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Gramschmidt and scalar product - Tonig00 - 05-08-2021 08:59 AM
Hello The explanation of this function says: " Given a basis of a vector subspace, and a function that defines a scalar product, returns an orthonormal basis for that function". I do not know how to it, I will put an example: Scalar product defined by: <(x1,x2),(y1,y2)> = x1y1-x1y2-x2y1+2x2y2 Given a basis such as: B={(1,2), (3,4)} How can I use the function to get an orthonormal basis? If I put gramschmidt([[1 2][3 4]],(a,b,c,d)->(a*b-a*d-b*c+2*b*d)) Gives me an error "Bad Argument Value" Is there a way to do this? Thanks in advance Toni [/i] RE: Gramschmidt and scalar product - parisse - 05-08-2021 10:34 AM
A scalar product takes 2 arguments (the vectors), not four (the coordinates). I recommend to define the scalar product separately. With indices starting at 0: Code:
RE: Gramschmidt and scalar product - roadrunner - 05-08-2021 12:31 PM
gramschmidt([1,1+x],(p,q)->2*p*q) returns: [1/sqrt(2),±∞*(-x-1+x+1)/sqrt(2)] am i doing something wrong? Why is there a ±∞ in there? -road RE: Gramschmidt and scalar product - Tonig00 - 05-08-2021 01:40 PM
Thanks very much for your fast answer. Very useful. It works. Regarding the p*q, I would say that the "x" is not defined so there you have ±inf. I suppose. You may do the definition before as Parise did with "sp:= …" Thanks again Toni RE: Gramschmidt and scalar product - parisse - 05-08-2021 06:50 PM
(05-08-2021 12:31 PM)roadrunner Wrote: gramschmidt([1,1+x],(p,q)->2*p*q)1+x is lineary dependent of 1, because your scalar product is defined on R. If you want to have something more interesting define a scalar product with an integral or by evaluating p and q at some points, like gramschmidt([1,1+x],(p,q)->subst(p,x=0)*subst(q,x=0)+subst(p,x=1)*subst(q,x=1)) RE: Gramschmidt and scalar product - roadrunner - 05-09-2021 11:17 AM
Thanks Tonig00 and parisse; i got it figured out now. -road |