Unit 9: Hyperplanes Juan Luis Melero ans Eduardo Eyras October 2018 1 Contents 1 Hyperplanes3 1.1 Definition.............................3 1.2 Hyperplane equations.......................4 1.3 Vector product..........................9 2 Tangent hyperplane 11 2.1 Directional derivative....................... 11 2.2 Theorems related to the tangent hyperplane.......... 12 2.2.1 Maximum value of a directional derivative....... 12 2.2.2 The gradient is orthogonal to the level curve...... 12 2.3 Definition of tangent hyperplane................. 14 2.4 Example of tangent hyperplanes and applications....... 15 2.4.1 Tangent hyperplanes for contour lines.......... 15 2.4.2 Lagrange multipliers................... 16 3 Exercises 18 2 1 Hyperplanes 1.1 Definition A hyperplane in an n−dimensional vector space Rn is defined to be the set of vectors: 0 1 x1 B . C u = @ . A xn satisfying the equation: a1x1 + ··· + anxn = b where a1; : : : ; an and b are real numbers with at least a1; : : : ; an non-zero. For example, in R2 a hyperplane is a line: Figure 1: Graphical representation of the hyperplane equation x + y = 4 And for example, in R3, a hyperplane is a 2D plane: 3 Figure 2: Graphical representation of the hyperplane equation 4x+y +z = 2 1.2 Hyperplane equations To define the hyperplane equation we need either a point in the plane and a unit vector orthogonal to the plane, two vectors lying on the plane or three coplanar points (they are contained in the hyperplane). Is is usually built from the combination of a point and a vector and it corresponds to a n − 1 dimensional vector subspace. For the first case, we will get the equation from a point and an orthogonal unit vector. The vector p − p0 lies on the plane, thus it is orthogonal to n: n ? (p − p0) =) hn; p − p0i = 0 This provides the vector equation of the plane: hn; p − p0i = 0. Given the vector: 0a1 n = @bA c 4 Figure 3: Representation of the points and orthogonal vectors of a R3 hyper- plane. p0 represents a vector indicating an specific known point in the plane. p represents a vector indicating any point in the plane. n represents a unit vector orthogonal to the plane. Performing the scalar product and taking common factors, we end up with the equation: a(x − x0) + b(y − y0) + c(z − z0) = 0 This is the scalar equation of the plane. It is also possible to write it in terms of a constant value D: D = ax0 + by0 + z0 =) ax + by + cz = D This is the general equation of the plane. As n defines the normal line to the hyperplane, we can define a normal line with the normal vector: 8 < x(t) = x0 + at r(t) = p0 + tn =) y(t) = y0 + bt : z(t) = z0 + ct 0a1 for n = @bA. c 5 Figure 4: Graphical representation of a line, which is a R2 hyperplane. v represents the vector in the direction of the hyperplane. p represents the point which the hyperplane goes through. Example: line in R2 in the direction of ~v = (1; 1) and going through the point p = (1; 0). Any point in the line is given by the equation: x = p + αv = 1 + α r = p + αv = 1 1 y = p2 + αv2 = 0 + α This is the parametric equation. We can isolate the parameter α and have the equation: x = p + αv = 1 + α r = p + αv = 1 1 ! y = x − 1 y = p2 + αv2 = 0 + α This is the intrinsic equation. We can build it as a hyperplane. Calculate a vector orthogonal to the line: p u1 1= 2 hn; vi = u1 + u2 = 0 ! n = ! n = p ; knk = 1 −u1 −1= 2 So, with the orthogonal vector and the specific known point p, we can build the equation for any point r of the hyperplane: hn; ri = hn; pi 6 p p x p p 1 1= 2 −1= 2 = 1= 2 −1= 2 ! y 0 1 1 ! p (x − y = p = y = x − 1 the same as before 2 2 If we want to build the normal line through the point, we use the normal vector and the parametric equation: p ( 1 1= 2 x(t) = x0 + t p n = p ! r(t) = p + tn ! 