NOTES WEEK 10 DAY 2. Unassigned HW: Let V and W be finite dimensional vector spaces and let x P V. Show, for all f, g : V W, that

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1 NOTES WEEK 10 DAY 2 SCOT ADAMS Unassigned HW: Let V and W be finite dimensional vector spaces and let x P V. Show, for all f, g : V W, that D x pf ` gq pd x fq ` pd x gq. Also, show, for all c P R, for all f : V W, that D x pc fq c pd x fq. We begin class today with a discussion about differentiating polynomials in the setting of finite dimensional vector spaces. Beginning of discussion. Let V and W be finite dimensional vector spaces and let P P P 4 pv, W q be a homogeneous polynomial function V Ñ W of degree 4. Let W : LpV, W q and let W : LpV, W q. Then DP : V LpV, W q W and DDP : V LpV, LpV, W qq W. Recall that BpV, V, W q denotes the set of all bilinear functions V ˆ V Ñ W. Recall that SBFpV, W q Ď BpV, V, W q denotes the set of symmetric bilinear functions V ˆ V Ñ W. Recall that there s a standard vector space isomorphism BpV, V, W q Ø LpV, LpV, W qq W. Recall that the composition of DDP : V W with the isomorphism W Ñ BpV, V, W q is, by definition, the second total derivative D 2 P : V BpV, V, W q. Our goal, in this discussion, is to show that domrdp s domrddp s domrd 2 P s V, and to compute DP : V Ñ LpV, W q W, DDP : V Ñ LpV, LpV, W qq W and D 2 P : V Ñ BpV, V, W q, Date: March 30, 2017 Printout date: April 1,

2 2 SCOT ADAMS and to show, for all q P V, that pd 2 P qpqq P SBFpV, W q. Through this discussion, keep in mind that, from our definitions, for all q P V, we have D q P pdp qpqq and D q DP pddp qpqq and D 2 qp pd 2 P qpqq. Let F : F P P SM 4 pv q denote the polarization of P. Then, for all q P V, we have P pqq F pq, q, q, qq. For all q, h P V, we have the binomial expansion F pq ` h, q ` h, q ` h, q ` hq rf pq, q, q, qqs ` 4rF pq, q, q, hqs ` 6rF pq, q, h, hqs ` 4rF pq, h, h, hqs ` rf ph, h, h, hqs, so, since P T q phq rp pq ` hqs rp pqqs, we get P T q phq rf pq ` h, q ` h, q ` h, q ` hqs rf pq, q, q, qqs 4rF pq, q, q, hqs ` 6rF pq, q, h, hqs ` 4rF pq, h, h, hqs ` rf ph, h, h, hqs. Fix q P V for this paragraph. To find pdp qpqq D q P P LpV, W q, we wish to linearize Pq T : V Ñ W. Put another way, we wish to linearize the mapping h ÞÑ Pq T phq : V Ñ W. For all h P V, we have a formula above for Pq T phq, and we want to find a linear expression L q phq of h such that h ÞÑ rpq T phqs rl q phqs : V Ñ W is sublinear. The first term 4rF pq, q, q, hqs is linear in h, whereas the others are quadratic, cubic and quartic in h. Since quadratics, cubics and quartics are all Op2q, these other terms are all sublinear. This suggests that D q P P LpV, W q comes from the first term. That is, this suggests that D q P is the linear mapping h ÞÑ 4rF pq, q, q, hqs : V Ñ W. Let s now implement these observations formally. In what follows, think of the letter L as meaning linear, and of the letter R as meaning remainder. For all q P V, let L q, R q : V Ñ W be defined by L q phq 4rF pq, q, q, hqs, R q phq 6rF pq, q, h, hqs ` 4rF pq, h, h, hqs ` rf ph, h, h, hqs. Then, for all q P V, we have P T q R q L q ` R q and L q P LpV, W q and P rp 2 pv, W qs ` rp 3 pv, W qs ` rp 4 pv, W qs. Recall that we have P 2 pv, W q Ď O 2 pv, W q Ď OV,W p1q. Also, recall that we have P 3 pv, W q Ď O 3 pv, W q Ď OV,W p1q. Finally, recall also that P 4 pv, W q Ď O 4 pv, W q Ď OV,W p1qq. P V, R q P OV W p1q. For all q P V, since L q P LpV, W q and Pq T L q R q P OV W p1q, we conclude that D q P L q, i.e., that pdp qpqq L q. This solves the

