We denote the derivative at x by DF (x) = L. With respect to the standard bases of R n and R m, DF (x) is simply the matrix of partial derivatives,

The derivative Let O be an open subset of R n, and F : O R m a continuous function We say F is differentiable at a point x O, with derivative L, if L : R n R m is a linear transformation such that, for y R n, small, () F (x + y) = F (x) + Ly + R(x, y) with (2) R(x, y) y 0 as y 0 We denote the derivative at x by DF (x) = L With respect to the standard bases of R n and R m, DF (x) is simply the matrix of partial derivatives, ( ) F / x F / x n Fj (3) DF (x) = = x, k F m / x F m / x n so that, if v = (v,, v n ) t, (regarded as a column vector) then ( F / x k )v k k (4) DF (x)v = ( F m / x k )v k k It will be shown below that F is differentiable whenever all the partial derivatives exist and are continuous on O In such a case we say F is a C function on O More generally, F is said to be C k if all its partial derivatives of order k exist and are continuous If F is C k for all k, we say F is C In (2), we can use the Euclidean norm on R n and R m This norm is defined by (5) x = ( x 2 + + x 2 ) /2 n for x = (x,, x n ) R n Any other norm would do equally well An application of the Fundamental Theorem of Calculus, to functions of each variable x j separately, yields the following If we assume F : O R m is differentiable in each variable separately, and that each F/ x j is continuous on O, then (6) F (x + y) = F (x) + n [ F (x + zj ) F (x + z j ) ] = F (x) + j= A j (x, y) = 0 F x j ( x + zj + ty j e j ) dt, n A j (x, y)y j, j=

2 where z 0 = 0, z j = (y,, y j, 0,, 0), and {e j } is the standard basis of R n Consequently, (7) F (x + y) = F (x) + R(x, y) = j= n j= 0 F x j (x) y j + R(x, y), n { F y j (x + z j + ty j e j ) F } (x) dt x j x j Now (7) implies F is differentiable on O, as we stated below (4) Thus we have established the following Proposition If O is an open subset of R n and F : O R m is of class C, then F is differentiable at each point x O As is shown in many calculus texts, one can use the Mean Value Theorem instead of the Fundamental Theorem of Calculus, and obtain a slightly sharper result Let us give some examples of derivatives First, take n = 2, m =, and set (8) F (x) = (sin x )(sin x 2 ) Then (9) DF (x) = ((cos x )(sin x 2 ), (sin x )(cos x 2 )) Next, take n = m = 2 and ( ) x x 2 (0) F (x) = x 2 x2 2 Then ( x2 x () DF (x) = 2x 2x 2 We can replace R n and R m by more general finite-dimensional real vector spaces, isomorphic to Euclidean space For example, the space M(n, R) of real n n matrices is isomorphic to R n2 Consider the function (2) S : M(n, R) M(n, R), S(X) = X 2 We have ) (3) (X + Y ) 2 = X 2 + XY + Y X + Y 2 = X 2 + DS(X)Y + R(X, Y ), with R(X, Y ) = Y 2, and hance (4) DS(X)Y = XY + Y X

3 For our next example, we take (5) O = Gl(n, R) = {X M(n, R) : det X 0}, which, as shown below, is open in M(n, R) We consider (6) Φ : Gl(n, R) M(n, R), Φ(X) = X, and compute DΦ(I) We use the following If, for A M(n, R), (7) A = sup{ Av : v R n, v }, then (8) A, B M(n, R) AB A B, hence Y M(n, R) Y k Y k Also (9) S k = I Y + Y 2 + ( ) k Y k Y S k = S k Y = Y Y 2 + Y 3 + ( ) k Y k+ (I + Y )S k = S k (I + Y ) = I + ( ) k Y k+, hence (20) Y < = (I + Y ) = so ( ) k Y k = I Y + Y 2, k=0 (2) DΦ(I)Y = Y Related calculations show that Gl(n, R) is open in M(n, R) In fact, given X Gl(n, R), Y M(n, R), (22) X + Y = X(I + X Y ), which by (20) is invertible as long as (23) X Y < One can proceed from here to compute DΦ(X) See the exercises We return to general considerations, and derive the chain rule for the derivative Let F : O R m be differentiable at x O, as above, let U be a neighborhood of z = F (x) in R m, and let G : U R k be differentiable at z Consider H = G F We have (24) H(x + y) = G(F (x + y)) = G ( F (x) + DF (x)y + R(x, y) ) = G(z) + DG(z) ( DF (x)y + R(x, y) ) + R (x, y) = G(z) + DG(z)DF (x)y + R 2 (x, y)

