ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS.

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1 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. Please review basic linear algebra, specifically the notion of spanning, of being linearly independent and of forming a basis as applied to a finite list of vectors. Also review the defintion of complement of a subspace, and direct sum of two subspaces D. In the context of linear algebra a plane is a two-dimensional real vector space. A basis for a plane consists of any two vectors E 1, E 2 which span the plane. Spanning means that any vector v in the plane can be written as v = ue 1 + ve 2 with u, v R. It is a theorem that for two vectors in the plane spanning is equivalent to being linearly independent. The linearly independent part implies that this expression for v is unique. Thus a choice of basis gives us coordinates on the plane. Any linear transformation T of the plane to itself is uniquely determined by where it sends a basis. Suppose that E 1 U 1 and E 2 U 2 Then ue 1 + ve 2 uu 1 + vu 2, i.e T (ue 1 + ve 2 ) = uu 1 + vu 2. Usually we fixate on a standard basis which we are taught to identify with (1, 0) and (0, 1) or i and j or e 1 and e 2. If U 1 = ae 1 + ce 2, U 2 = be 1 + de 2 then the matrix of this linear transformation is ( a b c d ) In other words: write out the image of the basis as column vectors. That gives the matrix of the transformation. Now we suppose our plane has a dot product and we write the dot product of two vectors V, W as V, W. Recall that a basis U 1, U 2 is called orthonormal if U 1, U 1 = 1, U 1, U 2 = 0 and U 2, U 2 = 0. Proposition 1. The linear transformation T R 2 R 2 of the plane is an isometry iff it takes an orthonormal basis E 1, E 2 to an orthonormal basis U 1, U 2. orthonormal Proof: We have all computed distances rel an orthonormal basis. Write P = xe 1 + ye 2 then P, P = x 2 + y 2 which is d(0, P ) 2, the distance of P squared from the origin. But d(0, T (P )) 2 = T (P ), T (P ) = xe 1 + ye 2, xe 1 + ye 2 = x 2 + y 2, again because U 1, U 2 are orthonormal. We leave it to the reader to verify that other distances are preserved, and that if U 1, U 2 is not orthonormal then distances are not preserved. 1. Higher dimensions. Quadratic forms. A quadratic form on a real vector space V is a map Q V R which is homogeneous quadratic: Q(λv) = λ 2 Q(v) whenever v V and λ R and, moreover, if 1

2 2 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. when written relative to a basis is a quadratic polynomial: (1) Q(Σx i E i ) = ΣQ ij x i x j EXERCISE. If Q is a quadratic polynomial rel to one basis, then it is a quadratic polynomial rel to any basis. EXERCISE: Verify that replace the matrix Q ij by the symmetric matrix 1 2 (Q ij + Q ji ) does not alter the quadratic form. This we may assume that the matrix Q ij of the quadratic form rel. our basis is a symmetric matrix. EXAMPLES.. Q(x, y) = xy is a quadratic form on V = R 2. Q(x, y) = ( x 4 + y 4 ) 1/2 is homogeneous quadratic but is not a quadratic form since it is not a quadratic polynomial. Here is the basic structure theorem of quadratic forms : Theorem 1. [BASIC STRUCTURE THEOREM aka Sylvester s theorem ] For any real quadratic form Q on the real vector space V there exists a basis U i, i = 1,..., n (ie a linear change of coordinates!) such that (2) Q(x 1 U 1 + x 2 + U x n U n ) = Σɛ i x 2 i where the constants ɛ i are either 1, 1 or 0. In other words, the theorem asserts that there is a linear invertible change of coordinates such that in the new coordinates the matrix of Q is diagonal with only 1 s, 1 s and 0 s on the diagonal. Definition 1. Call a basis U i as in the theorem an orthonormal basis, by abuse of notation. Then the number of 1 s, 0s and 1 s occuring in the normal form is independent of the choice of orthonormal basis. EXERCISE. Find the normal form and an orthonormal basis for the quadratic form Q = xy on the plane. SOLUTION. Set x = u + v, y = u v to convert Q to u 2 v 2. (What is the change of basis to an orthonormal basis? ) Definition 2. The number of 0 s is called the nullity. The number of 1 s is called the index. The form is called positive definite if all the ɛ i are +1 and it is called negative definite if all the ɛ i = 1. EXAMPLE. Thus the quadratic form xy on the plane has index 1 and nullity 0. Proposition 2. The nullity of Q is the dimension of the kernel of the matrix Q ij (rel. to any basis). The index of Q equals the maximum dimension, dim(s), over all subspaces S V for which the restriction of Q is negative definite. Sylvester s Law. Another was to say the basic structure theorem is that a real quadratic form Q on a real vector space V is determined up to linear isomorphism by 3 integers: dim(v), null(q), ind(q), the dimension of V and the nullity and index of the quadratic form. Thus any two quadratic forms (possibly on different vector spaces) for which these three integers are equal can be brought one to the other by an invertible linear map. Associated bilinear form. Let us pretend that Q(v) = v, v. Then Q(v + w) = Q(v)+2 v, w +Q(w) and Q(v w) = Q(v) 2 v, w +Q(w). Subtracting these two identities and dividing by 4 we find that v, w = 1 (Q(v + w) Q(v w)). We 4 can turn this pretention around.

