Algebra, lecture 8 M. Verbitsky Algebra and Geometry lecture 8: normal forms Misha Verbitsky Universit´ e Libre de Bruxelles December 1, 2016 1
Algebra, lecture 8 M. Verbitsky Jordan normal form DEFINITION: Let G be a group (typically, a Lie group such as GL ( V ) or SO ( V ) acting on a set M . Normal form is a subset Z ⊂ M which intersects each orbit in a finite, non-empty subset. EXAMPLE: The group GL ( V ) acts on the set End( V ) by g ( A ) = gAg − 1 (this is called adjoint action ). Jordan normal form is the set of matrices which have block form A λ 1 0 A λ 2 JNF := ... 0 A λ k with each block 1 0 λ i ... λ i A λ i = . ... 1 0 λ i Indeed, for each A ∈ End( C n ) there exists finitely many Jordan block matrices A ′ such that A ′ = gAg − 1 (finitely many because the blocks are not ordered, you can freely exchange A λ i and A λ j ). Therefore, the set Z of matrices of form JNF intersects with each orbit of the adjoint action in a nonempty, finite set. 2
Algebra, lecture 8 M. Verbitsky Normal form of orthogonal matrix (even dimension) THEOREM: Let V be a real vector space equipped with a positive definite scalar product. Consider the special orthogonal group SO ( V ) acting on SO ( V ) by g ( A ) = gAg − 1 (“adjoint action”; this action corresponds to a basis change from x 1 , ..., x n to g ( x 1 ) , ..., g ( x n )). Then for dim V even, for each A ∈ SO ( V ) there is a basis g ( x 1 ) , ..., g ( x n ) in which the matrix A has the following block form 0 A α 1 A α 2 gAg − 1 = ... 0 A α k with α i ∈ [0 , 2 π [ and A α i the corresponding rotation matrix, � � cos α i sin α i A α i = − sin α i cos α i This is called the normal form of orthogonal transform (for even dimen- sion). 3
Algebra, lecture 8 M. Verbitsky Normal form of orthogonal matrix (odd dimension) THEOREM: Let V be a real vector space equipped with a positive definite scalar product. Consider the orthogonal group SO ( V ) acting on SO ( V ) by g ( A ) = gAg − 1 (“adjoint action”; this action corresponds to a basis change from x 1 , ..., x n to g ( x 1 ) , ..., g ( x n )). For dim V odd, for some g ∈ SO ( V ) one has A α 1 0 A α 2 gAg − 1 = ... A α k 0 1 (there is an extra 1-dimensional unit block). This is called the normal form of orthogonal transform (for odd dimension). REMARK: Notice that eigenvalues of A α i are equal to cos α i ± √− 1 sin α i = e ±√− 1 α i , hence the block form is determined by the matrix A uniquely up to permutation. Therefore, it is indeed a normal form: in each adjoint orbit, there are only finitely many matrices of this form. 4
Algebra, lecture 8 M. Verbitsky Normal form of orthogonal matrix (proof, page 1) Proof. Step 1: Let us construct the normal form of orthogonal matrix A ∈ O ( V ) (orthogonal matrices, not necessarily preserving orientation). Consider an eigenvector v 1 ∈ V ⊗ R C , and let α 1 be the corresponding eigenvalue. Denote by h the scalar product h 0 on V , extended to V ⊗ R C by h ( x, y ) = h 0 (Re x, Re y ) + h 0 (Im x, Im y ). Clearly, h is SO ( V )-invariant. The quadratic form x �→ h ( x, x ) is positive definite and A -invariant. Since h ( v, v ) = h ( A ( v 1 ) , A ( v 1 )) = α 1 α 1 h ( v, v ), we have | α 1 | = 1: all eigenvalues α i of A ∈ O ( V ) satisfy | α i | = 1 Step 2: The eigenvector v 1 is real if and only if α 1 = ± 1. In the later case, consider the space V 1 = v ⊥ (orthogonal complement). For each w ∈ V 1 , 1 one has 0 = h ( v 1 , w ) = h ( A ( v 1 ) , A ( w )) = ± h ( v 1 , A ( w )) = 0, hence V 1 is A -invariant. Using induction, we may represent A in a block form as � � ± 1 0 gAg − 1 = A ′ , 0 where A ′ := g ′ A � V 1 ( g ′ ) − 1 is a block matrix in the normal form. � � 5
Algebra, lecture 8 M. Verbitsky Normal form of orthogonal matrix (proof, page 2) Step 3: To finish the proof, it remains to treat the case α 1 / ∈ R . In this case, v 1 and v 1 are eigenspaces, and A acts on a 2-dimensional real space � Re v, Im v � by rotation with eigenvalues α 1 , α 1 . The matric of such a rotation is � � cos α i sin α i A α i = . − sin α i cos α i To finish the proof, it would suffice to show that A preserves the space V 1 := � Re v 1 , Im v 1 � ⊥ . Then we would represent A ′ = g ′ A � V 1 ( g ′ ) − 1 as a block � � matrix in the normal form, using induction on dim V , and obtain cos α i sin α i 0 gAg − 1 = − sin α i cos α i 0 A ′ 0 Step 4: For each w ∈ V 1 , and each v ∈ � Re v 1 , Im v 1 � one has A ( v ) ∈ � Re v 1 , Im v 1 � , hence 0 = h ( A − 1 ( v ) , w ) = h ( v, A ( w )) = 0, and this implies that A ( w ) ∈ V 1 . 6