3 NOTES WEEK 10 DAY 2 3 problem of computing DP : V Ñ W, but we want a better solution, by understanding exactly how L q and F are related. We have F P SM 4 pv, W q. Recall that F P SM 3 pv, LpV, W qq is defined by pf pp, q, rqqphq F pp, q, r, hq. Then, for all q, h P V, we have pd q P qphq L q phq 4rF pq, q, q, hqs 4rpF pq, q, qqqphqs. It follows that D q P 4rF pq, q, qqs. Let F : F P SM 3 pv, LpV, W qq. Then, since W LpV, W q and since SM 3 is the same as STF, we see that F P STFpV, W q. Let P P P 3 pv, wq be the diagonal restriction of F ; that is, define P P P 3 pv, wq by P pqq F pq, q, qq. Then, for all q P V, we have pdp qpqq D q P 4rF pq, q, qqs 4rF pq, q, qqs 4rP pqqs. Then DP 4 P. If we think of P as being like the one-variable calculus expression x 4, and if think of P as being like the one-variable calculus expression x 3, then the formula DP 4 P is analogous to pd{dxqpx 4 q 4x 3. The formula DP 4 P is our solution to the problem of computing DP : V LpV, W q; note that domrdp s V. Thus, the familiar freshman calc assertion that polynomials are everywhere differentiable continues into the vector space setting. To better understand the formula DP 4 P, we will review all of the various transitions that we went through to go from P to P : P ÞÑ F ÞÑ F F ÞÑ P. The first transition P ÞÑ F is polarization P 4 pv, W q Ñ SM 4 pv, W q. The second F ÞÑ F is the map SM 4 pv, W q Ñ SM 3 pv, LpV, W qq, which I sometimes call reduction of (vector) inputs, because elements of SM 4 pv, W q accept, as input, four vectors from V, whereas elements of SM 3 pv, W q only accept three. The number of input vectors goes down, but the target space becomes more complicated, going from W to W LpV, W q. The next transition, from F to F isn t really a transition; we defined F to be F P SM 3 pv, LpV, W qq STFpV, W q, because we find the notation F to be more suggestive of the target space W. Finally, the last transition F Ñ P is restriction to the diagonal SM 3 pv, W q Ñ P 3 pv, W q. In summary, to go from P to P, we polarized, then reduced inputs, then restricted to the diagonal. We next repeat those three transitions, and go from P P P 3 pv, W q to P P P 2 pv, W q. We will end up with the formula DP 3 P and we will compute DDP Dr4 P s 4 rdp s 4 r3 P s 12 P. The first transition is to polarize P. Since P is the diagonal restriction of F, that step is easy: F is the polarization of P. Next, we reduce

4 4 SCOT ADAMS inputs: define F : pf q P SM 2 pv, LpV, W qq SBFpV, W q. Finally, we restrict to the diagonal: define P P P 2 pv, W q by P pqq F pq, qq. Let s next compute DP and DDP, as follows. For all q, h P V, P T q phq rp pq ` hqs rp pqqs rf pq ` h, q ` h, q ` hqs rf pq, q, qqs rf pq, q, qqs ` 3rF pq, q, hqs ` 3rF pq, h, hqs ` rf ph, h, hqs rf pq, q, qqs 3rF pq, q, hqs ` 3rF pq, h, hqs ` rf ph, h, hqs. Informally, one says that the first term 3rF pq, q, hqs is linear in h, while the others are sublinear. We leave it to the reader to formalize this, and prove, for all q, h P V, that pd q P qphq 3rF pq, q, hs. Then, for all q, h P V, we have pd q P qphq 3rF pq, q, hqs 3rpF pq, qqqphqs. Then, for all q P V, we have D q P 3rF pq, qqs. Then, for all q P V, we see that pdp qpqq D q P 3rF pq, qqs 3rP pqqs. Then DP 3 P. Then DDP Dr4 P s 4 rdp s 4 r3 P s 12 P, and this is our solution to the problem of computing DDP : V LpV, LpV, W qq; note that domrddp s V. Thus, the familiar freshman calc assertion that polynomials are everywhere twice-differentiable continues into the vector space setting. Last, we go from DDP : V Ñ W to D 2 P : V BpV, V, W q. We have a vector space isomorphism BpV, V, W q Ø LpV, LpV, W qq W, and D 2 P is, by definition, equal to the composite of DDP : V Ñ W followed by W Ñ BpV, V, W q. We wish to compute this composite. We have pf q pf q F P SM 2 pv, W q SBFpV, W q. Note that we reduced one input, and then reduced one onput again, to go from F to pf q. We can also reduce two inputs, to go from F to F P SM 2 pv, SM 2 pv, W qq SBFpV, SBFpV, W qq. WARNING: pf q is not quite the same as F, although they are closely related, as follows. Since SBFpV, W q Ď BpV, V, W q, we see that F : V ˆ V Ñ BpV, V, W q. We also have a vector space isomorphism BpV, V, W q Ø LpV, LpV, W qq W. We leave it as an unassigned exercise to show that if we compose F pf q : V ˆ V Ñ W followed by W Ñ BpV, V, W q, then that composite is F : V ˆ V Ñ BpV, V, W q. Recall that P is the restriction of F to the diagonal. Let