3 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. 3 Definition 3. The associated bilinear form to Q is defined by β(v, w) = 1 4 (Q(v + w) Q(v w)). Proposition 3. β is a bilinear symmetric form on V ; i.e. it has all the algebraic properties that an inner product has except possibly for positive-definiteness. Thus β V V R is linear in each V-slot, is symmetric: β(v, w) = β(w, v) and recovers Q by Q(v) = β(v, v). Sketch proof of Proposition. Expand out in a basis. Compute β(e i, E j ) = 1 2 (Q ij + Q ji ). Now a bilinear form β defines a map Φ β V V by defining Φ β (v) V R by Φ β (v)(w) = β(v, w) for all w V. RECALL: the dual space of V is the vector space consisting of all linear functions from V to R. EXERCISE. Q is nondegenerate if and only if Φ β is an isomorphism. EXERCISE. The rank of Q is the dimension of the kernel of Φ β. MATRIX COEFFICIENTS ARE DOT PRODUCTS. The bilinear form gives us a better understanding of the matrix coefficients in eq XXWW: Indeed Q(Σx i E i ) = β(σ i x i E i, Σ j x j E j ) = Σ ij x i x j β(e i, E j ) so read off that Q ij = β(e i, E j ) How the matrix of Q changes under change of coordinates. We will show that Q P QP t under a change of coordinates, where P is an invertible matrix encoding the change of coordinates. Proof. Let U i = ΣR j i E j be another basis of V so that the square matrix R j i is invertible. Let P j i be its inverse so that E i = ΣP j i U j, Since V i is a basis any vector can be uniquely expanded as v = Σy i U i. For book-keeping purposes it will work better if we use the standard convention of tensor calculus that coordinates have up indices and vectors down indices. So instead of x i in eq eq?? we use x i and instead of y i we use y i. Thus, v = Σy i U i = Σx i E i. Now (3) (4) (5) (6) v = Σy i U i = Σ i,j y i P j i E j = Σ j (Σy i P j i )E j = Σ j x j E j Since the coordinates of a vector are unique we get x j = Σ i P j i yi. Substituting this change of variables into our expansion eq?? we find that (7) (8) (9) or Q(v) = Σ i,j x i Q ij x j = (Σ i,j (Σ k P i ky k )Q ij Σ m P j my m = Σ k,m y k (Σ i,j P i kq ij P j m)y m Q(v) = Σ k,m y k Qk,m y m where the matrix in the new coordinates y is Q = P QP t

