My Math Forum "Simplification" of d/dlam(det(Y+lam*Z)) at lam = 0

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 January 16th, 2013, 07:53 AM #1 Newbie   Joined: Jan 2013 Posts: 4 Thanks: 0 "Simplification" of d/dlam(det(Y+lam*Z)) at lam = 0 Hello - I'm writing some code and need a numerical method of finding d/dlam(det(Y+lam*Z)) at lam = 0 (programming it symbolically gives a whole host of problems). I think I that I can say that each term that comes from the determinant and doesn't go to 0 when finding the derivative with respect to lam at lam = 0, will have one and only one term from Z - (any more and a "lam" will remain and be set to zero, any less and the derivative will set the term to zero). As each term in a derivative has one and only one term from each row I think I can say that if I replace a row of Y with the same row of Z then all the terms in the new derivative will be included in the above. (replace row 1 of Y with row 1 of Z or replace row 2 of Y with row 2 of Z) I'm pretty confident that if I sum all of the determinants formed in this way (if Y and Z are NxN there would be N determinants in the sum) that would exhaust all the terms that would come from d/dlam(det(Y+lam*Z)) at lam = 0. The problem is: I can't find a way to rigorously prove this, and I can't really "assume" it's right. The Question I'm asking is can anyone prove this? Also - if anyone can come up with another useful representation of the equation I'd be very grateful (not to mention impressed). In case anyone wishes to do this I'll mention that most but not all of the Y matrices I'm looking at are singular and I'll also say that if people know mathematics but not what forms will be useful for programming I'd be happy for any answers as they'll probably help give me ideas even if they're not directly applicable. Thanks in advance: Happle
 January 16th, 2013, 08:00 AM #2 Newbie   Joined: Jan 2013 Posts: 4 Thanks: 0 Re: "Simplification" of d/dlam(det(Y+lam*Z)) at lam = 0 Sorry - should have mentioned ( and can't see how to edit): Both Y and Z are symmetric. (it doesn't matter for my method but might if anyone else wanted a go...)
 January 18th, 2013, 04:27 AM #3 Newbie   Joined: Jan 2013 Posts: 4 Thanks: 0 Re: "Simplification" of d/dlam(det(Y+lam*Z)) at lam = 0 I think I have a proof so I thought I'd put it up in case anyone wanted to know the answer: If you expand det(A+B) for 2x2 matrices you can show that this is equal to det(A) + det(A with first row replaced by first row of B) + det(A with second row replaced by second row of B) + det(B). If you then look at det(A+B) for 3x3 matrices and expand by minors while using the above you can show that it goes to the sum of the determinants coming from the first row being from A or from B on the determinants of the previous 2x2 matrices. This means that overall you have the sum of the determinant of all possible combinations from replacing zero, one or more rows of A with B. You can't have offsets to ruin the order as each cofactor has two of the three terms set so they overlap. You can do that iteratively to show that det(A+B) is equal to the sum of determinants of matrices obtained by replacing all possible combinations of rows of A with B (or vice versa) for any size matrices. Once that's established, you know that any matrices with more or less than one row substituted from det(A+lam*B) would give a power of lam different to 1 in it's term, so the derivative with respect to lam at lam=0 would then go to zero. Hence the set I mentioned in the original post is exhaustive. (You could probably do the same with columns - I haven't checked but see no reason that you couldn't) If this is wrong I would greatly appreciate anyone pointing out the error - otherwise thanks for anyone that tried, and the request for alternate representations remains. Happle

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