric matrices are similar to each other if and only if they are congruent"
This is clearly false: take the 1-by-1 matrices A=1 and P=2, then A cannot be similar to B=4 since they fail to have the same determinant.
I removed the second sentence, as that seems to be the wrong one.--Patrick 10:58, 2 October 2005 (UTC)
[edit] Graphic request
I just added a graphic to this article. If you need something else, please be explicit. John Reid 22:24, 14 April 2006 (UTC)
[edit] Matrix similarity
This subject needs its own article and we need disambiguation page. Same with other subjects which just happen to have the same name, but nothing to do with the concept of similarity in Euclidean geometry. --345Kai 21:27, 19 April 2006 (UTC)
- After the article split, I'm wondering where two of the "See also" links should go, namely Hamming distance and Jaccard index. Both of them are similarity measures of vectors. I'm not sure they fit well in there given current content of that article, but they are definitely measures of similarity. Perhaps they might fit best in some third yet unwritten similarity article for vectors and strings (strings meaning from a computer science perspective)? Halcyonhazard 03:16, 25 September 2006 (UTC)
[edit] Similarity between parabolas
How is it that all parabolas are similar? After all, y = x2 and y = 4x2 differ by a scaling in one coordinate direction, not by a scaling in all directions equally. Ted 20:20, 13 March 2007 (UTC)
- Consider the part of the graph of
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- that is between x = −1 and x = 1. You could say that changing x to 2x squeezes this horizontally to
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- between x = −1/2 and x = 1/2, and that is NOT similar to the original graph, i.e. it is not similar to the part of the parabola between x = 1 and x = −1. However, the part of the new graph that is between x = −1/4 and x = 1/4 is similar to the part of the old graph between x = −1 and x = 1. Consider any point (a, b) on the graph of y = 4x2. Let c = 4a and d = 4b. Then d = 4b = 4(4a2) = 16a2 = (4a)2 = c2. So if the point (a, b) is on the graph of y = 4x2, then the point (4a, 4b) is on the graph of y = x2. The transformation
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- scales in all directions equally. Michael Hardy 22:17, 13 March 2007 (UTC)