4 4 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS.. A standard alternate way to write this is to think of the coordinates x as forming a column vector of length n and writing [Q] x for our symmetric matrix Q ij. Then we have Q(v) = x t [Q] x x, x = P t y, x t = y t P and so Q(v) = y t P [Q]P t y, showing that [Q] y = P [Q] x P t. Thus the theorem asserts that given any symmetric matrix Q we can find an invertible matrix P such that P QP t is diagonal with diagonal entries ±1 or Proof of the basic theorem, theorem??. We will first show that there is a matrix U i with respect to which the matrix of Q is diagonal Q(Σy i U i ) = Σd i (y i ) 2. This condition is equivalent to the assertion that there is a basis such that β(u i, U j ) = 0 whenever i j. (The diagonal entries are d i = β(u i, U i ). Definition 4. A basis U i for V such that β(u i, U j ) = 0 is called an orthogonal basis. If Q(v) = 0 for all v, then we are done: all the ɛ i = 0. Q is the zero matrix. So assume that there is a vector v with Q(v) 0. We will split V into Rv and v, where we use β to define the orthogonal complement to E 1. Definition 5. If v V then v = {w V β(v, w) = 0}. More generally if S V is a linear subspace then S = {v V β(v, s) = 0 for all s S}. Proposition 4. If β(v, v) 0 then V = Rv v. Recall the meaning of S 1 S 2 for S 1, S 2 V linear subsapces. We say that V = S 1 S 2 if the two conditions S 1 S 2 = {0} and S 1 + S 2 = V hold. These two conditions are equivalent to the condition that any vector v V can be expressed uniquely in the form v = s 1 + s 2 with s 1 S 1 and s 2 S 2. We leave the proof of the propositon up to you, reader. Now observe that if we set U 1 = v and let U i, i > 1 be a basis for v then β(u 1, U i ) = 0, i 1. If we can insure the U i basis is orthogonal we are done. The proof that we can diagonalize Q is now easily done by induction, since the process of taking v drops dimension by 1 as long as β(v, v) 0. See Artin, Algebra, chapter 7, section 2, esp. p To finish the proof of the theorem, normalize each of the U i for which d i = Q(U i ) 0 by setting E i = U i / d i if d i > 0 and by E i = U i / d i if d i < 0. (Leave those U i with Q(U i ) = 0 alone. They span the kernel of Q.) Observe that the new basis E i is orthonormal. 2. Isometries Let Q be a nondegenerate quadratic form. Definition 6. A linear isometry of Q is a linear map T V V such that Q(T (v)) = Q(v) for all v V.

5 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. 5 Lemma 1. (i) T V V is a linear isometry if and only if it preserves the corresponding bilinear form: β(t v, T w) = β(v, w) for all v, w V (ii) The linear isometries from a group under composition. (iii) If [Q] is the (symmetric) matrix of the form rel to a basis and [T ] is the matrix of the linear transformation R then T is a linear isometry of Q iff [T ][Q][T ] t = [Q]. Proof of (ii): If S, T are linear isometries and v V then Q(S(T (v)) = Q(T (v)) = Q(v) for all v V. Hence the linear isometries are closed under composition. And they are invertible by the matrix formulation: [T ][Q][T ] t = [Q]. Taking determinants of both sides we get det([t ]) 2 det[q] = det[q] and since det[q] 0 we have det[t ]) = ±1. Finally the inverse of an isometry is an isometry since we can write any w V as w = T v. Then Q(T 1 w) = Q(T 1 T v) = Q(v). But Q(v) = Q(T v) and Q(T v) = Q(w). so Q(T 1 w) = Q(w). QED. Notation. If n = dim(v) and Q is positive definite we write O(n) for the group of linear isometries of Q. It is called the orthogonal group of n-dimensional Euclidean space. On the other hand if q 0 is the index then there are p = n q plus signs amongst the ɛ i and q minus signs and the group of linear isometies of Q is then refered to as O(p, q). In the special case when q = 1 or when p = 1 we call O(p, q) a Lorentz group. Expressing linear isometries. Proposition 5. Fix, once and for all a standard orthonormal basis e 1,..., e n for V. a linear T V V is a linear isometry if and only T (e i ) = E i is also orthonormal (with the signs of the lengths of the e i and E i the same). PROOF. Indeed let v be arbitray and expand v in terms of the e i. Then Q(v) = Q(Σx i e i ) = Σɛ i (x i ) 2 where ɛ i = Q(e i ). But Q(T v) = Q(Σx i E i ) = Σx i x j β(e i, E j ) and the two expressions in the x i are equal for all n-tuples (x 1,..., x n ) if and only if β(e i, E j ) = 0 for i j and β(e i, E i ) = ɛ i = Q(e i ). In the case of O(n) we say that A O(n) is a rotation if det(a) = +1. EXAMPLE. Rotating the sphere. Theorem 2. There is a rotation taking any point and any direction on the 2-sphere to any other point and any other direction. The key here is to note that the pair (point, direction) form the first part of an orthonormal basis. Indeed, a point p on the unit sphere is a unit vector, and any direction v tangent to the sphere is perpindicular to the point p, since that point also serves as the normal vector to the sphere. Let us say this a bit more algebraically: The sphere is given by the equation Differentiating this identity we get P, P = 1 2 P, dp = 0 which we interpret as saying any vector dp tangent to the sphere is perpindicular to the point P. In other words: the tangent space T P S 2 which is the set of all derivatives of curves passing through P and lying on the sphere, is : T P S 2 = P.

