Linear Algebra. Paul Yiu. 6D: 2-planes in R 4. Department of Mathematics Florida Atlantic University. Fall 2011

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1 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall D: 2-planes in R 4

2 The angle between a vector and a plane The angle between a vector v R n and a subspace V is the angle between v and its orthogonal projection onto the subspace. (1) If e 1,..., e n is an orthonormal basis of R n such that e 1,..., e r span the subspace V, then the angle θ between v = n i=1 a ie i and V is given by r cos 2 i=1 θ = a2 i n. i=1 a2 i (2) The angle between v R 3 and the plane ax + by + cz = 0 is the complement of the angle between v and the normal of the plane, the vector (a, b, c). (3) The angle between two planes in R 3 is the angle between their normals. 1

3 Two 2-planes in general positions in R 4 We consider the problem of finding the angle(s) between two 2-planes A and B in R 4. It is enough to assume A B = {0}, for otherwise, the sum of two 2-planes is a 3-dimensional subspace of R 4. 2

4 Equation of a 2-plane in R 4 Let e 1, e 2, e 3, e 4 be the standard orthonormal basis of R 4, O = Span(e 1, e 2 ), and O = Span(e 3, e 4 ). For a point (x 1, x 2, x 3, x 4 ) R 4, we write ( ) x1 x = and y = x 2 The plane O has equation y = 0, while the plane O has equation x = 0. ( x3 x 4 ). 3

5 Equation of a 2-plane in R 4 Proposition. A 2-plane in R 4 which intersects O only in 0 has equation y = Ax for some 2 2 matrix A. Proof. Consider a 2-plane A in R 4 as the intersection of two hyperplanes We rewrite this in the form for 2 2 matrices C and D. If A O = {0}, then the system a 1 x 1 + a 2 x 2 + a 3 x 3 + a 4 x 4 = 0, b 1 x 1 + b 2 x 2 + b 3 x 3 + b 4 x 4 = 0. Cx + Dy = 0 Cx + Dy = 0, x = 0 has only zero solution, and D must be invertible. Therefore, the equation of A can be reorganized as y = D 1 Cx or simply y = Ax for some 2 2 matrix A. 4

6 Orthogonal projections Consider two 2-planes with equations A : y = Ax and B : y = Bx for 2 2 matrices A and B. Proposition. The orthogonal projection of a vector (u, Au) A in the n-plane B is the vector (v, Bv) with v = (I + B t B) 1 (I + B t A)u. Proof. The orthogonal projection (v, Bv) of (u, Au) is such that (u, Au) (v, Bv) = (u v, Au Bv) is orthogonal to every vector in B, namely, (w, Bw) for arbitrary 2 1 matrix w, This means (w, Bw), (u v, Au Bv) = 0. 0 = w t (u v) + (Bw) t (Au Bv) = w t (u v + B t (Au Bv)) for arbitrary w. Therefore, u v + B t (Au Bv) = 0 = (I + B t A)u (I + B t B)v = 0. The matrix I + B t B is positive definite, and must be invertible. From this the result follows. 5

7 Angle between two 2-planes Proposition. The orthogonal projection of a vector (u, Au) A in the n-plane B is the vector (v, Bv) with v = (I + B t B) 1 (I + B t A)u. Corollary. The angle θ between (u, Au) A and its projection (v, Bv) B is given by cos 2 θ = ut Qu u t Pu, where Q = (I + A t B)(I + B t B) 1 (I + B t A) and P = I + AA t. Proof. This is cos 2 θ = vt (I + B t B)v u t (I + A t A)u, and the result follows from substitution. 6

8 Advanced calculus We consider the critical values of a ratio of two quadratic forms: f(u) = ut Qu u t Pu where P and Q are symmetric 2 2 matrices, and P positive definite. Note that f(u) is homogeneous of degree 0: f(λu) = f(u) for any nonzero λ R. Therefore, we may regard f as a differentiable real valued function on the unit circle. It has a maximum and a minimum. 7

9 Advanced calculus Lemma. Differentiation of u t (Q fp)u = 0 gives ) 2(Q fp)u u t Pu ( f u 1 f u 2 = 0. Proof. Differentiating u t (Q fp)u = 0 with respect to u 1 using the product rule, we have ( ) u1 (1 0)(Q fp) u 2 ( ) ( ) f u1 (u 1 u 2 ) P u 1 u 2 ( ) 1 + (u 1 u 2 )(Q fp) = 0. 0 Since each of these terms is a scalars, the first and the last terms on the left side are equal, and we have ( ) ( ) ( ) u1 f u1 2(1 0)(Q fp) (u 1 u 2 ) P = 0. u 1 u 2 u 2 8

10 ... 2(1 0)(Q fp) ( u1 u 2 ) ( ) ( ) f u1 (u 1 u 2 ) P = 0. u 1 u 2 Similarly, 2(0 1)(Q fp) ( u1 u 2 Combining these two equations, we have 2 ) ( ) ( ) f u1 (u 1 u 2 ) P = 0. u 2 ( ) 1 0 (Q fp)u u t Pu 0 1 ( f u 1 f u 2 ) u 2 = 0. 9

11 Critical values and critical points Lemma. Differentiation of u t (Q fp)u = 0 gives ) 2(Q fp)u u t Pu ( f u 1 f u 2 = 0. Proposition. A critical value f of f(u) := ut Qu u t Pu is a root of det(q xp), and this occurs at u for which (Q fp)u = 0. Proof. At a critical point u, f u 1 = 0 and f u 2 = 0. Therefore, 2(Q fp)u = 0. This is possible (with a nonzero u) only if det(q fp) = 0, and the corresponding u is a nontrivial solution of the linear homogeneous system represented by the matrix Q fp. 10

12 Back to two 2-planes in R 4 There are in general two critical angles between two 2-planes in R 4 in general positions. For ( the angles ) between the two planes O : y = 0 and A : y = Ax where 1 2 A =, 2 1 note that P = I and (( ) 1 0 Q = ( ) ( )) = 2 1 ( ) = ( ) The squared cosines of the two angles between the two 2-planes are the eigenvalues of ( ) 1 3 2, which are 1 5 and 1. The two critical angles are arccos and arccos

13 Isoclinic 2-planes in R 4 When the two critical values of f(u) = ut Qu u t Pu coincide, f is a constant function. In this case, the angle between a nonzero vector (u, Au) A and its orthogonal projection in B is constant. We say that the two 2-planes are isoclinc. Proposition. if and only if The two 2-planes A : y = Ax and B : y = Bx are isoclinic for some f R. (I + A t B)(I + B t B) 1 (I + B t A) = f(i + A t A) 12

14 Isoclinic 2-planes in R 4 When the two critical values of f(u) = ut Qu u t Pu coincide, f is a constant function. In this case, the angle between a nonzero vector (u, Au) A and its orthogonal projection in B is constant. We say that the two 2-planes are isoclinc. What are the 2-planes isoclinic with O : y = 0? A : y = Ax is isoclinic with O if and only if (I 2 + A t A) 1 = fi 2 = A t A = (f 1 1)I 2. Therefore, A is a scalar multiple of an orthogonal 2 2 matrix. ( ) ( ) cosα sin α cosα sin α A = k or A = k sinα cosα sin α cosα for some k, α R. 13