NOTES WEEK 04 DAY 1 SCOT ADAMS

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1 NOTES WEEK 0 DAY 1 SCOT ADAMS DEFINITION 01 Let m, n P N, B P BpR m, R n q Let e 1,, e m be the standard basis of R m Let f 1,, f n be the standard basis of R n Then we define rbs P R nˆm by rbs ji Bpe i, f j q NOTE TO SELF: Next year, consider changing this to rbs P R mˆn and rbs ij Bpe i, f j q Then in Definition 03 below, we d need to write u rl A pvqs, and the would be the dot product R m ˆ R m Ñ R This would probably be a little easier to remember, although we d have to stress that: L rbs : R n Ñ R m REMARK 02 For all m, n P N, the map is linear B ÞÑ rbs : BpR m, R n q Ñ R nˆm Proof Unassigned HW In the expression rl A puqs v appearing in Definition 03 below, the is the dot product R n ˆ R n Ñ R DEFINITION 03 Let m, n P N, A P R nˆm Then we define the bilinear form B A P BpR m, R n q by B A pu, vq rl A puqs v Assigned HW#15 REMARK 0 For all m, n P N, the map is linear A ÞÑ B A : R nˆm Ñ BpR m, R n q Proof Unassigned HW Assigned HW#16 Date: February 7, 2017 Printout date: February 13,

2 2 SCOT ADAMS DEFINITION 05 Let V and W be vector spaces Then we define SBFpV, W q : tb P BpV, V, W v P V, Bpu, vq Bpv, uqu A function B : V ˆV Ñ W is called a W -valued symmetric bilinear form on V if B P SBFpV, W q DEFINITION 06 Let Z be a vector space and Y Ď Z Then Y is a subspace of Z means both y P Y, x ` y P Y ) and P P Y, cy P Y ) Any subspace of a vector space is a vector space, after restricting the linear operations, leaving 0 unchanged, and restricting negation For any two vector spaces V and W, SBFpV, W q is a subspace of the vector space BpV, V, W q, so SBFpV, W q is a vector space DEFINITION 07 For any vector space V, SBFpV q : SBFpV, Rq; a function B : V ˆV Ñ R is called a symmetric bilinear form on V if B P SBFpV q Let n P N Recall that R nˆn sym : ta P R nˆn A t Au denotes the is a subspace of the is a vector space set of symmetric n ˆ n real matrices Then R nˆn sym vector space R nˆn, so R nˆn sym Assigned HW#17 REMARK 08 For all n P N, the maps B ÞÑ rbs : SBFpR n q Ñ Rsym nˆn and A ÞÑ B A : Rsym nˆn Ñ SBFpR n q are inverses of one another, and are both linear, and so they are both vector space isomorphisms Proof Unassigned HW The content of Remark 08 is that, from the point of view of linear algebra, there s no difference between SBFpR n q and Rsym nˆn, and the maps B ÞÑ rbs and A ÞÑ B A provide a dictionary between them Many of us find R nˆn sym easier to think about than SBFpR n q The dictionary lets us leverage our understanding of Rsym nˆn to an understanding of SBFpR n q DEFINITION 09 Let V and W be vector spaces, B P SBFpV, W q Then we define Q B : V Ñ W by Q B pvq Bpv, vq