5 NOTES WEEK 10 DAY 2 5 P be the restriction of F to the diagonal. Then the composite of P : V Ñ W followed by W Ñ BpV, V, W q, is P : V Ñ BpV, V, W q. So, since DDP 12 P, we conclude, from the definition of D 2 P, that D 2 P 12 P. This is our solution to the problem of computing D 2 P : V BpV, V, W q; note that domrd 2 P s V. Thus, the familiar freshman calc assertion that polynomials are everywhere twice-differentiable continues into the vector space setting. One last comment: We observed that F : V ˆV Ñ BpV, V, W q. Recall that SBFpV, W q Ď BpV, V, W q and that F P SBFpV, SBFpV, W qq. Then, in fact, we have F : V ˆ V Ñ SBFpV, W q; that is, F actually takes values in SBFpV, W q. Restricting to the diagonal, we see that P : V Ñ SBFpV, W q. Then D 2 P 12 P : V Ñ SBFpV, W q. Then, for all q P V, we have pd 2 Qqppq P SBFpV, W q. End of discussion. Let V and W be finite dimnensinoal vector spaces, f : V W. Then D 2 f : V BpV, V, W q. We eventually want to prove, for any q P domrd 2 fs, that pd 2 fqpqq P SBFpV, W q. This will allow us to write D 2 f : V SBFpV, W q. In the one last comment of the discussion above, we saw this for f P P 4 pv, W q, but the argument generalizes to all polynomials. In fact, that argument shows: THEOREM 0.1. Let V and W be finite dimensional vector spaces. Let l P N 0. Let P P P l pv, W q. Let 0 : 0 V,SM k pv,w q : V Ñ SM kpv, W q be the zero map, defined by 0pV q 0 SMk pv,w q 0 V W. Then, for all k P N, we have both ( if k ď l, then D k P P P l k pv, SM k pv, W qq ) and ( if k ą l, then D k P 0 ). Proof. In the discussion above, we handled the cases l 4, k P t1, 2u. The other cases are similar, and we leave the details to the reader. Theorem 0.1 is the vector space analogue of the familiar freshman calc result that the kth derivative of an lth degree polynomial is: ( a polynomial of degree l k, if k ď l ) and ( the zero function, if k ď l ). In the vector space setting, the degree, if positive, goes down by when we differentiate. What s different (from single variable calc) is that the target space of the derivative increases in complexity each time we

6 6 SCOT ADAMS differentiate. In the discussion this increasing complexity was indicated by the nummber of overbars on the W s, starting with W, and moving to W LpV, W q, and then to W LpV, LpV, W qq. We move on to a discussion about differentiating matrix inversion: Beginning of discussion. Let n P N, let V : R nˆn and let U : ta P V deta 0u. Then U is a dense open subset of V. Define F : U Ñ V by F paq A 1. Then F : V V, and we seek to calculate DF : V Ñ LpV, V q. In freshman calc, the computation of pd{dxqpx 1 q is analogous. We compute: px ` hq 1 x 1 1 x ` h 1 x px ` hq x px ` hqx h px ` hqx. We divide by h, then let h Ñ 0, and get pd{dxqp1{xq 1{x 2. We seek to upgrade from scalars x, h to matrices X, H, but, when A and B are square matrices, an expression like A is ambiguous, because it B might mean AB 1 or it might mean B 1 A. Going back to freshman calc, we might write the computation a little differently: px ` hq 1 x 1 rx px ` hqsrpx ` hqxs 1 hrpx ` hqxs 1. If we upgrade to matrices X and H, we have a problem: There exist X, H P V such that X P U and X ` H P U, but px ` Hq 1 X 1 rx px ` HqsrpX ` HqXs 1. The fundamental problem is that matrices don t commute. With matrices, when we try to find a common denominator, experience has taught us that we need to put one factor of the denominator on the left and the other factor on the right. In this case, we ll put px ` Hq 1 on the left and X 1 on the right, as follows: px ` Hq 1 X 1 rpx ` Hq 1 srx px ` HqsrX 1 s. This equation holds for all X P U and all H P U X. Then px ` Hq 1 X 1 rpx ` Hq 1 srx px ` HqsrX 1 s rpx ` Hq 1 shrx 1 s. We begin with the first factor, namely: rpx ` Hq 1 s. Recall that p q 1 : U Ñ H is a quotient of two polynomials, the adjugate map

