Linear and Bilinear Algebra (2WF04) Jan Draisma

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1 Linear and Bilinear Algebra (2WF04) Jan Draisma

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4 CHAPTER 3 The minimal polynomial and nilpotent maps 3.1. Minimal polynomial Throughout this chapter, V is a finite-dimensional vector space of dimension n over K, and φ is a linear map from V to itself. We write φ 0 for the identity map I : V V, φ 2 for φ φ, φ 3 for φ φ φ, etc. Let p = p(t) K[t] be a polynomial in one variable t, namely p = c 0 + c 1 t c d t d. Then we define p(φ) := c 0 I + c 1 φ c d φ d. We say that φ is a root of p if p(φ) is the zero map V V. A straightforward calculation shows that the map K[t] L(V ), p p(φ) is an algebra homomorphism, i.e., it satisfies (p + q)(φ) = p(φ) + q(φ) and (pq)(φ) = p(φ)q(φ) = q(φ)p(φ), where the multiplication in K[t] is multiplication of polynomials and the multiplication in L(V ) is composition of linear maps. Since L(V ) has finite dimension over K, the linear maps I, φ, φ 2,... cannot be all linearly independent. Hence the linear map p p(φ) must have a non-zero kernel I. This kernel is an ideal in K[t]: if q, r I, which means that q(φ) = r(φ) = 0, then also q + r I and pq I for all p K[t]. Let p min = p min (φ) be a non-zero polynomial of minimal degree in I that has leading coefficient equal to 1, i.e., which is monic. Every polynomial p I is a multiple of p min. Indeed, by division with remainder we can write p = qp min + r with deg r < deg p min. But then r = p qp min lies in I, and hence must be zero since p min had the least degree among non-zero elements of I. Hence p = qp min. In particular, it follows that p min I with the required properties (minimal degree and monic) is unique, and this is called the minimal polynomial of φ. The minimal polynomial of a square matrix is defined in exactly the same manner. Exercise Write a Mathematica function MinimalPolynomial that takes as input a square integer matrix A and a number p that is either 0 or a prime, and that outputs the minimal polynomial of A over Q if p = 0 and the minimal polynomial of A over K = Z/pZ if p is a prime. Exercise Show that p min is an invariant of φ under the conjugation action of GL(V ) on L(V ) Chopping up space using p min The minimal polynomial will guide us in finding the so-called rational Jordan normal form of φ. The first step is the following lemma. Lemma Suppose that φ is a root of p q, where p and q are coprime polynomials. Then V splits as a direct sum ker p(φ) ker q(φ). 19

5 20 3. THE MINIMAL POLYNOMIAL AND NILPOTENT MAPS Perhaps a warning is in order here: (pq)(φ) = 0 does not imply that one of p(φ), q(φ) must be zero. Proof. Write 1 = ap + bq for suitable a, b K[t]; these can be found using the extended Euclidean algorithm for polynomials. Then for v V we have v = a(φ)p(φ)v + b(φ)q(φ)v. The first term lies in ker q(φ), because q(φ)a(φ)p(φ)v = a(φ)(p(φ)q(φ))v = 0 (while linear maps in general do not commute, polynomials evaluated in φ do check this if you haven t already done so). Similarly, the second term b(φ)q(φ)v lies in ker p(φ). This shows that ker p(φ) + ker q(φ) = V. Finally, if v is in the intersection ker p(φ) ker q(φ), then the above shows that v = = 0. Hence the sum in ker p(φ) + ker q(φ) is direct, as claimed. Exercise Prove that, in the setting of the lemma, im p(φ) = ker q(φ) (and vice versa). In the lemma, both subspaces V 1 := ker p(φ) and V 2 := ker q(φ) are stable under φ, which means that φ maps V i into V i for i = 1, 2. Indeed, if v V 1, then p(φ)φv = φp(φ)v = 0 so that φv V 1, as well. Now factor p min as p m1 1 p mr r where the p i are distinct, monic, irreducible polynomials in K[t], and the exponents m i are strictly positive. Write q i := p mi i. The polynomial q 1 is coprime with q 2 q r. Hence we may apply the lemma and find V = ker q 1 (φ) ker(q 2 q r )(φ). Now let W be the second space on the right-hand side. This space is stable under φ, and the restriction φ W on it is a root of the polynomial q 2 q r. Moreover, q 2 is coprime with q 3 q r. Hence we may apply the lemma with V replaced by W, φ replaced by φ W, p replaced by q 2 and q replaced by q 3 q r to find W = ker q 2 (φ W ) ker(q 3 q r )(φ W ). Now ker q 2 (φ) V and ker(q 3 q r )(φ) V are both contained in W, so that we may as well write W = ker q 2 (φ) ker(q 3 q r )(φ), and hence V = ker q 1 (φ) ker q 2 (φ) ker(q 3 q r )(φ). Continuing in this fashion we find that V = ker q i (φ); in what follows we write V i := ker q i (φ) and φ i : V i V i for the restriction of φ to V i. Now recall that we are trying to find invariants and normal forms for linear maps under conjugation. Suppose that we have a second element ψ L(V ) and whant to find out whether it is in the same orbit as φ. First, we have already seen in Exercise that both minimal polynomials must be the same. Moreover, if φ = gψg 1, then for each i = 1,..., r, the linear map g maps ker q i (ψ) isomorphically onto V i. This proves that the dimensions of these spaces are invariants. Conversely, if these dimensions are the same for φ and for ψ, then one can choose arbitrary linear isomorphisms h i : ker q i (ψ) V i. These together give a linear isomorphism

