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Matrix Similarity Matrix similarity Isak Johansson and Jonas Bederoff Eriksson Supervisor: Tilman Bauer Department of Mathematics 2016 1 Abstract This thesis will deal with similar matrices, also referred to as matrix conju- gation. The first problem we will attack is whether or not two given matrices are similar over some field. To solve this problem we will introduce the Ratio- nal Canonical Form, RCF. From this normal form, also called the Frobenius normal form, we can determine whether or not the given matrices are sim- ilar over any field. We can also, given some field F , see whether they are similar over F or not. To be able to understand and prove the existence and uniqueness of the RCF we will introduce some additional module theory. The theory in this part will build up to finally prove the theorems regarding the RCF that can be used to solve our problem. The next problem we will investigate is regarding simultaneous conjugation, i.e. conjugation by the same matrix on a pair of matrices. When are two pairs of similar matrices simultaneously conjugated? Can we find any necessary or even sufficient conditions on the matrices? We will address this more complicated issue with the theory assembled in the first part. 2 Contents 1 Introduction 4 2 Modules 5 2.1 Structure theorem for finitely generated modules over a P.I.D, existence . 8 2.2 Structure theorem for finitely generated modules over a P.I.D, uniqueness . 10 3 Similar matrices and normal forms 12 3.1 Linear transformations . 12 3.2 Rational Canonical Form . 12 3.3 Similiar Matrices . 14 3.4 Smith Normal Form . 16 3.5 Finding a conjugating matrix . 18 4 Simultaneous Conjugation 25 3 1 Introduction Similarity between matrices is a well studied problem in the area of linear algebra. Most of the theory regarding matrix similarity has been known for a very long time. Ferdinand Georg Frobenius, who named the normal form we will use in this report, lived over a hundred years ago. Besides the purely mathematical reasons for this report to be interesting, it is interesting in the way that it can be used to find similarity between objects that is not immediately visible. If we have two linear transformations in different bases, how can we know if they are in fact the same? Or given a linear transformation, which is the "nicest" way of representing it as a matrix? Is it always possible to write it in this "nice" form? These questions can be answered after reading this report. We will assume that the reader is familiar with some basic abstract algebra. This includes the theory of groups, rings, principal ideal domains, fields and modules. The structure of the report will be to define new concepts and then use them to prove useful theorems. We will also give some examples so the reader will have the opportunity to apply the theory and understand it in a better way. Our sources to this paper was mainly the book Abstract algebra by David S. Dummit and Richard M. Foote, but we have also been reading a couple of reports similar to this one. From now on we let R denote a commutative unitary ring. 4 2 Modules Definition 1. Let M be an R-module. The torsion of M is all the elements of M for which there exists an r 2 R − 0 such that their product is zero i.e. T or(M) = fm 2 M j 9r 2 R − 0 ; rm = 0g: If M = T or(M), M is said to be a torsion module. Definition 2. Let M be an R-module and let A be a subset of M. If for every non-zero element x 2 M there exist unique r1; r2; :::; rn 2 R − 0 and a1; a2; :::; an 2 A; ai 6= aj when i 6= j, for some n 2 Z≥0 such that x = r1a1 +r2a2 +:::+rnan then M is a free module on the set A. The set A is called the basis for M, and the rank of M is the number of basis elements i.e. the cardinality of A. Definition 3. Let M be an R-module. The annihilator of M is all the elements of R for which the product with every element in M is zero