4 with R 2 (x, y) y Thus G F is differentiable at x, and 0 as y 0 (25) D(G F )(x) = DG(F (x)) DF (x) Another useful remark is that, by the Fundamental Theorem of Calculus, applied to ϕ(t) = F (x + ty), (26) F (x + y) = F (x) + 0 DF (x + ty)y dt, provided F is C For a typical application, see (66) For the study of higher order derivatives of a function, the following result is fundamental Proposition 2 Assume F : O R m is of class C 2, with O open in R n Then, for each x O, j, k n, (27) x j F x k (x) = F (x) x k x j Proof It suffices to take m = We label our function f : O R For j n, we set (28) j,h f(x) = h( f(x + hej ) f(x) ), where {e,, e n } is the standard basis of R n The mean value theorem (for functions of x j alone) implies that if j f = f/ x j exists on O, then, for x O, h > 0 sufficiently small, (29) j,h f(x) = j f(x + α j he j ), for some α j (0, ), depending on x and h Iterating this, if j ( k f) exists on O, then, for x O, h > 0 sufficiently small, (30) k,h j,h f(x) = k ( j,h f)(x + α k he k ) = j,h ( k f)(x + α k he k ) = j k f(x + α k he k + α j he j ), with α j, α k (0, ) Here we have used the elementary result (3) k j,h f = j,h ( k f) We deduce the following

Proposition 3 If k f and j k f exist on O and j k f is continuous at x 0 O, then (32) j k f(x 0 ) = lim h 0 k,h j,h f(x 0 ) Clearly (33) k,h j,h f = j,h k,h f, so we have the following, which easily implies Proposition 2 Corollary 4 In the setting of Proposition 3, if also j f and k j f exist on O and k j f is continuous at x 0, then (34) j k f(x 0 ) = k j f(x 0 ) 5 We now describe two convenient notations to express higher order derivatives of a C k function f : Ω R, where Ω R n is open In one, let J be a k-tuple of integers between and n; J = (j,, j k ) We set (35) f (J) (x) = jk j f(x), j = x j We set J = k, the total order of differentiation As we have seen in Proposition 2, i j f = j i f provided f C 2 (Ω) It follows that, if f C k (Ω), then jk j f = lk l f whenever {l,, l k } is a permutation of {j,, j k } Thus, another convenient notation to use is the following Let α be an n-tuple of non-negative integers, α = (α,, α n ) Then we set (36) f (α) (x) = α α n n f(x), α = α + + α n Note that, if J = α = k and f C k (Ω), (37) f (J) (x) = f (α) (x), with α i = #{l : j l = i} Correspondingly, there are two expressions for monomials in x = (x,, x n ): (38) x J = x j x jk, x α = x α xα n n, and x J = x α provided J and α are related as in (37) Both these notations are called multi-index notations We now derive Taylor s formula with remainder for a smooth function F : Ω R, making use of these multi-index notations We will apply the one variable formula (052) (053), ie, (39) ϕ(t) = ϕ(0) + ϕ (0)t + 2 ϕ (0)t 2 + + k! ϕ(k) (0)t k + r k (t),

6 with (40) r k (t) = k! t 0 (t s) k ϕ (k+) (s) ds, given ϕ C k+ (I), I = ( a, a) (See the exercises, and also Appendix D for further discussion) Let us assume 0 Ω, and that the line segment from 0 to x is contained in Ω We set ϕ(t) = F (tx), and apply (39) (40) with t = Applying the chain rule, we have (4) ϕ (t) = Differentiating again, we have (42) ϕ (t) = n j F (tx)x j = F (J) (tx)x J j= J =, K = J = F (J+K) (tx)x J+K = J =2 F (J) (tx)x J, where, if J = k, K = l, we take J + K = (j,, j k, k,, k l ) Inductively, we have (43) ϕ (k) (t) = F (J) (tx)x J Hence, from (39) with t =, or, more briefly, F (x) = F (0) + J = (44) F (x) = where (45) R k (x) = k! J =k F (J) (0)x J + + k! J k J =k+ J =k J! F (J) (0)x J + R k (x), ( 0 F (J) (0)x J + R k (x), ) ( s) k F (J) (sx) ds x J This gives Taylor s formula with remainder for F C k+ (Ω), in the J-multi-index notation We also want to write the formula in the α-multi-index notation We have (46) F (J) (tx)x J = ν(α)f (α) (tx)x α, where J =k α =k (47) ν(α) = #{J : α = α(j)},

