Math H110 How to Recognize a Quadratic Form October 31 2000 6 47 am What is a Quadratic Form It is a scalar valued functional v vv obtained from a Symmetric Bilinear Form uv a functional that is both symmetric uv vu and linear u v w uv uw for all real scalars and and all vectors u v and w in a real vector space If column vectors u and v represent vectors u Bu and v Bv in some basis B then uv uTYv vTYu for some symmetric matrix Y YT and v vTYv Symmetric bilinear operator maps the space of vectors u and v linearly to its dual space u is a linear functional in the space dual to vectors v and uv vu is its scalar value v is a linear functional in the space dual to vectors u and uv vu is its scalar value See the notes on Least Squares Approximation and Bilinear Forms Matrix Y represents this operator in the basis B representing u by uTY and Y v by vTY Changing B to a new basis BC 1 changes u s representative u to Cu in order to keep u BC 1 Cu and similarly changes v to Cv and then to keep uv Cu T CT 1YC 1 Cv unchanged Y changes to CT 1YC 1 In other words the Congruent matrices Y and CT 1YC 1 represent the same symmetric bilinear form and the same quadratic form in different coordinate systems bases A quadratic form v Zvv is obtainable from any non symmetric bilinear form Zuv Zvu represented by a nonsymmetric matrix Z thus v vTZv even though uTZv vTZu No good purpose is served this way depends upon only the symmetric part of Z defined thus uv Zuv Zvu 2 and so Y Z ZT 2 Moreover only the symmetric bilinear form can be recovered from via the identity uv u v u v 4 Exercise Suppose Q is a matrix perhaps neither square nor real that satisfies Qx T Qx xTx for all real column vectors x of the right dimension Must QTQ I Why Another way to recover a symmetric bilinear form from its quadratic form is by differentiation d v v dv 2 vdv In other words the derivative of a quadratic form v is a linear functional v of course that is also linear in the form s argument v and symmetric in so far as v u u v In matrix terms v 2vTY and v u 2vTYu 2uTYv u v Conversely if y v satisfies y o 0 and if its derivative y v 2 v is linear in v then y v 2 is a constant symmetric bilinear operator symmetric because all continuous second derivatives are symmetric and y v ov y u du vv is a quadratic form In matrix terms y v u 2vTYu is linear in v y v uw 2wTYu 2uTYw y v wu because Y YT must be symmetric and y v ov 2uTYdu vTYv regardless of the path of integration In short a quadratic form can be recognized as such by determining whether its derivative depends linearly on the form s argument Quadratic forms can be defined over complex vector spaces but in two ways There are complex quadratic forms that take complex scalar values algebraically the same as above but quite different from what follows There are real valued quadratic forms very much like what follows but each obtained from an Hermitian bilinear form Huv that is linear in one argument say u and Conjugate Linear in the other Hu v w Huv Huw where and are the complex conjugates of and Bilinear operator H is Hermitian just when Hvu Huv In matrix terms Huv v Hu with an Hermitian matrix H H its complex conjugate transpose H is congruent to C 1HC 1 The Hermitian quadratic form v Hvv v Hv obtained from H is real for all complex v and consequently slightly more complicated than real quadratic forms on real spaces to which this note is confined Prof W Kahan Page 1 2 This document was created with FrameMaker 4 0 4 Math H110 How to Recognize a Quadratic Form October 31 2000 6 47 am Every quadratic form satisfies an identity called the Parallelogram Law x y x y 2 x 2 y This law is easy to deduce when is obtained from a given bilinear form do so The law gets its name from the case v v of ordinary length defined in an Euclidean space by the Pythagorean formula Conversely Theorem Any real continuous scalar function that satisfies the parallelogram law for all vectors x and y in a real linear space must be a quadratic form This seems plausible setting x y o implies o 0 and then setting x o and letting y o in y y 0 implies o oT if the derivative exists and then doing it again with x y 2 x x y 2 y suggests that x is independent of x But a suggestion is not a proof The Theorem s proof found by C Jordan and J von Neumann early in the 20th century is unusual enough to be worth reproducing here First we need a Lemma If a continous scalar functional x satisfies o 0 and x y x y 2 x for all vectors x and y in a real linear space x must be a linear functional x cTx Proof Start by discovering that x y x y 2 x y 2 x y 2 x y 2 since o 0 2 x y 2 by hypothesis x y 2 x y 2 x y 2 x y 2 by hypothesis x y for all vectors x and y Next for positive integers n 1 2 3 in turn use this discovery to verify by induction that nx n 1 x x n 1 x x n x Then from 0 nx nx nx nx verify that nx n x for all x And from x nx n n x n infer that x n x n Similarly m n x m n x for all rational m n Since every real number is a limit of rational numbers and since is continuous x x for every real and every vector x This ensures that x y x y confirming that must be a linear functional x cTx for some cT End of lemma s proof To prove the Theorem take any given scalar function that satisfies the parallelogram law above and from construct the functional x y x y x y 4 Evidently x o 0 and x x 2x 4 x and y x x y via the Parallelogram Law Moreover 4 z x y 4 z x y z x y z x y z x y z x y 2 z x 2 y 2 z x 2 y Parallelogram Law 8 z x for all vectors x y and z Now identify z x with the x of the lemma to deduce that is a linear functional of its second argument and because x …
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