6 3.3. THE PARTITION ASSOCIATED TO A NILPOTENT MAP 21 h : V V with the property that for ψ := hψh 1 we have ker q i (ψ ) = V i. Write ψ i for the restriction of ψ to V i. Of course, ψ and φ are GL(V )-conjugate if and only if ψ and φ are, and this is true if and only if each ψ i is GL(V i)-conjugate to each φ i. Indeed, if φ i = g i ψ i g 1 i, then the g i together define an element g of GL(V ) conjugating ψ into φ. Conversely, a g conjugate ψ into φ necessarily leaves each V i stable, and the restrictions g i to the V i conjugate ψ i into φ i. Now replace φ by a single φ i, and V by V i. The point of all this is that we have thus reduced the search for invariants and normal forms to the case where φ L(V ) is a root of a polynomial p m with p irreducible and monic. Then it follows that the minimal polynomial of φ itself must also be a power of p; assume that we have taken m minimal, so that p min = p m. In the next chapter, we will split such a linear map φ into two commuting parts: a semi-simple part φ s whose mimimal polynomial is p itself, and a nilpotent part φ n. The remainder of the present chapter deals with the latter types of matrices The partition associated to a nilpotent map Starting afresh, let V be a finite-dimensional vector space of dimension n, and let φ L(V ) be a linear map such that φ m = 0 for some sufficiently large m; such linear maps are called nilpotent. Note that any linear map conjugate to φ is then also nilpotent. Let m be the smallest natural number with φ m = 0, so that p min (φ) = t m. Exercise In the setting preceding this exercise, fix a non-zero vector v V and let p 0 be the largest exponent for which φ p v is non-zero. Prove that v, φv,..., φ p v are linearly independent. Conclude that m n. If p = n 1, give the matrix of φ with respect to the basis v, φv,..., φ n 1 v. Write K i V for ker φ i. Then we have 0 = K 0 K 1... K m = V because if φ i v = 0 then certainly φ i+1 v = φ(φ i v) = 0. Moreover, then also φ i 1 (φv) = 0, so that φ maps K i into K i 1. Hence φ induces linear maps φ i : K i+1 /K i K i /K i 1 mapping v + K i to φv + K i 1. Note that each φ i is injective: if φv already lies in K i 1, then φ i v = φ i 1 φv = 0, so that v K i and v + K i = 0. This shows that dim(k 1 /K 0 ) dim(k 2 /K 1 )... dim(k m /K m 1 ). Write λ i := dim(k i /K i 1 ) = dim K i dim K i 1 for i = 1,..., m. Then λ 1 + λ λ m (= dim K m dim K 0 ) = n and λ 1 λ 2... λ m > 0. A sequence (λ 1,..., λ m ) with these two properties is called a partition of n with parts λ 1,..., λ m. We call λ the partition associated to φ though this is non-standard terminology: usually the partition associated to φ is defined as the transpose of λ in the following sense. Definition Given any partition λ = (λ 1,..., λ m ) of n, we define a second partition λ = (λ 1,..., λ p) of n by p := λ 1 and λ k := {i {1,..., m} λ i k}. The partition λ is called the transpose of λ.

7 22 3. THE MINIMAL POLYNOMIAL AND NILPOTENT MAPS λ = (5, 5, 2, 1) λ = (4, 3, 2, 2, 2) Figure 1. A Young diagram of a partition and its transpose. Why this is called the transpose is best illustrated in terms of Young diagrams depicting λ and λ ; see Figure 1. Exercise Prove that λ is, indeed, a partition of n, and that (λ ) = λ. A straightforward check shows that the associated partition is an invariant under conjugation. The following theorem states that it is a complete invariant for nilpotent linear maps. Theorem Two nilpotent linear maps φ, ψ L(V ) are in the same orbit of the conjugation action of GL(V ) if and only if the partitions of n as constructed above from φ and ψ are the same. Morever, every partition of n occurs in this manner. The proof of the first part of this theorem will take up most of the remainder of this chapter. For the second part, it suffices to exhibit an n n-matrix giving rise to any partition λ = (λ 1,..., λ m ) of n. Let λ = (λ 1,..., λ p) be the transpose of λ. Let A be the block-diagonal n n-block matrix J... where J i =......, J p with λ i rows and columns (the block consists of a single zero if λ i equals 1). We claim that A gives rise the partition λ in the manner above. Indeed, J i contributes 1 to dim ker A k dim ker A k 1 if λ i k, and zero otherwise. Hence dim ker Ak dim ker A k 1 is the number of indices i for which λ i is at least k. This gives the transpose of λ, which is λ itself again Completeness of the associated partition Given a nilpotent linear map φ L(V ) with associated partition λ = (λ 1,..., λ m ) whose transpose is λ = (λ 1,..., λ p), we will prove that there exists a basis of V with respect to which φ has the matrix A constructed above. This then proves that any other nilpotent ψ L(V ) with the same associated partition λ is conjugate to φ. Let us first discuss the proof informally. To find such a basis, we need only find the right-most vector of each Jordan block ; the other vectors in each block are then found by repeatedly applying φ. The right-most vector in a Jordan block of size