and we define the relation α = α(j) to hold provided the condition (37) holds, or equivalently provided x J = x α Thus ν(α) is uniquely defined by (48) α =k ν(α)x α = J =k x J = (x + + x n ) k One sees that, if α = k, then ν(α) is equal to the product of the number of combinations of k objects, taken α at a time, times the number of combinations of k α objects, taken α 2 at a time, times the number of combinations of k (α + + α n ) objects, taken α n at a time Thus ( )( ) ( ) k k α k α α n (49) ν(α) = = α In other words, for α = k, (50) ν(α) = k! α!, where α! = α! α n! Thus the Taylor formula (44) can be rewritten α 2 (5) F (x) = where (52) R k (x) = α =k+ α k k + α! α n α! F (α) (0)x α + R k (x), ( 0 ) ( s) k F (α) (sx) ds x α k! α!α 2! α n! The formula (5) (52) holds for F C k+ It is significant that (5), with a variant of (52), holds for F C k In fact, for such F, we can apply (52) with k replaced by k, to get (53) F (x) = with (54) R k (x) = α k α =k k α! α! F (α) (0)x α + R k (x), ( 0 ) ( s) k F (α) (sx) ds x α We can add and subtract F (α) (0) to F (α) (sx) in the integrand above, and obtain the following 7

8 Proposition 5 If F C k on a ball B r (0), the formula (5) holds for x B r (0), with (55) R k (x) = α =k k α! ( 0 ( s) k [ F (α) (sx) F (α) (0) ] ) ds x α Remark Note that (55) yields the estimate (56) R k (x) α =k x α α! sup F (α) (sx) F (α) (0) 0 s The term corresponding to J = 2 in (44), or α = 2 in (5), is of particular interest It is (57) 2 J =2 F (J) (0) x J = 2 n j,k= 2 F x k x j (0) x j x k We define the Hessian of a C 2 function F : O R as an n n matrix: ( (58) D 2 2 ) F F (y) = (y) x k x j Then the power series expansion of second order about 0 for F takes the form (59) F (x) = F (0) + DF (0)x + 2 x D2 F (0)x + R 2 (x), where, by (56), (60) R 2 (x) C n x 2 sup 0 s, α =2 F (α) (sx) F (α) (0) In all these formulas we can translate coordinates and expand about y O For example, (59) extends to (6) F (x) = F (y) + DF (y)(x y) + 2 (x y) D2 F (y)(x y) + R 2 (x, y), with (62) R 2 (x, y) C n x y 2 sup 0 s, α =2 F (α) (y + s(x y)) F (α) (y) Example If we take F (x) as in (8), so DF (x) is as in (9), then ( ) D 2 sin x sin x F (x) = 2 cos x cos x 2 cos x cos x 2 sin x sin x 2

9 The results (6) (62) are useful for extremal problems, ie, determining where a sufficiently smooth function F : O R has local maxima and local minima Clearly if F C (O) and F has a local maximum or minimum at x 0 O, then DF (x 0 ) = 0 In such a case, we say x 0 is a critical point of F To check what kind of critical point x 0 is, we look at the n n matrix A = D 2 F (x 0 ), assuming F C 2 (O) By Proposition 2, A is a symmetric matrix A basic result in linear algebra is that if A is a real, symmetric n n matrix, then R n has an orthonormal basis of eigenvectors, {v,, v n }, satisfying Av j = λ j v j ; the real numbers λ j are the eigenvalues of A We say A is positive definite if all λ j > 0, and we say A is negative definite if all λ j < 0 We say A is strongly indefinite if some λ j > 0 and another λ k < 0 Equivalently, given a real, symmetric matrix A, (63) for some C > 0, all v R n, and (64) A positive definite v Av C v 2, A negative definite v Av C v 2, A strongly indefinite v, w R n, nonzero, such that v Av C v 2, w Aw C w 2, for some C > 0 In light of (44) (45), we have the following result Proposition 6 Assume F C 2 (O) is real valued, O open in R n Let x 0 O be a critical point for F Then (i) D 2 F (x 0 ) positive definite F has a local minimum at x 0, (ii) D 2 F (x 0 ) negative definite F has a local maximum at x 0, (iii) D 2 F (x 0 ) strongly indefinite F has neither a local maximum nor a local minimum at x 0 In case (iii), we say x 0 is a saddle point for F The following is a test for positive definiteness Proposition 7 Let A = (a ij ) be a real, symmetric, n n matrix For l n, form the l l matrix A l = (a ij ) i,j l Then (65) A positive definite det A l > 0, l {,, n} Regarding the implication, note that if A is positive definite, then det A = det A n is the product of its eigenvalues, all > 0, hence is > 0 Also in this case, it follows from (65) that each A l must be positive definite, hence have positive determinant, so we have The implication is easy enough for 2 2 matrices If A is symmetric and det A > 0, then either both its eigenvalues are positive (so A is positive definite) or both are negative (so A is negative definite) In the latter case, A = (a ) must be negative, so we have in this case We prove for n 3, using induction The inductive hypothesis implies that if det A l > 0 for each l n, then A n is positive definite The next lemma then guarantees that A = A n has at least n positive eigenvalues The hypothesis that det A > 0 does not allow that the remaining eigenvalue be 0, so all the eigenvalues of A must be positive Thus Proposition 7 is proven, once we have the following