Algebra, lecture 8 M. Verbitsky Normal form of orthogonal matrix (proof, page 3) Step 5: We proved that any A ∈ O ( V ) can be represented in block form with the blocks either 1-dimensional and equal to ± 1 or 2-dimensional � � cos α i sin α i and equal to . For A ∈ SO ( V ), there is an even number of − sin α i cos α i � � − 1 0 − 1-blocks, which can be grouped together to block matrices = A π . 0 − 1 The number of 1-blocks is even when dim V is even and odd when it � � 1 0 is odd. We can group them together pairwise into matrices = A 0 . 0 1 7
Algebra, lecture 8 M. Verbitsky Orthogonal matrices are exponents COROLLARY: Let A ∈ SO ( V ) be a matrix written in block form as A α 1 0 A α 2 A = ... 0 A α k with � � cos α i sin α i A α i = − sin α i cos α i Then A = e B , where B is written in the same basis as 0 B α 1 B α 2 B = ... 0 B α k with � � 0 α i B α i = − α i 0 8
Algebra, lecture 8 M. Verbitsky Normal form of a bilinear symmetric form Recall that the group GL ( V ) acts on bilinear forms by g ( h )( · , · ) = h ( g − 1 ( · ) , g − 1 ( · )) , where g ∈ GL ( V ), h ∈ Bil( V ) = V ∗ ⊗ V ∗ . This action corresponds to a basis change. We consider an action of GL ( V ) on pairs of bilinear symmetric forms, and find a normal form of this action on the set of pairs of forms. DEFINITION: Let h ∈ Sym 2 V ∗ be a bilinear symmetric form, and x 1 , ..., x n a basis. This basis is called orthogonal if h ( x i , x j ) = 0 for i � = j , and orthonormal if in addition h ( x i , x i ) = ± 1. REMARK: Previously, we proved that any non-degenerate bilinear sym- metric form on R n admits an orthonormal basis. This result can be understood as providing the normal form of a non-degenerate bilinear symmetric form. 9
Algebra, lecture 8 M. Verbitsky Normal form for a pair of bilinear symmetric forms (1) The following result is proven at the end of this lecture. Let V = R n , and h, h ′ ∈ Sym 2 V ∗ be two bilinear symmetric Theorem 1: forms, with h positive definite. Then there exists a basis x 1 , ..., x n which is orthonormal with respect to h , and orthogonal with respect to h ′ . REMARK: In this basis, h ′ is written as diagonal matrix, with eigenvalues α 1 , ..., α n independent from the choice of the basis. Indeed, consider h, h ′ as Then h 1 h − 1 is an endomorphism with maps from V to V ∗ , h ( v ) = h ( v, · ). eigenvalues α 1 , ..., α n . This implies that Theorem 1 gives a normal form of the pair h, h ′ . 10
Algebra, lecture 8 M. Verbitsky Finding principal axes of an ellipsoid REMARK: Theorem 1 implies the following statement about ellipsoids: for any positive definite quadratic form q in R n , consider the ellipsoid S = { v ∈ V | q ( v ) = 1 } . The group SO ( n ) acts on R n preserving the standard scalar product. Then for some g ∈ SO ( n ) , g ( S ) is given by equation � a i x 2 i = 1 , where a i > 0 . This is called finding principal axes of an ellipsoid . 11
Algebra, lecture 8 M. Verbitsky Compactness DEFINITION: Recall that a subset Z ⊂ R n is called (sequentially) compact if any sequence x 1 , ..., x n , ... ⊂ Z has a converging subsequence. THEOREM: A subset Z ⊂ R n is sequentially compact if and only if Z is closed and bounded (that is, contained in a ball of finite diameter). EXERCISE: Prove this theorem. EXERCISE: Let f be a continuous function on a compact Z . Prove that Z is bounded and attains its supremum on Z . COROLLARY: Let f be a continuous function on a sphere S n ⊂ R n +1 . Then f is bounded, and attains its supremum. Further on, we need the following lemma. LEMMA: Let V = R n , and h, h ′ ∈ Sym 2 V ∗ be two bilinear symmetric forms, h positive definite, and q ( v ) = h ( v, v ) , q ′ ( v ) = h ′ ( v, v ) the corresponding quadratic forms. Consider q ′ as a function on a sphere S = { v ∈ V | q ( v ) = 1 } , and let x ∈ S be the point where q ′ attains maximum. Denote by x ⊥ h and x ⊥ h ′ the orthogonal complement with respect to h, h ′ . Then x ⊥ h = x ⊥ h ′ . 12
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