Generalisations of Singular Value Decomposition to Dual-Numbered

Home , Dual number, Laguerre transformations

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M = UΣV T where U T U = V T V = I, and Σ is a diagonal matrix.

Theorem (Dual -SVD). Every square dual-numbered matrix M Mn(D) can be decomposed as ∗ ∈ T M = UΣV T T where UU = V V = I, and Σ is a block-diagonal matrix where each block is ′ σi ǫσi ′ ′ of one of the forms σi , ′ − or ǫσi , where each σi and σi is real, ǫσ σi i and σ′ =0. i 6 1.1 Dual -SVD ∗ Our study of -SVD is motivated by Yaglom’s 1968 book Complex numbers in geometry ([10]).∗ Yaglom considers the group of Laguerre transformations, which is analogous to the group of Moebius transformations over the complex numbers. az+b The Laguerre transformations are the group of functions of the form z cz+d where a,b,c,d are elements of D, z is a variable over D, and ad bc is not7→ a zero divisor. It can easily be seen that every Laguerre transformation− z az+b can 7→ cz+d a b be represented as the 2 2 matrix . × c d Yaglom classifies the elements of this group in a geometric way. We restate the classification in the language of matrices. Yaglom argues that every invertible 2 2 matrix over the dual numbers can be expressed in exactly one of the following× two ways: T T T • UΣV where UU = V V = I and Σ is a diagonal matrix with real- valued entries. T σ ǫσ′ • UΣ where UU = I,Σ= − , and both σ and σ′ are real (and ǫσ′ σ σ′ is non-zero). The first of these forms resembles Singular Value Decomposition. What we propose in this paper is a generalisation of Singular Value Decomposition to square dual-numbered matrices which includes both forms as special cases.

1.2 Dual T -SVD Our study of T -SVD is motivated by recent research [8, 2, 3, 7] in mechanics, where the authors consider either a form of SVD that is essentially T -SVD, or a form of Polar Decomposition that is essentially T -Polar Decomposition. For completeness, we describe T -Polar Decomposition below.

2 Corollary (Dual T -Polar Decomposition). Every square dual-numbered matrix M can be expressed in the form M = UP where U T U = I and P is symmetric. One of the main applications of the T -SVD is in ﬁnding the Moore-Penrose generalised inverse of a dual matrix (whenever it exists). This has applications in kinematic synthesis (see [7]). Very recently, Udwadia et al. [9] studied the existence of the Moore-Penrose generalised inverse for dual-numbered matrices and showed that unlike in the case of real and complex matrices, not all dual-numbered matrices have such inverses. In contrast to their result, we prove that the T -SVD does exist for all dual-numbered matrices. Our result is new as none of the authors above have proved that the T -SVD exists in general.

1.3 Structure of the paper Section 2 contains preliminaries. In section 3, we will prove the /T -spectral theorems. In section 4, we will show that every square dual-number∗ matrix has a /T -SVD. ∗ 2 Preliminaries 2.1 Dual numbers The dual numbers are the ring R[ǫ]/(ǫ2). In other words the dual numbers are pairs of real numbers, usually written as a + bǫ, with the following operations defined on them: • (a + bǫ) + (c + dǫ) = (a + c) + (b + d)ǫ. • (a + bǫ)(c + dǫ)= ac + (ad + bc)ǫ. • (a + bǫ)−1 = a−1 ǫba−2. − Given a polynomial with real coefficients F (X) R[X], evaluating F on a dual number gives F (a + bǫ) = F (a)+ ǫbF ′(a) where∈ F ′ denotes the derivative of F . Multiples of the dual number ǫ are sometimes referred to as “infinitesimal” with the intuition being that ǫ is “so small” that it squares to 0. The dual numbers contrast with the complex numbers. The complex numbers are defined as the ring R[i]/(i2+1). Like the dual numbers, the complex numbers are pairs of real numbers with certain operations defined on them. The difference is that while the complex numbers are defined by adjoining an element i such that i2 = 1, the dual numbers are defined by adjoining an element ǫ such that ǫ2 = 0. − Unlike the complex numbers, the dual numbers do not form a field. As such, instead of talking about vector spaces over dual numbers, one must talk about modules over dual numbers. Modules over dual numbers don’t necessarily have

3 bases: For instance, the module ǫD (that is, multiples of the dual number ǫ) doesn’t have a basis. For this reason, one cannot in general talk about the “rank” of a matrix over the dual numbers. A module over the dual numbers (or generally any ring) that does have a basis is referred to as a free module.

2.2 Spectral theorem over symmetric real matrices The spectral theorem over symmetric matrices states that a symmetric matrix A over the real numbers can be orthogonally diagonalised. In other words, if A is a real matrix such that A = AT , then there exists a matrix P such that P T P = I and A = P DP T for some diagonal matrix D.