i.e. Ann(M) = fr 2 R j rm = 0 8m 2 Mg: Definition 4. An R-module M is said to be finitely generated if it is generated by some finite subset, i.e. if there is some finite subset A = fa1; a2; :::; ang of M such that M = RA = Ra1+Ra2+:::+Ran = fr1a1+r2a2+:::+rnan j ri 2 R; 1 ≤ i ≤ ng: Definition 5. An R-module M is said to be cyclic if it is generated by a single element, i.e. if there is some element x 2 M such that M = Rx = frx j r 2 Rg: Theorem 6. Let R be a P.I.D, M be a finitely generated free R-module of rank n and N be a submodule of M. Then the following statements are true. i) N is free of rank m, m ≤ n. ii) There exists a basis fy1; y2 : : : yng for M such that fa1y1; a2y2 : : : amymg is a basis for N, where ai 2 R − f0g and ai j ai+1. Proof. The theorem holds for N = 0, so assume N is not equal to zero. Let Φ be any homomorphism from M into R. Then Φ(N) will be an ideal in R. 5 Since R is a P.I.D there exists an element aΦ in R such that Φ(N) = (aΦ). Now we want to collect all the ideals in R obtained from mapping N in M into R. Σ = f(aΦ) j Φ 2 HomR(M; R)g Since R is a P.I.D. there exists an element in Σ, which is not properly contained in any other element of Σ[1]. This implies that there exists an φ 2 HomR(M; R) such that φ(N) = (aφ). Let a1 = aφ and y be an element of N such that φ(y) = a1. Next we show that a1 divides Φ(y) for every Φ 2 HomR(M; R). Let d be a divisor of both a1 and Φ(y) such that d = r1a1 + r2Φ(y) for some r1; r2 2 R, then (a) ⊆ (d) and (Φ(y)) ⊆ (d). Now define a homomorphism ν such that ν = r1φ + r2Φ then ν(y) = r1a1 + r2Φ(y) = d. So (a1) ⊆ (ν(y)) = (d). But from how we defined a1 we must have the equality (a1) = (ν(y)) = (d) and in particular (Φ(y)) ⊆ (a1). Applying this knowledge to the natural projection homomorphism πi on the basis fx1; x2 : : : xng for M we obtain πi(y) = a1bi; bi 2 R; 1 ≤ i ≤ n: Now define y1 2 M as n X y1 = bixi; i=1 but then the equality a1y1 = y holds and φ(y1) = 1, since a1 = φ(y) = φ(a1y1) = a1φ(y1). We will now prove that y1 can be taken as an element in a basis for M and a1y1 can be taken as an element in a basis for N, i.e. a) M = Ry1 ⊕ ker(φ) b) N = Ra1y1 ⊕ (N \ ker(φ)) For a) we can take any element x 2 M and add and subtract φ(x)y1 to it, x = φ(x)y1 + (x − φ(x)y1): Note that x − φ(x)y1 is an element of ker(φ) since φ(x − φ(x)y1) = φ(x) − φ(x)φ(y1) = 0 6 Therefore x can be written as x = Ry1 +ker(φ). To see that we have a direct sum suppose ry1 is an element of ker(φ), then 0 = φ(ry1) = rφ(y1) = r =) r = 0: To prove b) we first recall that a1 divides φ(z) for all z 2 N. In a similiar way as before we rewrite z as z = φ(z)y1 + (z − φ(z)y1); and evaluating φ at z gives z = ba1y1 + (z − ba1y1): So clearly, N = Ra1y1 + (N \ ker(φ)): This is just a special case of a) so the sum is direct. We will prove the first part of the theorem by induction. If m = 0, N must be a torsion module, but since N is also a free module, N = 0. The theorem holds for N = 0 as noted earlier. By a) we see that m = 1 + rank(N \ kerφ) =) rank(N \ kerφ) = m − 1 So by induction N \ kerφ is a free R-module with rank m − 1 and N is a free module of rank m. We will again use induction and the identities a) and b) to prove the second part of the theorem. Applying a) to M we get that ker(φ) is a free R-module of rank n − 1. If we then continue with replacing M in a) with ker(φ) and N in b) with N \kerφ we get that fy2; y3; : : : ; yng is a basis for ker(φ) and that fa2y2; a3y3; : : : ; amymg is a basis for N \kerφ. Here ai 2 R, 1 ≤ i ≤ m and we have the divisibility relation a2 j a3 j · · · j am. If we again use the identities we see that fy1; y2; : : : ; yng is a basis for M and fa1y1; a2y2; : : : ; amymg is basis for N, since the sums are direct. If we can show that a1 divides a2, we are done. Let be the projection homomorphism from M to R, such that (y1) = (y2) = 1 and (yi) = 0 for all 3 ≤ i ≤ n.
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