0 Lemma 8 In the setting of Proposition 7, if A n is positive definite, then A = A n has at least n positive eigenvalues Proof Since A is symmetric, R n has an orthonormal basis v,, v n of eigenvectors of A; Av j = λ j v j If the conclusion of the lemma is false, at least two of the eigenvalues, say λ, λ 2, are 0 Let W = Span(v, v 2 ), so w W = w Aw 0 Since W has dimension 2, R n R n satisfies R n W 0, so there exists a nonzero w R n W, and then w A n w = w Aw 0, contradicting the hypothesis that A n is positive definite Remark Given (65), we see by taking A A that if A is a real, symmetric n n matrix, (66) A negative definite ( ) l det A l > 0, l {,, n} We return to higher order power series formulas with remainder and complement Proposition 5 Let us go back to (39) (40) and note that the integral in (40) is /(k + ) times a weighted average of ϕ (k+) (s) over s [0, t] Hence we can write r k (t) = (k + )! ϕ(k+) (θt), for some θ [0, ], if ϕ is of class C k+ This is the Lagrange form of the remainder; see Appendix D for more on this, and for a comparison with the Cauchy form of the remainder If ϕ is of class C k, we can replace k + by k in (39) and write (67) ϕ(t) = ϕ(0) + ϕ (0)t + + (k )! ϕ(k ) (0)t k + k! ϕ(k) (θt)t k, for some θ [0, ] Pluging (67) into (43) for ϕ(t) = F (tx) gives (68) F (x) = J k J! F (J) (0)x J + k! J =k F (J) (θx)x J, for some θ [0, ] (depending on x and on k, but not on J), when F is of class C k on a neighborhood B r (0) of 0 R n Similarly, using the α-multi-index notation, we have, as an alternative to (53) (54), (69) F (x) = α k α! F (α) (0)x α + α =k α! F (α) (θx)x α,

for some θ [0, ] (depending on x and on α, but not on α), if F C k (B r (0)) Note also that (70) 2 J =2 F (J) (θx)x J = 2 n j,k= 2 F x k x j (θx)x j x k = 2 x D2 F (θx)x, with D 2 F (y) as in (58), so if F C 2 (B r (0)), we have, as an alternative to (59), (7) F (x) = F (0) + DF (0)x + 2 x D2 F (θx)x, for some θ [0, ] We next complemant the multi-index notations for higher derivatives of a function F by a multi-linear notation, defined as follows If k N, F C k (U), and y U R n, set (72) D k F (y)(u,, u k ) = t tk F (y + t u + + t k u k ), t = =t k =0 for u,, u k R n For k =, this formula is equivalent to the definition of DF given at the beginning of this section For k = 2, we have (73) D 2 F (y)(u, v) = u D 2 F (y)v, with D 2 F (y) as in (58) Generally, (72) defines D k F (y) as a symmetric, k-linear form in u,, u k R n We can relate (72) to the J-multi-index notation as follows We start with (74) t F (y + t u + + t k u k ) = and inductively obtain (75) t tk F (y + Σt j u j ) = hence (76) D k F (y)(u,, u k ) = In particular, if u = = u k = u, J = = J k = J = = J k = J = F (J) (y + Σt j u j )u J, F (J + +J k ) (y + Σt j u j )u J uj k k, F (J + +J k ) (y)u J uj k k (77) D k F (y)(u,, u) = F (J) (y)u J J =k

