1 Matrix representations of canonical matrices


 Samuel Hoover
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1 1 Matrx representatons of canoncal matrces 2d rotaton around the orgn: ( ) cos θ sn θ R 0 = sn θ cos θ 3d rotaton around the xaxs: R x = cos θ sn θ 0 sn θ cos θ 3d rotaton around the yaxs: cos θ 0 sn θ R y = sn θ 0 cos θ 3d rotaton around the zaxs: cos θ sn θ 0 R z = sn θ cos θ d rotaton around the xy plane: 4d rotaton around the yz plane: R xy = 0 0 cos θ sn θ 0 0 sn θ cos θ R yz = 4d rotaton around the zw plane: cos θ 0 0 sn θ sn θ 0 0 cos θ cos θ sn θ 0 0 sn θ cos θ 0 0 R zw =
2 4d rotaton around the xw plane: 4d rotaton around the yw plane: 4d rotaton around the xz plane: cos θ sn θ 0 R xw = 0 sn θ cos θ cos θ 0 sn θ R yw = sn θ 0 cos θ cos θ 0 sn θ R xz = sn θ 0 cos θ In general, the dagonals alternate sgns, wth the man dagonal and subdagonal postve 2 Constructng arbtrary rotaton matrces To construct an arbtrary 4D rotaton matrx around the uv plane, where u and v are orthonormal, let M be a rotaton that transforms the xy plane nto the uv plane Then R uv = MR xy M 1 We are left to construct M The choce of the frst two columns of M s clear: they should be the mages of ˆx and ŷ under M, namely, u and v themselves The remanng two columns, z and w, must be chosen such that {u, v, z, w } s parwse orthonormal To solve for z and w, start from the fact that and u r = 0 (1) v r = 0 (2) for every r on the plane perpendcular to both u and v We want two such vectors r, and we want them to be orthonormal So we have two equatons n four unknowns, whch means the solutons fll a plane (that s good) Let u = ( a 1 b 1 c 1 d 1 ) and v = ( a 2 b 2 c 2 d 2 ) 2
3 After some algebra (there must be a better way) we can fnd at least one soluton: ( r 1 = 0, 1, b 1d 2 d 1 b 2, 1 ) (b 2 + c 2 z) c 1 d 2 d 1 c 2 d 2 Ths vector was arrved at by frst solvng for w n Equaton 2 and pluggng the result nto Equaton 1 The resultng equaton was solved for z Then I pcked 0 and 1 for x and y to get r 1 If we let z = r 1 / r 1, then z wll be of unt length and wll be orthogonal to both u and v We stll need w But ths s smple: t wll just be w = u % v % z Note that the crossproduct s negated because of the clam (stll more or less unjustfed) that n a rghthanded system, we should generate each new dmenson by takng the crossproduct of the prevous three, negated whenever gong to an even number of dmensons Thus M = u v z w 3 Rotaton by means of normal vector (3) Here I propose a method for the rotaton of a quadrc surface va ts normal vector If n(x) the normal vector to a gven surface at pont x, and R s a rotaton matrx, then at pont Rx (whch s on the rotated surface), t s reasonable to guess that the normal vector wll be Rn(x) Snce the normal vector s aqured va dfferentaton, I am led to beleve that ntegraton should provde a means for fndng the equaton of the rotated surface At the very least, some surface corresponds to the rotated normal feld, snce the rotated normal feld s stl a gradent feld That s, g st g = Rn There are two man obstacles here The frst s the computaton of g, whch s unque up to a constant term The second s the choce of that constant term The followng notaton wll be used: f(x) = Ax + B j x j + C x + D (4) =1 That s, the A are the coeffcents of quadratc terms; B j are coeffcents of crossterms; and C are coeffcents of lnear terms 31 Computaton of g Suppose f : R n R Then the solutons of f(x) = 0 le on some ndmensonal hypersurface, and n(x) = f(x) gves a vector normal to the hypersurface at pont x 3
4 Let R : R n R n be a rotaton Specfcally, let r 1 r 2 R = r n Then Rn = R f = r 1 f r 2 f r n f Next, we ntegrate the result, the th component wth respect to the th coordnate That s, r 1 f dx 1 = g 1 + h 1 (x 2, x 3, x 4,, x n ) r 2 f dx 2 = g 2 + h 2 (x 1, x 3, x 4,, x n ) r k f dx k = g k + h k (x 1, x 2,, x k 1, x k+1,, x n 1 ) r n f dx n = g n + h n (x 1, x 2, x 3,, x n 1 ) Also worth notng s that, snce g s assumed to be welldefned, all the RHS s n the above system of n equatons n n unknowns happen to be equal Thus the dervatve of any term g k + h k ( ) wrt x m s known: t s nothng other than r m f: g = r m f x m We can derve the followng formula: g k = r km x m 2A k x k + = 1 k B k x + C k + = 1 k r k B k x (5) Refer to Eqn 4 for the meanngs of the coeffcents The next step s to fnd a smlar expresson for / x m (g k +h k ) (whch must be equal to ether sde of Eqn 5), cancel out terms, and arrve at an equaton for h k, whch hopefully we can solve to fnd a unque (up to a constant term) expresson for g 4
5 32 Choce of the constant term 4 Implementaton of translaton 41 Alternate representaton for quadrc hypersurfaces that mght make translatons easer Frst, I ll derve another representaton for f(x) = 0 Group the terms of f n the followng way (usng the same notaton as n Eqn 4): f(x) = A 1 x C 1 x 1 + B 12 x 1 x 2 + B 13 x 1 x 3 + B 1n x 1 x n +A 2 x C 2 x 2 + B 23 x 2 x 3 + B 24 x 2 x 4 + B 2n x 2 x n + + A n x 2 n + C n x n + D Now we can complete the square nvolvng the A and C terms: ( A x 2 + C x = A x 2 + C ) x A [ ( = A x + C ) 2 C2 2A 4A 2 Fnally we can wrte f as a sum: f(x) = ] ( = A x + C ) 2 C2 2A 4A ( A x + C ) 2 + B j x x j C2 (6) 2A 4A }{{} }{{} }{{} square and lnear terms constant terms cross terms =1 Note how ths s almost a descrpton of a quadrc hypersurface that has been translated by the vector c = (C 1 /2A 1, C 2 /2A 2,, C n /2A n ) The way n whch ths sn t the case s n the cross terms; we d lke to see somethng lke (x c ) sadly ths s not the case Can we force t to be the case? Let s try: B j (x j c j ); (x c )(x j c j ) = x x j c x c j x j + c c j Then the cross terms secton of Eqn 6 would be equvalent to the followng: B j x x j B j (c x + c j x j + c c j ) (7) } {{ } error terms 5
6 These error terms are effectvely changes that need to be made to the lnear and constant terms At the moment, however, I can t seem to wrap my mnd around ths Does t just mean we change C s and D? But then we d be usng dfferent values n the square and lnear terms sectons than we orgnally planned, whch means we re effectvely dong a dfferent translaton hmm I see no way to reconcle ths 6
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