Nov 25, 2018 - Now elements of a would be a family of diagonalizable matrices that commute ... So the effect of conjugation is permuting the diagonal entries.
Numerical Linear Algebra From Lie Theory Qiuye Jia November 25, 2018
1
Introduction
The convention of notations would be: M (n, C): The n × n matrices whose entries are complex numbers. And we would simply write M (n) for short. GL(n): gl(n): Lie algebra of GL(n). Generally letters of this kind of typeface (Fraktur script, usually in lower case) would refer to lie algebras and letters of ordinatry typeface (usually in upper case) would refer to Lie groups. And the corresponding Lie algebra and Lie group would use the same letters. Then the classical matrix decompositions1 SVD, QR, and LU correspond to the Cartan, Iwasawa and Bruhat decompositions of Lie groups. The task of following sections is exlain this in detail. The common point of these decompositions is that the do not impose any requirement on the matrix. Although it would be reasonable to dive into real number cases since we are heading for the numerical goal, whereas a more vital aspect is that we are using Lie theory, whose real number version is much more complicated than the complex version. So we would replace ‘othogonal’ by ‘unitary’, ‘symmetric’ by ‘Hermitian’, and ‘transpose’ by ‘conjugate transpose’. And we would denote conjugate transpose of matrix A by AH . It is a well-known result that: Proposition 1 g := gl(n) = M (n).
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SVD and Cartan Decomposition
From now on, G := GL(n, C) Cartan decomposition entails a involution θ of g. We take the involution to be θ(A) := −AH . This is not only a involution, it is also a Cartan involution. Thant is to say, let B(x, y) 1
be the Killing form of g, then Bθ (x, y) := −B(x, θ(y)) is a positive definite bilinear form in x and y. So the Cartan decomposition is like g = k⊕p, where k is the eigenspace of θ with eigenvalue 1 and p is that with engenvalue −1 (these two are only possible eigenvalues since θ2 = id). In our situation, for an element x to lie in k, we require xH = −x, that is, skew-Hermitian, which corresponds the Lie algebra of U (n), the unitary group. Next we look at p. The equation for x ∈ p would be x = xH , that is, being Hermitian. Moreover, now let K be the subgroup (by subgroup, we always mean Lie subgroup) of G with Lie algebra k, so K = U (n). p can be written as ∪k∈K Ad(k)(a).
(1)
Here a is the maximal Abelian subalgebra of p. Now elements of a would be a family of diagonalizable matrices that commute with each other, which should be simultaneously diagonalizable due to a well known fact in elementary linear algebra. (This is actually a special case of (1), which says the maximal Abelian subalgebra is unique up to an adjoint action). Now according to Cartan decomposition G = KP.
(2)
In conjunction with above refinement of P , we see that G = K(KAK) = KAK (K is a group, multiplication and taking inverse would not change it). Now since we can find an U0 ∈ K = U (n), such that U0H AU0 = D, where D is the group of (invertible) diagonal matrices. Absorbing the U0 factor into K, we get G = KAK,
(3)
which is exactly the SVD decomposition.
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QR and Iwasawa Decomposition
The setting for k, a. Now a new n is introduced. The decomposition reads: g = k ⊕ a ⊕ n.
(4)
We have illustrated in the previous section what does k and a looks like. Now we turn to n. It is the nilpotnet matrices. Or, more explicitly, strictly upper triangular matirces. After we take exponent to get the corresponding Lie group N , it would be the upper triangular matirces with all diagonal entries being 1. The corresponding decomposition of group is G = KAN. Recall K = U (n), A = D. Absorbing A factor into N , we get the QR decomposition. 2
(5)
4
LPL(UPU) and Bruhat Decomposition
The general Bruhat decomposition tells us: a
G = BW B =
BwB.
(6)
w∈W
Here W is the Weyl group of G. To distinguish from the Weyl group of Lie algebras, we spell it out: Let T be a maximal torus of B, and N (T ) its normalizer, Z(T ) its centralizer, i.e. N (T ) = {x ∈ B|xT x−1 ⊂ T },
(7)
Z(T ) = {x ∈ B|xtx−1 = t, ∀t ∈ T }
(8)
Then the Weyl group W would be N (T )/Z(T ). In our case, the Borel subgroup B can be taken to be the upper-triangular matrices. And T can be taken to be diagonal matrices. All of them are assumed to be subgroup of G = GL(n). So now, for x ∈ N (T ), we need xtx−1 to be diagonal if t is diagonal. Since these are two diagonal matrices similar to each other, they have same eigenvalues, hence the same entries. So the effect of conjugation is permuting the diagonal entries. And since the action of Z(T ) is keeping t fixed, we know that N (T )/Z(T ) should be Sn , the n−th symmetry group. To be more specific, there is a permutation P do the same job as x: xtx−1 = P tP −1 . So P −1 x ∈ Z(T ). This gives N (T ) = Sn o Z(T ) and the desired quotient result. We can also view Z(T ) as the kernel of the adjoint representation (restricted to B) and consider the cokernel. We can also write Z(T ) out explicitely. We consider termwise equations. Let x = (xij ), etc. Suppose t = diag{t1 , t2 , ..., tn }. Then x ∈ Z(T ) means xt = tx. So we have xij tj = ti xij . Since t is arbitrary, we know that xij = 0 if i 6= j. So x ∈ Z(T ) are diagonal. And conversely, it is obvious that any diagonal matrix would be in Z(T ). So Z(T ) consists of diagonal matirces. Going back to our decomposition G = BW B. For a matrix A, we can write it as A = U1 P U2 , where P is a permutation matrix. One way to look at this is P = U1−1 AU2−1 . That is, any matrix A could be transformed into a permutation matrix through upper triangular matrices.
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We need to recall that, LU decomposiiton is actually a ‘P LU ’ decomposiiton. That is, we need a permutation at the left. Otherwise, we need to impose a seemly unnatural hypothesis that ‘all the leading minors of A are non-singular’. Now let w0 ∈ W be the permutation that interchanges 1 with n, 2 with n − 1, and so on. For a upper-triangular matrices U1 ∈ B, we know that L1 := w0 U1 w0 is lower-triangular.
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