2.3 Spectral theorem over skew-symmetric real matrices The spectral theorem over skew-symmetric matrices states that a skew- symmetric matrix A over the real number can be orthogonally block- diagonalized, where every block is a skew-symmetric 2 2 block except possibly for one block, which has dimensions 1 1 and whose only× entry equals 0. × 2.4 Notation and terms By a + bǫ, we mean a bǫ. Given a dual number a + bǫ, we call a the standard part and b the infinitesimal− part. We sometimes denote the standard and infinitesimal parts of a dual number z by st(z) and (z) respectively. We also use the term infinitesimal to describe a dual numberℑ whose standard part is zero, and appreciable to describe a dual number whose standard part is non-zero. We sometimes call a dual number whose infinitesimal part is zero a real number. All the above terms generalise to matrices and vectors over dual numbers in the obvious way. Likewise, the operations infinitesimal part and standard part generalise to sets of dual-numbered vectors by applying these operations to each element of the set. T We sometimes write M ∗ for M . We use (u, v) to denote uT v, and u, v to denote uT v. Two vectors are consid- ered T -orthogonal if (u, v)=0, andh -orthogonali if u, v = 0. ∗ h i A dual matrix U which satisfies UU¯ T = U¯ T U = I is called unitary. Similarly, one which satisfies U T U = UU T = I is called T -orthogonal. A dual matrix A such that A = A¯T is called Hermitian. Similarly, one which satisfies A = AT is called symmetric. A vector v is called an eigenvector of a dual-numbered matrix A if v is appreciable and Av = λv for some λ D. ∈

4 3 Dual-number /T -spectral theorem ∗ Theorem (Dual -spectral). Given a Hermitian dual number matrix M ∗ ∈ Mn(D), we can decompose the matrix as:

M = V ΣV ∗ where V is unitary, and Σ is a block-diagonal matrix where each block is either of the form:

• σi ,

′ σi ǫσi • or ′ − ǫσi σi ′ ′ where each σi and σ is real, and σ =0. i i 6 Proof. Let M be a Hermitian matrix. We ﬁnd a real matrix S such that S st(A)ST is diagonal. We let M ′ = SMST , which we write as a block matrix λ1I + ǫB11 ǫB12 ǫB1n · · · .  T .. .  ǫB λ2I + ǫB22 . . − 12 .  . .. ..   . . . ǫBn−1,n   T T   ǫB ǫB λnI + ǫBnn  − 1n ··· − n−1,n  where each Bii is skew-symmetric. We let

I ǫB12 ǫB1n λ1−λ2 · · · λ1−λn T . .  ǫB12 .. .  λ −λ I . P = 1 2 ,  . . . −   . .. .. ǫBn 1,n  λn−1−λn  T T   ǫB ǫB −   1n n 1,n I   λ1−λn · · · λn−1−λn  and let M ′′ = PM ′P ∗. We end up with M ′′ being equal to a direct sum of ′′ matrices: M = (λ1I + ǫB11) (λ2I + ǫB22) (λnI + ǫBnn). We ﬁnally ⊕ ⊕···⊕ use the spectral theorem for skew-symmetric matrices to ﬁnd matrices Qi such T that QiBiiQi is equal to a direct sum of skew-symmetric 2 2 blocks except × ′′ for potentially one zero block. We thus get that (Q1 Q2 Qn)M (Q1 T ⊕ ⊕···⊕ ⊕ Q2 Qn) is a block-diagonal matrix in the desired form. ⊕···⊕ Theorem (Dual T -spectral). Every symmetric matrix can be orthogonally diagonalised.

5 Proof. The proof is the same as for the -spectral case except that ∗ λ1I + ǫB11 ǫB12 ǫB1n · · · .  T .. .  ǫB λ2I + ǫB22 . . M ′ = 12 ,  . .. ..   . . . ǫBn−1,n   T T   ǫB ǫB λnI + ǫBnn  1n · · · n−1,n  and I ǫB12 ǫB1n λ1−λ2 · · · λ1−λn T . .  ǫB12 .. .  λ −λ I . P = − 1 2 ,  . . . −   . .. .. ǫBn 1,n  λn−1−λn  T T   ǫB ǫB −   1n n 1,n I  − λ1−λn ··· − λn−1−λn  and M ′′ = P MP T . And instead of using the spectral theorem for skew- symmetric matrices, we use the one for symmetric matrices.

The following theorem can be used to show that the /T -spectral decompo- sitions are unique. Note that in order to show that the∗ -spectral decompo- ∗ σ ǫσ′ sition is unique, one must ﬁrst fully diagonalise the matrix − to ǫσ′ σ σ + iǫσ′ 0 by working over C D. 0 σ iǫσ′ ⊗ − Theorem (Uniqueness of eigenvalues). An eigenbasis of a linear endomorphism T : Dn Dn corresponds to a unique multi-set of eigenvalues. →

Proof. Let e1,e2,...,en be a basis made up of eigenvectors of T . Let the corresponding eigenvalues be λ1, λ2,...,λn respectively.

