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Algebra, lecture 8

• M. Verbitsky

Algebra and Geometry

lecture 8: normal forms Misha Verbitsky

Universit´ e Libre de Bruxelles December 1, 2016

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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 JNF :=

   

Aλ1 Aλ2 ... Aλk

   

with each block Aλi =

   

λi 1 λi ... ... 1 λi

    .

Indeed, for each A ∈ End(Cn) 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 Aλi and Aλj). Therefore, the set Z of matrices of form JNF intersects with each orbit

• f the adjoint action in a nonempty, finite set.

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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 x1, ..., xn to g(x1), ..., g(xn)). Then for dim V even, for each A ∈ SO(V ) there is a basis g(x1), ..., g(xn) in which the matrix A has the following block form gAg−1 =

    

Aα1 Aα2 ... Aαk

    

with αi ∈ [0, 2π[ and Aαi the corresponding rotation matrix, Aαi =

• cos αi

sin αi − sin αi cos αi

• 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 x1, ..., xn to g(x1), ..., g(xn)). For dim V odd, for some g ∈ SO(V ) one has gAg−1 =

       

Aα1 Aα2 ... Aαk 1

       

(there is an extra 1-dimensional unit block). This is called the normal form

• f orthogonal transform (for odd dimension).

REMARK: Notice that eigenvalues of Aαi are equal to cos αi ± √−1 sin αi = e±√−1 αi, 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 v1 ∈ V ⊗R C, and let α1 be the corresponding eigenvalue. Denote by h the scalar product h0 on V , extended to V ⊗R C by h(x, y) = h0(Re x, Re y) + h0(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(v1), A(v1)) = α1α1h(v, v), we have |α1| = 1: all eigenvalues αi

• f A ∈ O(V ) satisfy |αi| = 1

Step 2: The eigenvector v1 is real if and only if α1 = ±1. In the later case, consider the space V1 = v⊥

1

(orthogonal complement). For each w ∈ V1,

• ne has 0 = h(v1, w) = h(A(v1), A(w)) = ±h(v1, A(w)) = 0, hence V1 is

A-invariant. Using induction, we may represent A in a block form as gAg−1 =

• ±1

A′,

• where A′ := g′A
• V1(g′)−1 is a block matrix in the normal form.

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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 α1 / ∈ R. In this case, v1 and v1 are eigenspaces, and A acts on a 2-dimensional real space Re v, Im v by rotation with eigenvalues α1, α1. The matric of such a rotation is Aαi =

• cos αi

sin αi − sin αi cos αi

• .

To finish the proof, it would suffice to show that A preserves the space V1 := Re v1, Im v1⊥. Then we would represent A′ = g′A

• V1(g′)−1 as a block

matrix in the normal form, using induction on dim V , and obtain gAg−1 =

  

cos αi sin αi − sin αi cos αi A′

  

Step 4: For each w ∈ V1, and each v ∈ Re v1, Im v1 one has A(v) ∈ Re v1, Im v1, hence 0 = h(A−1(v), w) = h(v, A(w)) = 0, and this implies that A(w) ∈ V1. 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 and equal to

• cos αi

sin αi − sin αi cos αi

• . For A ∈ SO(V ), there is an even number of

−1-blocks, which can be grouped together to block matrices

• −1

−1

• = Aπ.

The number of 1-blocks is even when dim V is even and odd when it is odd. We can group them together pairwise into matrices

• 1

1

• = A0.

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Algebra, lecture 8

• M. Verbitsky

Orthogonal matrices are exponents COROLLARY: Let A ∈ SO(V ) be a matrix written in block form as A =

    

Aα1 Aα2 ... Aαk

    

with Aαi =

• cos αi

sin αi − sin αi cos αi

• Then A = eB, where B is written in the same basis as

B =

    

Bα1 Bα2 ... Bαk

    

with Bαi =

• αi

−αi

• 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 ∈ Sym2 V ∗ be a bilinear symmetric form, and x1, ..., xn a basis. This basis is called orthogonal if h(xi, xj) = 0 for i = j, and

• rthonormal if in addition h(xi, xi) = ±1.

REMARK: Previously, we proved that any non-degenerate bilinear sym- metric form on Rn 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. Theorem 1: Let V = Rn, and h, h′ ∈ Sym2 V ∗ be two bilinear symmetric forms, with h positive definite. Then there exists a basis x1, ..., xn 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 α1, ..., αn independent from the choice of the basis. Indeed, consider h, h′ as maps from V to V ∗, h(v) = h(v, ·). Then h1h−1 is an endomorphism with eigenvalues α1, ..., αn. This implies that Theorem 1 gives a normal form

• f the pair h, h′.

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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 Rn, consider the ellipsoid S = {v ∈ V | q(v) = 1}. The group SO(n) acts on Rn preserving the standard scalar product. Then for some g ∈ SO(n), g(S) is given by equation aix2

i = 1, where ai > 0.

This is called finding principal axes of an ellipsoid. 11

Algebra, lecture 8

• M. Verbitsky

Compactness DEFINITION: Recall that a subset Z ⊂ Rn is called (sequentially) compact if any sequence x1, ..., xn, ... ⊂ Z has a converging subsequence. THEOREM: A subset Z ⊂ Rn 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 Sn ⊂ Rn+1. Then f is bounded, and attains its supremum. Further on, we need the following lemma. LEMMA: Let V = Rn, and h, h′ ∈ Sym2 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

Algebra, lecture 8

• M. Verbitsky

Maximum of a quadratic form on a sphere LEMMA: Let V = Rn, and h, h′ ∈ Sym2 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 complements with respect to h, h′. Then x⊥h = x⊥h′. Proof: Let us rescale q, q′ in such a way that q q′, with equality on x. Suppose that v ∈ x⊥h. Then q(x + εv) = q(x) + ε2q(v). However, q′(x + εv) = q(x) + ε2q′(v) + 2εh′(v, x). This gives q(x) + ε2q(v) q(x) + ε2q′(v) + 2εh′(v, x) cancelling q(x) and dividing by ε > 0, obtain ε(q(v) − q′(v)) 2h′(v, x). for all ε > 0. This implies that 0 2h′(v, x) for all v ∈ x⊥h. Since v → h′(v, x) is a linear form on v, inequality 0 h′(v, x) implies that h′(v, x) = 0. 13

Algebra, lecture 8

• M. Verbitsky

Normal form for a pair of bilinear symmetric forms (2) Theorem 1: Let V = Rn, and h, h′ ∈ Sym2 V ∗ be two bilinear symmetric forms, with h positive definite. Then there exists a basis x1, ..., xn which is orthonormal with respect to h, and orthogonal with respect to h′. Proof: Let 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 x1 ∈ S be the point where q′ attains maximum. Then x⊥h

1

= x⊥h′

1

. Using induction, we may assume that on x⊥h

1 , Theorem 1 is already proven, and there exists a

basis x2, ..., xn orthonormal for h and orthogonal for h′. Then x1, x2, ..., xn is

• rthonormal for h and orthogonal for h′.

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