3 NOTES WEEK 0 DAY 1 3 The set : tpv, vq v P V u is sometimes called the diagonal in V ˆ V We sometimes say that Q B is the diagonal restriction of B, even though, strictly speaking, it is not the restriction of B to DEFINITION 010 Let V and W be vector spaces Then a function Q : V Ñ W is said to be quadratic if there exists B P SBFpV, W q such that Q Q B In Definition 010, instead of quadratic function, some would say homogeneous quadratic function In this course, the default meaning of quadratic is homogeneous quadratic, just as the default meaning of linear is homogeneous linear A quadratic function V Ñ W is often called a W -valued quadratic form on V DEFINITION 011 Let V and W be vector spaces Then we define QpV, W q : tq B B P SBFpV, W qu For any set S, for any vector space W, the set W S of maps S Ñ W is a vector space, under the usual notions of adding and scalar multiplying functions For any vector spaces V and W, QpV, W q is a subspace of the vector space W V, so QpV, W q is a vector space DEFINITION 012 For any vector space V, QpV q : QpV, Rq A quadratic function V Ñ R is often called a quadratic form on V We worked through an example where students proposed numbers A, B, C, D, E, F P R and we studied Q : R 3 Ñ R defined by Qpx, y, zq Ax 2 ` By 2 ` Cz 2 ` Dxy ` Exz ` F yz I showed how to polarize this homogeneous polynomial into a function B : R 3 ˆ R 3 Ñ R defined by Bppx, y, zq, px 1, y 1, z 1 qq Axx 1 ` Byy 1 ` Czz 1 ` j j j xy1 ` x 1 y xz1 ` x 1 z yz1 ` y 1 z D ` E ` F We checked that B is symmetric and bilinear, and that Q B Q, thus proving that Q P QpR 3 q More generally, for any Q P QpV, W q, we can construct its polarization B Q : V ˆ V Ñ W as follows:

4 SCOT ADAMS DEFINITION 013 Let V and W be vector spaces, Q P QpV, W q Then B Q : V ˆ V Ñ W is defined by B Q pu, vq rqpu ` vqs rqpu vqs Our next goal is to show that B Q P SBFpV, W q and that the maps Q ÞÑ B Q and B ÞÑ Q B give a dictionary between quadratic forms and symmetric bilinear forms REMARK 01 Let V and W be vector spaces, B P SBFpV, W q Let Q : Q B Then B B Q Proof For all u, v P V, define u v : Bpu, vq For all v P V, we have Qpvq Q B pvq Bpv, vq v v Let u, v P V be given We wish to show that Bpu, vq B Q pu, vq That is, we wish to show that B Q pu, vq u v We calculate B Q pu, vq rqpu ` vqs rqpu vqs rpu ` vq pu ` vqs rpu vq pu vqs rpu uq`2pu vq`pv vqs rpu uq 2pu vq`pv vqs pu vq u v, as desired REMARK 015 Let V and W be vector spaces, Q P QpV, W q Let B : B Q Then B P SBFpV, W q and Q Q B Proof Since Q P QpV, W q, choose C P SBFpV, W q such that Q Q C Then, by Remark 01 (with B replaced by C), we get C B Q Then B B Q C Then B C P SBFpV, W q and Q Q C Q B REMARK 016 For any vector spaces V and W, the maps B ÞÑ Q B : SBFpV, W q Ñ QpV, W q and Q ÞÑ B Q : QpV, W q Ñ SBFpV, W q are inverses of one another, and are both linear, and so they are both vector space isomorphisms

5 NOTES WEEK 0 DAY 1 5 Proof The fact that Q ÞÑ B Q takes values in SBFpV, W q is part of Remark 015 The fact that the two maps are inverses is part of Remark 015, together with Remark 01 Linearity of the two maps is an unassigned exercise Imagine, in a time before computers, we needed to calculate, say 3986 ˆ 267 Multiplcation px, yq ÞÑ xy : R 2 Ñ R is a symmetric bilinar form and the corresponding quadratic form is the squaring function: x ÞÑ x 2 : R Ñ R Polarization tells us that we can recover multiplication from the squaring function Broadly, we might say, If you can square numbers, then you can multiply More specifically, we have a polarization formula for y P R, xy px ` yq2 px yq 2 Let s say we have access to what might be called a book of quarter squares ; this book begins with the equation 1 2 { 025, then, below it, one finds 2 2 { 100, then, below it, 3 2 { 225, etc On each page there are 100 such equations, and the book spans 100 pages It ends with the equation { at the bottom of the 100th page By polarization, we know that 3986 ˆ 267 p3986 ` 267q2 p q 2 We compute 3986 ` On page 2 of the book, we look up { We compute On page 15 of the book, we look up { We compute We conclude that 3986 ˆ THE BIG IDEA: We have 1-1 correspondences QpR n q Ø SBFpR n q Ø R nˆn sym Colloquially, we might say, Quadratic algebra is the same as symmetric bilinar algebra which is a part of linear algebra In studying freshman calculus, one tries, first to do linear approximations, aka finding the tangent line Moving on, one looks for quadratic approximations, then cubic approximations, then quartic approximations, etc, and all of these come under the heading Taylor polynomials When we try to do all this in higher dimensions, studying functions R m Ñ R n the

