Math. H110 How to Recognize a Quadratic Form October 31, 2000 6:47 amProf. W. Kahan Page 1/2 What is a Quadratic Form ? It is a scalar-valued functional ƒ( v ) = ¥vv obtained from a Symmetric Bilinear Form ¥uv , a functional that is both symmetric , ¥uv = ¥vu , and linear : ¥u ( µ v + ß w ) = µ ¥uv + ß ¥uw for all real scalars µ and ß and all vectors u , v and w in areal vector space. If column vectors u and v represent vectors u = B u and v = B v in some basis B , then ¥uv = u T Yv = v T Yu for some symmetric matrix Y = Y T , and ƒ( v ) = v T Yv .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 u T Y and Y _ v by v T Y . Changing B to a new basis B C –1 changes u ’s representative u to Cu in order to keep u = B C –1 (Cu) , and similarly changes v to Cv ; and then, to keep ¥uv = (Cu) T (C T–1 YC –1 )(Cv) unchanged, Y changes to C T–1 YC –1 . In other words, the Congruent matrices Y and C T–1 YC –1 represent the same symmetric bilinear form ¥ and the same quadratic form ƒ(…) in different coordinate systems ( bases ).A quadratic form ƒ( v ) = Zvv is obtainable from any non-symmetric bilinear form Zuv ≠ Zvu represented by a nonsymmetric matrix Z thus: ƒ( v ) = v T Zv even though u T Zv ≠ v T Zu . No good purpose is served this way; ƒ(…) depends upon only the symmetric part ¥ of Z defined thus: ¥uv := ( Zuv + Zvu )/2 and so Y = (Z + Z T )/2 . Moreover, only the symmetric bilinear form ¥ can be recovered from ƒ(…) via the identity ¥uv = ( ƒ( u + v ) – ƒ( u – v ) )/4 . Exercise: Suppose Q is a matrix ( perhaps neither square nor real ) that satisfies (Qx) T (Qx) = x T x for all real column vectors x of the right dimension. Must Q T Q = I ? Why? Another way to recover a symmetric bilinear form from its quadratic form is by differentiation: dƒ( v ) = ƒ ' ( v )dv = 2¥vdv . In other words, the derivative of a quadratic form ƒ(v) is a linear functional ƒ'(v) ( of course ) that is also linear in the form’s argument v and symmetric in so far as ƒ'(v)u = ƒ'(u)v . In matrix terms ƒ'(v) = 2vTY and ƒ'(v)u = 2vTYu = 2uTYv = ƒ'(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(µv + ßw) = µHuv + ßHuw where µ and ß are the complex conjugates of µ and ß . 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 ƒ(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.This document was created with FrameMaker404Math. H110 How to Recognize a Quadratic Form October 31, 2000 6:47 amProf. W. Kahan Page 2/2Every quadratic form ƒ(…) satisfies an identity called the Parallelogram Law :ƒ(x + y) + ƒ(x – y) = 2ƒ(x) + 2ƒ(y) .This law is easy to deduce when ƒ(…) is obtained from a given bilinear form ¥ ; do so! The law gets its name from the case √ƒ(v) = ||v|| of ordinary length defined in an Euclidean space by the Pythagorean formula. Conversely …Theorem: Any real continuous scalar function ƒ(…) that satisfies the parallelogram lawfor all vectors x and y in a real linear space must be a quadratic form.This seems plausible; setting x = y = o implies ƒ(o) = 0 , and then setting x = o and letting y —› o in ƒ(y) – ƒ(–y) = 0 implies ƒ'(o) = oT if the derivative exists, and then doing it again with ƒ(x + y) – 2ƒ(x) + ƒ(x – y) = 2ƒ(y) suggests that ƒ"(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 ç(x) satisfies ç(o) = 0 and ç(x+y) + ç(x–y) = 2ç(x) for all vectors x and y in a real linear space, ç(x) must be a linear functional ç(x) = cTx .Proof: Start by discovering thatç(x+y) = ç( (x+y)/2 + (x+y)/2 ) + ç( (x+y)/2 – (x+y)/2 ) … since ç(o) = 0= 2ç( (x+y)/2 ) … by hypothesis= ç( (x+y)/2 + (x-y)/2 ) + ç( (x+y)/2 – (x-y)/2 ) … by hypothesis= ç(x) + ç(y) for all vectors x and y .Next, for positive integers n = 1, 2, 3, ... in turn, use this discovery to verify by induction thatç(nx) =