7 NOTES WEEK 10 DAY 2 7 V Ñ V and the determinant map V Ñ R. It follows that the function p q 1 : U Ñ H is continuous. (Meaning: p q 1 is continuous at each point of its domain, U.) For a fixed X, H ÞÑ px ` Hq 1 is the composition of the two continuous functions H ÞÑ X ` H and p q 1, and is therefore continuous. Matrix negation A ÞÑ A : V Ñ V is also continuous, and, composing with that, we see, for each X, that H ÞÑ rpx ` Hq 1 s is continuous; this map sends the n ˆ n zero matrix, denoted 0 nˆn, to rpx ` 0 nˆn q 1 X 1. Then, for each X, H ÞÑ rpx ` Hq 1 s can be approximated by the constant map H ÞÑ X 1, with remainder in OV V. Next, we look at the second factor in rpx ` Hq 1 shrx 1 s, namely H. The map H ÞÑ H is linear and therefore in O V V p1q. Finally, the third factor is X 1. For each X, the map H ÞÑ X 1 is constant, and therefore in O V V. From this analysis, for each X, the mapping H ÞÑ rpx ` Hq 1 shrx 1 s can be approximated by the linear map H ÞÑ rx 1 shrx 1 s, with remainder in rov V sro V V p1qsro V V s Ď OV V p1q. Since the remainder is sublinear, we see, for each X, H, that pd X F qphq rx 1 shrx 1 s. End of discussion. We write this discussion up, crossing our eyes and dotting our tees: THEOREM 0.2. Let n P N, V : R nˆn, U : ta P V deta 0u. Define F : U Ñ V by F paq A 1. Then, for all X P U, for all H P V, we have pd X F qphq rx 1 shrx 1 s. Proof. Let X P U be given. We wish to show, for all H P V, that pd X F qphq rx 1 shrx 1 s. Define the linear function L P LpV, V q by LpHq rx 1 shrx 1 s. We wish to show, for all H P V, that pd X F qphq LpHq. That is, we wish to show D X F L. That is, we wish to show that F T X L P O V V p1q. We have domrf T X s pdomrf sq X U X. Recall: V Rnˆn. Let P BpV, V, V q denote matrix multiplication, defined by X Y XY. Then rov V s ro V V p1qs Ď OV V p1q. Let ε : F T X : U X Ñ V. We define α P LpV, V q by αphq HrX 1 s. For all H P U X, we have F T X phq rf px ` Hqs rf pxqs px ` Hq 1 X 1, and so we see

8 8 SCOT ADAMS that εphq px ` Hq 1 ` X 1. For all H P U X, we have F T XpHq px ` Hq 1 X 1 rpx ` Hq 1 srx px ` HqsrX 1 s rpx ` Hq 1 shrx 1 s, and we have rlphqs rx 1 shrx 1 s, and so rf T XpHqs rlphqs rpx ` Hq 1 shrx 1 s ` rx 1 shrx 1 s r px ` Hq 1 ` X 1 shrx 1 s rεphqsrαphqs rεphqs rαphqs. Then FX T L ε α. So, since ro V V s ro V V p1qs Ď OV V p1q and since α P LpV, V q Ď O V V p1q, it suffices to show that ε P OV V. Since X P U and since U is open in V, we see that U is a nbd in V of X. Then U X is a nbd in V of 0 V. So, as domrεs U X, we need only show both that εp0 V q 0 V and that ε is continuous at 0 V. By definition of FX T, we get FX T p0 V q 0 V, and so εp0 V q rfx T p0 V qs 0 V. It remains only to show that ε is continuous at 0 V. Since ε FX T, it suffices to show that FX T is continuous at 0 V. We have FX T pxq rf px ` qs rcf V s, so, since the constant function C F pxq V : V Ñ V is continuous on V, it suffices to show that F px ` q is continuous at 0 V. Recall: V R nˆn. Let adj : V Ñ V and det : V Ñ R denote the adjugate and determinant maps. Then adj P P n 1 pv, V q and det P P n pv, Rq, so adj and det are both polynomials, and are therefore continuous on V. So, since F adj{det, we see that F : U Ñ V is continuous on U. Then F px ` q : U X Ñ V is continuous on U X. So since 0 V X X P U X, we conclude that F px ` q is continuous at 0 V, as desired.