6 6 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. Now we have agreed to represent directions in R 3 by unit vectors. So v is a unit vector orthogonal to p. Thus set E 1 = p, E 2 = v and complete this basis to obtain an orthonormal basis E 1, E 2, E 3 of R 3. There are two such completions: E 3 = ±p v. To prove the theorem then, let (p, v) and (q, w) be a pair of point, direction on the sphere. Complete them as above to form two orthonormal bases E 1, E 2, E 3 and U 1, U 2, U 3 of R 3. Now define T R 3 R 3 to be that linear transformationg taking E i to U i. If det(t ) 1 then det(t ) = 1 and replacing U 3 by U 3 will yield that det(t ) = 1 so that T is a rotation. 3. The sphere. Part 1. Distance on the sphere. We imagine ourselves bugs crawling on the surface of the sphere. (Maybe we are??!) We cannot bury through the sphere or fly above it. To get from point A to point B we must travel along curves constrained to the sphere. Now each sufficiently nice path c has a length l(c). We then define d(a, B) = inf{l(c) c a path lying on the sphere and joininga to B} We recall from real analysis that inf means infimum and is just a fancy way to say the minimum of a collection of real numbers when this collection might not have any minimum. For example, the infimum of the collection of all positive fractions of the form 1/n, n = 1, 2, 3,... is 0, although their is no minimum. We recall from vector calculus how to compute lengths of paths. Assuming that the paths are p.w. differentiable. Then c(t) = (x(t), y(t), z(t)) with some bounds a t b on the path parameter t. We have l(c) = ds = In other words, the integrand for computing l is We expand out the integrand: dc dt dt = a b dc/dt dt dc/dt = Q(dc/dt)(!) dx dt (t)2 + dy dt (t)2 + dy dt (t)2 dt = dx 2 + dy 2 + dz 2 so, as an convenient and useful shorthand we put. ds 2 = dx 2 + dy 2 + dz 2 Now, if the curve lies on the sphere then x 2 + y 2 + z 2 = 1. Computation. Switch to spherical coordinates. Verify that ds 2 = dr 2 + r 2 (dφ 2 + sin 2 φdθ 2 )s. Conclude that on the unit sphere, since r = 1 and thus dr = 0 we have ds 2 = (dφ 2 + sin 2 φdθ 2 ) Here is another way to say what is going on. We will think of (p, v) as part of a basis, now labelled E 3, E 1, E 2 Consider the map F (φ, θ) = cos(φ)e 3 + sin(φ)(cos(θ)e 1 + sin(θ)e 2 )

7 ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. 7 Verify that φ, θ sweeps out the sphere, and that this map is locally one-to-one and onto as long as φ 0, π. It follows that we can express any smooth curve on the sphere in the form: Please verify that: c(t) = F (φ(t), θ(t)) 2 dc/dt 2 = dφ + sin 2 (φ(t)) dθ dt dt COROLLARY. Distance formula: the distance between two points A and B on the sphere is realized by the shortest arc of the great circle obtained by intersecting the plane spanned by A, B and the origin with the sphere. That distance is d(a, B) = cos 1 ( A, B ) Proof. We first consider the case B A. Apply Graham-Schmidt to A, B to get an orthonormal frame E 3 = A, E 1, E 2 with E 1 a linear combination of A and B. Then in the corresponding spherical coordinates E 3 is given by φ = 0 and A = cos(φ 1 )E 3 + sin(φ 1 )E 1 with cos(φ 1 ) = A, B. Now let φ(t), θ(t) represent any path joining A to B, say φ(0) = 0, φ(b) = φ 1, θ(b) = 0. Since sin 2 φ > 0 we have that dφ2 + sin 2 φdθ 2 dφ with equality if and only if dθ = 0. Thus l(c) dφ = φ 1 with equality if and only if the curve c follows along the curve θ = 0 which is the equation in our spherical coordinates for an arc of a great circle from A to B. QED. 2.