6 6 SCOT ADAMS linear approximations are linear maps R m Ñ R n, which can be studied through a subject called: linear algebra One perspective on linear algebra is that we can convert any linear map R m Ñ R n to a matrix in R nˆm, and we learn many tricks for understanding matrices Moving on, one looks for quadratic approximations, then cubic approximations, then quartic approximations, etc, and all of these come under the heading Taylor polynomials In higher dimensions, the study of quadratic maps R m Ñ R n might, at fist blush, seem to require us to have a whole new subject: quadratic algebra Fortunately, we can get at quadratic maps through linear algebra, as follows Polarize the quadratic map to a symmetric bilinear map B : R m ˆ R m Ñ R n Next, there are n projection maps p 1,, p n : R n Ñ R defined P t1,, nu, p i px 1,, x n q x i To study B, we break it apart in to n symmetric bilinar maps p 1 B,, p n B : R m ˆ R m Ñ R Then we form the matrix of each one of these, obtaining rp 1 Bs,, rp n Bs P R mˆm sym So, if we can understand symmetric matrices, we have no need for a subject called quadratic algebra The study of symmetric matrices is a part of linear algebra, and the most important theorem about them is called the Finite Dimensional Spectral Theorem Look it up! What happens when we go on to cubic approximations, quartic approximations, etc, all applied to a function R m Ñ R n? We developed ideas that reduced quadratic algebra to a part of linear algebra Can they be extended, say, to cubic algebra? The answer is yes, and we ll finish today s notes giving a bird s eye view of how that s done Keep in mind that, in this course, we won t be spending much time beyond the quadratic level However, here are some of the basic definitions and results you would want (You re not responsible for learning these) For any vector spaces V, W, X, Y, let T pv, W, X, Y q be the set of trilinear maps V ˆ W ˆ X Ñ Y, ie, maps T : V ˆ W ˆ X Ñ Y st for all v P V, for all w P W, T pv, w, q : X Ñ Y is linear, for all v P V, for all x P X, T pv,, xq : W Ñ Y is linear and for all w P W, for all x P X, T p, w, xq : V Ñ Y is linear; these are called Y -valued trilinear forms on V ˆ W ˆ X Let V and Y be vector spaces Then STFpV, Y q denotes the set of all T P T pv, V, V, Y q such that, for all v, w, x P V, T pv, w, xq T pv, x, wq T pw, v, xq T pw, x, vq T px, v, wq T px, w, vq;

