Excellent, thank you! On Thu, Oct 15, 2015 at 12:22 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Mike, surely the relation you're looking for is the fact that the derivative of the characteristic polynomial of M is the sum of the characteristic polynomials of its principal submatrices M_1, M_2,..., M_n, where M_i is what you get by dropping the ith row and column from M.
Oh hey look, proof here: http://math.stackexchange.com/questions/978815/is-the-derivative-of-the-char...
--Michael
On Thu, Oct 15, 2015 at 2:59 PM, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
Solve for the roots of the derivative polynomial. These are the eigenvalues of M'. String them along the diagonal, and make an arbitrary similarity transformation. If M' has repeated eigenvalues, Jordan blocks are also possible.
-- Gene
From: Mike Stay <metaweta@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, October 15, 2015 9:36 AM Subject: [math-fun] Characteristic polynomial question
What operation on a matrix M gives a new matrix M' whose characteristic polynomial is the derivative of the char. poly. of M? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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