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Oct 9, 2010 at 10:24 vote accept Vagabond
Oct 9, 2010 at 10:22 comment added Vagabond Yes, so that settles it. I realise one can pose it as a problem of linkage, if 4 rods are linked to form a quadrilateral then its not rigid so there would be a lot of solution. The question about when the matrix is further assumed to be generalized vandermonde type under the hypotheses that all entries has to be algebraic integers is what I need, I guess I should pose it as a separate question but I will try some more before that.
Oct 9, 2010 at 9:00 comment added Denis Serre Right. But, as I evoqued in my answer, the construction as a lot of flexibility. Actually, there is a full open set in the parameter set for which the construction works.
Oct 9, 2010 at 8:08 comment added Vagabond The $N$ you took if you eliminate the first row and column we get [-1 1; 1 -1] which is singular and we have the condition that none of the sub matrices are singular. I am not sure though if that affects the rest of the construction.
Oct 9, 2010 at 7:00 comment added Denis Serre I make a correcion, because I forgot a factor $i$ in $N^{-1}e$. This does not change the rest.
Oct 9, 2010 at 6:59 history edited Denis Serre CC BY-SA 2.5
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Oct 8, 2010 at 19:48 history edited Denis Serre CC BY-SA 2.5
An example of singular $A$ for $n=4$.
Oct 8, 2010 at 18:23 comment added Vagabond How do we proceed from here ? I would be eagerly waiting for the rest of the argument.
Oct 8, 2010 at 18:01 history edited Denis Serre CC BY-SA 2.5
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Oct 8, 2010 at 18:00 comment added Denis Serre Correct! I wanted to write $\hat M$. I make the correction immediately.
Oct 8, 2010 at 17:09 comment added Vagabond When I do what you suggested I get $e^T M^{-1} e = 1$
Oct 8, 2010 at 16:10 history answered Denis Serre CC BY-SA 2.5