We intend to show that for any eigenvalue λ of T , the number of “ei”s in our basis with eigenvalue equal to λ is equal to dim(st(Vλ)). Since the quantity dim(st(Vλ)) depends only on T , it implies that every eigenbasis has the same amount of eigenvectors with eigenvalue equal to λ.

Assume that e1,e2,...,ek all have some eigenvalue λ, and no other ei has eigenvalue λ. We will show that k = dim(st(Vλ)) by showing that st(e1), st(e2),..., st(ek) form a basis of st(Vλ).

Claim 1. st(e1), st(e2),..., st(ek) form a spanning set of st(Vλ) (the standard part of the eigenspace corresponding to λ).

Proof of claim 1. Given a vector in st(Vλ), we wish to express it as a linear combination of the above vectors. Our vector is equal to st(e) for some e Vλ. n ∈ We can express e as e = i=1 αiei. From this we can conclude that st(e) = n i=1 st(αi) st(ei). But noticeP that this sum includes vectors st(ek+1) to st(en), whichP we don’t want in our linear combination. We thus need to show that

6 n st(αi) = 0 for all i > k. To this end, notice that T (e)= λe = λ ( i=1 αiei) and n n T (e) = i=1 αiT (ei) = i=1 αiλiei. Since we know that thePei are linearly independent,P we have thatP αi(λi λ) = 0 for all i. From αi(λi λ) = 0, − − one can check that if λ = λi then αi is inﬁnitesimal. Since for all i > k, we 6 have that λi = λ, we must have that αi is inﬁnitesimal for i > k. Recalling k 6 n that e = i=1 αiei + i=k+1 αiei, taking standard parts of both sides gives k st(e)= Pi=1 st(αi) st(ePi) + 0, which is what we sought to show. P Claim 2. st(e1), st(e2),..., st(ek) forms a linearly independent set.

Proof of claim 2. Assume that there exist (αi) (where each αi is real) such that k k αi st(ei) = 0. We then have that st αiei =0. If αj = 0 for some i=1 i=1 6 P k P j then we get that i=1 ǫαiei = 0, which contradicts the linear independence of e1,e2,...,en. ThereforeP all αi equal 0.

4 Dual-number /T -Singular Value Decomposi- ∗ tion

These are the theorems we will prove in this section.

Theorem (Dual -SVD). Given a square dual number matrix M Mn(D), we can decompose the∗ matrix as: ∈

M = UΣV ∗ where U and V are unitary, and Σ is a block-diagonal matrix where each block ′ σi ǫσi ′ ′ is either of the form σi , ′ − or ǫσi , where each σi and σi is real, ǫσ σi i and σ′ =0. i 6 Theorem (Dual T -SVD). Given a square dual number matrix M Mn(D), we can decompose the matrix as: ∈

M = UΣV T where U and V are T -orthogonal, and Σ is a diagonal matrix. We will start by proving these theorems for invertible matrices. Theorem (SVD for invertible matrices). Every invertible matrix has a T / - SVD. ∗

Proof. We shall prove this for the T -SVD, but the argument is the same for the -SVD. ∗ Let M be an arbitrary dual matrix. Observe that M T M is a symmetric matrix. As such, by the T -spectral theorem, we have that M T M = V ΣV T for some

7 orthogonal matrix V and diagonal Σ. Also observe that the standard part of M T M is positive-deﬁnite. From this, it follows that the standard part of Σ is positive. It follows that √M T M exists and is equal to V √ΣV T . Observe also that M(√M T M)−1 is an orthogonal matrix which we shall call U. Finally, we have that (UV )√ΣV T is the T -SVD of M.

Theorem (Dual -SVD). Given a square dual number matrix M Mn(D), we can decompose the∗ matrix as: ∈