7 NOTES WEEK 0 DAY 1 7 these are Y -valued symmetric trilinear forms on V ˆ W ˆ X For all T P STFpV, Y q, we define C T : V Ñ Y by C T pvq T pv, v, vq; to get C T, we restrict T to the diagonal (This colloquialism is not exactly accurate, but does carry the basic idea) A mapping C : V Ñ Y is said to be cubic (or homogeneous cubic) if there exists T P STFpV, Y q such that C C T The only other key point is cubic polarization Given a cubic C : V Ñ Y is there a unique T P STFpV, Y q such that C C T, and, also, is there a formula for T in terms of C? The uniqueness is not hard, once you see the development of the formula, and we ll leave it as an exercise Let s focus on the formula We start with the case where V and Y are Euclidean spaces, and, to keep things simple, let s take V R 2 and Y R Ask a friend to pick four numbers a, b, c, d P R Define a function C : R 2 Ñ R by Cpx, yq ax 3`bx2 y`cxy 2`dy3 Then C is cubic and its symmetric trilinear polarization T : R 2 ˆ R 2 ˆ R 2 Ñ R is given by j xx T ppx, yq, px 1, y 1 q, px 2, y 2 qq axx 1 x 2 1 y ` xx 2 y ` x 1 x 2 y ` b ` 3 j xyy1 ` xyy 2 ` xy 1 y 2 c ` dyy 1 y 2 3 Here s a way to think about this: Focus first the cubic function px, yq ÞÑ x 2 y : R 2 Ñ R; call this F There are many trilinear maps R 2 ˆ R 2 ˆ R 2 Ñ R whose diagonal restriction is F, but there are three that stand out: ppx, yq, px 1, y 1 q, px 2, y 2 qq ÞÑ xx 1 y 2 : R 2 ˆ R 2 ˆ R 2 Ñ R, ppx, yq, px 1, y 1 q, px 2, y 2 qq ÞÑ xy 1 x 2 : R 2 ˆ R 2 ˆ R 2 Ñ R and ppx, yq, px 1, y 1 q, px 2, y 2 qq ÞÑ yx 1 x 2 : R 2 ˆ R 2 ˆ R 2 Ñ R None of these is symmetric, but, if we average them together, we get a symmetric trilinear form whose diagonal restriction is F One says the polarization of x 2 y is pxx 1 y 2 ` xy 1 x 2 ` yx 1 x 2 q{3 The terms in the numerator are obtained by taking x, x, y, leaving one of them unchanged, putting a prime on another of them, and putting a double prime on the third, then multiplying the results together There are three ways to do this, and we do this in all three possible ways, and then average We think of our T px, y, zq as a linear combination of x 3, x 2 y, xy 2, y 3, with coefficients a, b, c, d Through the process described above, we polarize

8 8 SCOT ADAMS each of x 3, x 2 y, xy 2, y 3, and then form the a, b, c, d linear combination of the results to get the polarization of T px, yq One sometimes says polarization is linear to convey the idea that the polarization of a linear combination is the linear combination (with the same coefficients) of the polarizations We d also like to have a polarization formula, for any vector spaces V and Y of any cubic T : V Ñ Y To simplify notation, let s focus on finding a way of computing 2766 ˆ 388 ˆ 9217, given a 100 page big book of cubes whose first formula says and whose last formula says We define v 2766, w 388 and x 9217 We wish to compute vwx Let t : v ` w and u : v w From quadratic polarization, we have vwx pv ` wq2 pv wq 2 j x t2 x u2 x We seek formulas for t 2 x and u 2 x, that are easily computable from the big book of cubes We ll focus on t 2 x We have Subtracting, we get pt ` xq 3 t 3 ` 3t 2 x ` 3tx 2 ` x 3 pt xq 3 t 3 3t 2 x ` 3tx 2 x 3 pt ` xq 3 pt xq 3 6t 2 x ` 2x 3 Solving for t 2 x, we get t 2 x pt ` xq3 pt xq 3 2x 3 Similarly, we 6 have u 2 x pu ` xq3 pu xq 3 2x 3 Then 6 vwx t2 x u2 x pt ` xq3 pt xq 3 2x 3 2 pu ` xq3 pu xq 3 2x 3 2 pt ` xq3 pt xq 3 pu ` xq 3 ` pu xq 3 2 Recalling that t v ` w and u v w, we get vwx p1{2q p pv ` w ` xq 3 pv ` w xq 3 pv w ` xq 3 ` pv w xq 3 q

9 NOTES WEEK 0 DAY 1 9 A big book of one twenty-fourth cubes is what we need Recall: v 2766, w 388, x 9217 Then vwx and p1{2q p pv ` w ` xq 3 pv ` w xq 3 pv w ` xq 3 ` pv w xq 3 q Success! For any vector spaces V, Y, for any cubic form C : V Ñ Y, a similar analysis shows that the polarization T : V ˆ V ˆ V Ñ Y is given by: T pv, w, xq p1{2q p rcpv ` w ` xqs rcpv ` w xqs rcpv w ` xqs ` rcpv w xqs q In each degree, there are many polarization formulas, but the one given above works perfectly well for degree 3 With your endless free time, I recommend working out a polarization formula in degree Enjoy!