2 0 y(t) = y − t p1 −1= 2 0 2 Using the same point p0 = (1; 0) as before, we get: ( ) x(t) = 1 + t p1 2 ! y = 1 − x y(t) = −t p1 2 Figure 5: Graphical representation of the normal line of an R2 hyperplane. v represents the vector in the direction of the hyperplane. p represents the point which the hyperplane and the normal line go through. y = 1 − x is the equation of the normal line. 3 Example in R : consider the normal vector ~n and the vector p0 representing a given point in the plane: 0 4 1 0 0 1 ~n = @−1A ; p0 = @ 1 A 6 −7 7 We can write the equation of the hyperplane defined by these two vectors: hn; p − p0i = 0 ! ! a(x − x0) + b(y − y0) + c(z − z0) = 0 ! ! 4(x − 0) − (y − 1) + 6(z + 7) = 0 ! ! 4x − y + 6z = −43 where p represents a vector for a generic point in the plane: 0x1 p = @yA z As we said before: a(x−x0)+b(y−y0)+c(z−z0) = 0 $ ax+by+cz = D; with D = ax0+by0+cz0 There is no point (x; y; z) can fulfill two conditions: ax + by + cz = D ax + by + cz = D0 D 6= D0 So there are no common points, i.e. no intersection ! they are parallel. A hyperplane in Rn can be represented by a single equation: a1x1 + a2x2 + ··· + anxn = D 0 1 a1 B . C where n = @ . A is a vector orthogonal to the hyperplane as we chose a an point p0 in the plane such that hn; p0i = D. All planes of that similar form, for any value of D are parallel to each other. In particular, for D = 0 we have a1x1 + a2x2 + ··· + anxn = 0, which can be re-written in temps of a 1 × n matrix AT = n: 0 1 x1 a : : : a B . C 1 n @ . A = 0 ! Au = 0 xn 8 Hence, the hyperplane is the kernel of the matrix A: Ker(A). The hyper- plane is a vector subspace. In addition, considering that rank(A) = 1 and dim(Rn) = n then we see that the hyperplane has dimension n − 1: rank(A) + dim(Ker(A)) = n ! dim(Ker(A)) = n − 1 1.3 Vector product Vector product or the cross product is an operation that allows to find a vector perpendicular to two known vectors. This is useful here because to build the hyperplane equation we may need it. 3 3 Given two vectors in R , u; v 2 R and the canonical basis E3 = fe1; e2; e3g, we define the vector product of these two vectors as: 0 1 e1 e2 e3 u × v = det @u1 u2 u3A = (we solve it with the adjoint method) v1 v2 v3 u2 u3 u1 u3 u1 u2 = det e1 − det e2 + det e3 v2 v3 v1 v3 v1 v2 Properties: 1. u × v is orthogonal to both vectors 2. If u and v are linearly dependent, then u × v = 0 3. u × v = −v × u Proof of property 1: hu; u × vi = hu1e1; (u2v3 − u3v2)e1i+hu2e2; (u3v1 − u1v3)e2i+hu3e3; (u1v2 − u2v1)e3i = u1(u2v3 − u3v2) he1; e1i + u2(u3v1 − u1v3) he2; e2i + u3(u1v2 − u2v1) he3; e3i = = u1(u2v3 − u3v2) + u2(u3v1 − u1v3) + u3(u1v2 − u2v1) = = u1u2v3 − u1u3v2 + u2u3v1 − u2u1v3 + u3u1v2 − u3u2v1 = 0 Similar for hv; u × vi. 9 Proof of property 2: (recall the properties of determinants) 0 1 0 1 e1 e2 e3 e1 e2 e3 u × v = det @u1 u2 u3A = det @ u1 u2 u3 A = 0 v1 v2 v3 αu1 αu2 αu3 Proof of property 3: (recall the properties of determinants) 0 1 0 1 e1 e2 e3 e1 e2 e3 u × v = det @u1 u2 u3A = −det @v1 v2 v3A = −v × u v1 v2 v3 u1 u2 u3 Example: determine the equation of the plane in R3 that contains the points: P = (1; −2; 0) = OP Q = (3; 1; 4) = OQ R = (0; −1; 2) = OR First, we calculate two vectors in the plane: 031 0 1 1 021 PQ = OQ − OP = @1A − @−2A = @3A 4 0 4 0 0 1 0 1 1 0−11 PR = OR − OP = @−1A − @−2A = @ 1 A 2 0 2 Now we use the vector product of these two vectors to calculate an orthogonal vector: 0 1 e1 e2 e3 n = PQ × PR = det @ 2 3 4 A −1 1 2 3 4 2 4 2 3 = det − det + det = 2e − 8e + 5e 1 2 −1 2 −1 1 1 2 3 0 2 1 So the vector n = @−8A is orthogonal to the plane. It can be verified with 5 the scalar product: 021 hn; P Qi = 2 −8 5 @3A = 0 4 10 The same for PR. Now we can write the equation of the plane: hn; ri = hn; pi hn; p − p0i = 0 ! a(x − x0) + b(y − y0) + c(z − z0) = 0 Here we choose the normal vector n calculated before and any of the three initial points, since they all are contained in the plane. We will choose the 0 1 1 point P = @−2A, but any of the points would have been correct. So we get 0 the equation: 2(x − 1) − 8(y + 2) + 5(z − 0) = 0 ! 2x − 8y + 5z = 18 2 Tangent hyperplane 2.1 Directional derivative Partial derivatives, and overall the gradient vector, can be interpreted as the derivatives in the direction to the axis (the canonical basis vectors).