M = UΣW ∗ where U and W are unitary, and Σ is a block-diagonal matrix where each block ′ σi ǫσi ′ ′ is either of the form σi , ′ − or ǫσi , and each σi and σi is real. ǫσ σi i Proof. Let T : V V be a linear endomorphism over a finite-dimensional free D-module. → The operator T ∗T is Hermitian. Thus, apply the -spectral theorem to it to ∗L get an orthonormal basis B = b1,b2,...,bn . Let V be the span of the set of vectors in B that get mapped to{ appreciable} vectors by T . Let V R be the span of the set of vectors in B that get mapped to infinitesimal vectors by T .1 Observe that T is injective over V L. Let I be the image of V L under T . Observe that I has the same dimension as V L. Thus, there exists a unitary operator U such that U(I)= V L. Clearly, UT (V L)= V L. Observe that UT (V R) V R. ⊆ Since UT L is a linear endomorphism, we may take its -SVD by theorem Dual |V ∗ -SVD for invertible matrices. We also observe that UT V R is an infinitesimal ∗map (meaning that it maps every argument to infinitesimals),| and can therefore be expressed as ǫT ′. We can thus take the SVD of T ′. Taking the direct sum of the -SVDs of UT L and UT R yields the -SVD of UT . ∗ |V |V ∗ ′ σi ǫσi The block types σi and ′ − come from UT V L , and the block type ǫσi σi | ′ ǫσ comes from UT R . i |V 1 Finally, we multiply the SVD of UT on the left by U − , and we are done.

Lemma 1. The following claims made in the above proof are true: 1. The Gram-Schmidt process is a valid algorithm for producing an orthonormal basis for some free D-module W given a basis for W . 2. T is injective over V L. 3. There exists a unitary map U such that UT (V L)= V L.

1It so happens that V L is equal to the image of T ∗T , and that V R is equal to both ker(T ∗T ) and ker(T tT ) (where T t denotes the tranpose of T ). These are easy to prove, and we won’t do so here.

8 4. UT (V L) UT (V R). ⊥ 5. UT (V R) V R. ⊆ Proof. We proceed to prove each claim. Proof of claim 1. The proof is by induction. Assume that dim(W ) = 0 and the basis is . Then Gram-Schmidt finishes immediately and produces the orthonor- ∅ mal basis . Now assume that given a set of vectors v1, v2,...,vn, vn+1 , the first n iterations∅ of Gram-Schmidt succeed in producing{ an orthonormal} set of vectors e1,e2,...,en whose span is equal to the span of v1, v2,...,vn . { } n { } vn+1− hei,vn+1iei i=1 We perform another iteration and get en+1 = n where vn+1− hei,vn+1iei Pi=1 | | u means st(u) . We need to check that the division isP sound. Assuming it | | | | n isn’t, we would have that vn+1 i=1 ei, vn+1 ei = 0, which implies that n | − h i | ǫ (vn+1 =1 ei, vn+1 ei) = 0, whichP contradicts the linear independence − i h i of v1, v2,...,vP n, vn+1 . The linear independence, spanning property and { } orthonormality of e1,e2,...,en,en+1 are easily shown. { } Proof of claim 2. It’s easily seen that T ∗T is injective over V L (based on the definition of V L). So then assume that T (u)= T (v) for some u and v in V L. We then have that T ∗T (u)= T ∗T (v). By injectivity of T ∗T , we have that u = v. Proof of claim 3. We begin by constructing an orthonormal basis for T (V L). L Clearly, given an orthonormal basis b1,b2,...,bk of V , we get a basis L { } T (b1),T (b2),...,T (bk) of T (V ). But this latter basis may not be orthonor- {mal. We perform Gram-Schmidt} on this latter basis to get an orthonormal basis L c1,c2,...,ck of T (V ). We extend the orthonormal basis b1,b2,...,bk of { L } { } V to an orthonormal basis b1,b2,...,bk,bk+1,...,bn of the whole space, { L } and the orthonormal basis c1,c2,...,ck of T (V ) to an orthonormal basis { } c1,c2,...,ck,ck+1,...,cn of the whole space. We leave it to the reader to { } verify that this extension is possible. We now define U(ci) = bi for all i, and observe that this U is unitary, and maps T (V L) to V L. Proof of claim 4. We take the inner product of the spaces UT (V L) and UT (V R). We get UT (V L),UT (V R) = T (V L),T (V R) = T ∗T (V L), V R = V L, V R = 0. We’veh used the fact thati Th∗T (V L)= V L,i whichh should be easyi toh verify.i L Proof of claim 5. Let b1,b2,...,bk be basis of V , and bk+1,bk+2,...,bn be a basis of V R. Their{ union is a basis} for the whole space.{ Let v UT (V R}). n R L ∈ Express v as i=1 λibi. Since UT (V ) V , we have that for i k, λi =0. It ⊥ ≤ R follows that vPis in the span of b1,b2,...,bk , and therefore that v is in V . { }

Theorem (Dual T -SVD). Given a square dual number matrix M Mn(D), we can decompose the matrix as: ∈ M = UΣV T where U and V are T -orthogonal, and Σ is a diagonal matrix.

9 Proof. Essentially the same as for the -SVD, except that the role of the - spectral theorem is changed to the T -spectral∗ theorem. ∗

5 Acknowledgements

I am thankful to Gregory Gutin and Ilya Spitkovsky for helpful comments and suggestions. The software Sympy v1.6 ([6]) was helpful for ﬁnding some of the proofs.

References

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