2 Hence (68) yields (78) F (x) = F (0)+DF (0)x+ + (k )! Dk F (0)(x,, x)+ k! Dk F (θx)(x,, x), for some θ [0, ], if F C k (B r (0)) In fact, rather than appealing to (68), we can note that ϕ(t) = F (tx) = ϕ (k) (t) = t tk ϕ(t + t + + t k ) = D k F (tx)(x,, x), and obtain (78) directly from (67) We can also use the notation D j F (y)x j = D j F (y)(x,, x), t = =t k =0 with j copies of x within the last set of parentheses, and rewrite (78) as (79) F (x) = F (0) + DF (0)x + + Note how (78) and (79) generalize (7) (k )! Dk F (0)x (k ) + k! Dk F (θx)x k Exercises Consider the following function f : R 2 R: f(x, y) = (cos x)(cos y) Find all its critical points, and determine which of these are local maxima, local minima, and saddle points 2 Let M(n, R) denote the space of real n n matrices Assume F, G : M(n, R) M(n, R) are of class C Show that H(X) = F (X)G(X) defines a C map H : M(n, R) M(n, R), and DH(X)Y = DF (X)Y G(X) + F (X)DG(X)Y 3 Let Gl(n, R) M(n, R) denote the set of invertible matrices Show that Φ : Gl(n, R) M(n, R), Φ(X) = X is of class C and that DΦ(X)Y = X Y X 4 Identify R 2 and C via z = x+iy Then multiplication by i on C corresponds to applying ( ) 0 J = 0

Let O R 2 be open, f : O R 2 be C Say f = (u, v) Regard Df(x, y) as a 2 2 real matrix One says f is holomorphic, or complex-analytic, provided the Cauchy-Riemann equations hold: u x = v y, Show that this is equivalent to the condition Generalize to O open in C m, f : O C n u y = v x Df(x, y) J = J Df(x, y) 5 Let f be C on a region in R 2 containing [a, b] {y} Show that, as h 0, 3 [ ] f f(x, y + h) f(x, y) (x, y), h y uniformly on [a, b] {y} Hint Show that the left side is equal to h f (x, y + s) ds, h 0 y and use the uniform continuity of f/ y on [a, b] [y δ, y + δ]; cf Proposition A5 6 In the setting of Exercise 5, show that d b b f f(x, y) dx = (x, y) dx dy a a y 7 Considering the power series show that f(x) = f(y) + f (y)(x y) + + f (j) (y) (x y) j + R j (x, y), j! R j y = j! f (j+) (y)(x y) j, R j (x, x) = 0 Use this to re-derive (053), and hence (22) (23) We define big oh and little oh notation: f(x) = O(x) (as x 0) f(x) C x as x 0, f(x) = o(x) (as x 0) f(x) 0 x as x 0

4 8 Let O R n be open and y O Show that f C k+ (O) f(x) = α k f C k (O) f(x) = α k α! f (α) (y)(x y) α + O( x y k+ ), α! f (α) (y)(x y) α + o( x y k ) Exercises 9 deal with properties of the determinant, as a differentiable function on spaces of matrices Foundational work on determinants is developed in an accompanying set of auxiliary exercises on determinants 9 Let M n n be the space of complex n n matrices, det : M n n C the determinant Show that, if I is the identity matrix, D det(i)b = Tr B, ie, d dt det(i + tb) t=0 = Tr B 0 If A(t) = ( a jk (t) ) is a curve in M n n, use the expansion of (d/dt) det A(t) as a sum of n determinants, in which the rows of A(t) are successively differentiated, to show that, for A M n n, D det(a)b = Tr ( Cof(A) t B), where Cof(A) is the cofactor matrix of A Suppose A M n n is invertible Using show that det(a + tb) = (det A) det(i + ta B), D det(a)b = (det A) Tr(A B) Comparing the result of Exercise 0, deduce Cramer s formula: (det A)A = Cof(A) t 2 Assume G : U O, F : O Ω Show that F, G C = F G C (Hint Use (7)) Show that, for any k N, F, G C k = F G C k

5 3 Show that the map Φ : Gl(n, R) Gl(n, R) given by Φ(X) = X is C k for each k, ie, Φ C Hint Start with the material of Exercise 3 Auxiliary exercises on determinants If M n n denotes the space of n n complex matrices, we want to show that there is a map (79) det : M n n C which is uniquely specified as a function ϑ : M n n C satisfying: (a) ϑ is linear in each column a j of A, (b) ϑ(ã) = ϑ(a) if à is obtained from A by interchanging two columns (c) ϑ(i) = Let A = (a,, a n ), where a j are column vectors; a j = (a j,, a nj ) t Show that, if (a) holds, we have the expansion (80) det A = a j det (e j, a 2,, a n ) = j = j,,j n a j a jn n det (e j, e j2,, e jn ), where {e,, e n } is the standard basis of C n 2 Show that, if (b) and (c) also hold, then (8) det A = (sgn σ) a σ() a σ(2)2 a σ(n)n, σ S n where S n is the set of permutations of {,, n}, and (82) sgn σ = det (e σ(),, e σ(n) ) = ± To define sgn σ, the sign of a permutation σ, we note that every permutation σ can be written as a product of transpositions: σ = τ τ ν, where a transposition of {,, n} interchanges two elements and leaves the rest fixed We say sgn σ = if ν is even and sgn σ = if ν is odd It is necessary to show that sgn σ is independent of the choice of such a product representation (Referring to (82) begs the question until we know that det is well defined)

6 3 Let σ S n act on a function of n variables by (83) (σf)(x,, x n ) = f(x σ(),, x σ(n) ) Let P be the polynomial (84) P (x,, x n ) = Show that j<k n (85) (σp )(x) = (sgn σ) P (x), and that this implies that sgn σ is well defined (x j x k ) 4 Deduce that there is a unique determinant satisfying (a) (c), and that it is given by (8) 5 Show that (8) implies (86) det A = det A t Conclude that one can replace columns by rows in the characterization (a) (c) of determinants Hint a σ(j)j = a lτ(l) with l = σ(j), τ = σ Also, sgn σ = sgn τ 6 Show that, if (a) (c) hold (for rows), it follows that (d) ϑ(ã) = ϑ(a) if à is obtained from A by adding cρ l to ρ k, for some c C, where ρ,, ρ n are the rows of A Re-prove the uniqueness of ϑ satisfying (a) (d) (for rows) by applying row operations to A until either some row vanishes or A is converted to I 7 Show that (87) det (AB) = (det A)(det B) Hint For fixed B M n n, compare ϑ (A) = det(ab) and ϑ 2 (A) = (det A)(det B) For uniqueness, use an argument from Exercise 6 8 Show that (88) det a 2 a n 0 0 0 a 22 a 2n = det 0 a 22 a 2n = det A 0 a n2 a nn 0 a n2 a nn

where A = (a jk ) 2 j,k n Hint Do the first identity by the analogue of (d), for columns Then exploit uniqueness for det on M (n ) (n ) 9 Deduce that det(e j, a 2,, a n ) = ( ) j det A j where A kj is formed by deleting the k th column and the j th row from A 0 Deduce from the first sum in (80) that (89) det A = More generally, for any k {,, n}, (90) det A = n ( ) j a j det A j j= n ( ) j k a jk det A kj j= This is called an expansion of det A by minors, down the k th column By definition, the cofactor matrix of A is given by Cof(A) jk = ( ) j k det A kj 7 Show that (9) det Hint Use (88) and induction a a 2 a n a 22 a 2n a nn = a a 22 a nn Auxiliary exercises on the cross product If u, v R 3, show that the formula (92) w (u v) = det w u v w 2 u 2 v 2 w 3 u 3 v 3 for u v = Π(u, v) defines uniquely a bilinear map Π : R 3 R 3 R 3 Show that it satisfies i j = k, j k = i, k i = j,

8 where {i, j, k} is the standard basis of R 3 2 We say T SO(3) provided that T is a real 3 3 matrix satisfying T t T = I and det T > 0, (hence det T = ) Show that (93) T SO(3) = T u T v = T (u v) Hint Multiply the 3 3 matrix in Exercise on the left by T 3 Show that, if θ is the angle between u and v in R 3, then (94) u v = u v sin θ Hint Check this for u = i, v = ai + bj, and use Exercise 2 to show this suffices 4 Show that κ : R 3 Skew(3), the set of antisymmetric real 3 3 matrices, given by (95) κ(y, y 2, y 3 ) = satisfies 0 y 3 y 2 y 3 0 y y 2 y 0 (96) Kx = y x, K = κ(y) Show that, with [A, B] = AB BA, (97) κ(x y) = [ κ(x), κ(y) ], Tr ( κ(x)κ(y) t) = 2x y