8 3.4. COMPLETENESS OF THE ASSOCIATED PARTITION 23 i lies in ker φ i but not in ker φ i 1 (since then it would generate a block of smaller size) and neither in φ ker φ i+1 (since then it would lie in larger Jordan blocks ). This motivates the following construction: For i = 1,..., m let U i be a vector space complement in ker φ i of the span ker φ i 1 + φ ker φ i+1. This U i will play the role of span of the right-most vectors in Jordan blocks of size i. Then φ j is injective on U i for all j = 0,..., i 1 indeed, if φ j u = 0 for u U i, then certainly φ i 1 u = 0, and hence u = 0 since ker φ i 1 U i = {0} by choice of U i. We claim that in the φ-stable subspace W i := U i + φu i φ i 1 U i of V the sum is in fact direct. Indeed, suppose that u 0 + φu φ i 1 u i 1 = 0, where all u s are elements of U i. Applying φ i 1 to this yields that u 0 is zero. Applying φ i 2 then yields that u 1 is zero, etc. Hence u 0,..., u i 1 are all zero. Note that, so far, we have only used that U i intersects ker φ i 1 trivially. After choosing any basis C i of U i, we obtain a basis B i := i 1 j=0 φ j C i of W i, and after ordering B i suitably, we obtain an ordered basis such that the matrix of the restriction of φ to W i is a block diagonal matrix with dim U i -many Jordan blocks of size i. Next we set out to prove that V is the direct sum of all W i with i = 1,..., m; here we will use that U i intersects im φ + ker φ i 1 trivially. This follows from the second condition on U i, namely, if u U i can be written as φv + x with v V and x ker φ i 1, then we have 0 = φ i u = φ i+1 v + 0, so that v ker φ i+1. Hence u φ ker φ i+1 + ker φ i 1, and since U was required to intersect this space trivially, we find u = 0. Thus U intersects im φ+ker φ i 1 trivially, as claimed. Consequently, we have (1) φ j U i im φ j+1 = {0} for j = 0,..., i 1. Indeed, if φ j+1 v = φ j u with u U i and v V, then φ j (u φv) = 0, and hence in particular u φw ker φ i 1, so that u im φ + ker φ i 1, hence u = 0. Now suppose that 0 =φ 0 u 1,0 +φ 0 u 2,0 + φ 1 u 2, φ 0 u m 2, φ m 3 u m 2,m 3 +φ 0 u m 1,0 + φ 1 u m 1, φ m 2 u m 1,m 2 +φ 0 u m,0 + φ 1 u m,1 + φ 2 u m, φ m 1 u m,m 1 where u ij U i. Applying φ m 1 we find that φ m 1 u m,0 = 0, so u m,0 = 0. Then applying φ m 2 we find that φ m 2 u m 1,0 + φ m 1 u m,1 = 0.

9 24 3. THE MINIMAL POLYNOMIAL AND NILPOTENT MAPS By (1) with j = m 2 and i = m 1 we find that both terms must be zero. Using injectivity of φ j on U i with i > j, the first term yields u m 1,0 = 0. Similarly, we find u m1 = 0. Then applying φ m 3 we find, φ m 3 u m 2,0 + φ m 2 u m 1,1 + φ m 1 u m,2 = 0. This implies that the first term lies in im φ m 2, and by (1) applied with j = m 3 and i = m 2 we find that u m 2,0 = 0. Then the second term lies in im φ m 1, and applying (1) with j = m 2 and i = m 1 we find that u m 1,1 = 0. Then, finally, also the last term is zero. Repeating this reasoning for all of the diagonals in the expression above, we find that all u i,j are zero, and hence that the sum in W W m is direct. Finally, this sum is also all of V, as you are asked to prove below. Exercise By a dimension count, prove that V is spanned by the W i. Here you will need that U i is a vector space complement in ker φ i of ker φ i 1 +φ ker φ i+1, rather than just a vector space intersecting the latter space trivially (the latter is also true for the zero space, for instance). Exercise Write a Mathematica program that takes as input a nilpotent matrix and computes the associated partition. Apply this program to the matrix Exercise Let n = 3 and let K = F q be a field with q elements. (1) How many orbits does GL 3 (K) have by conjugation on the set of nilpotent 3 3-matrices? For each orbit, give a representative matrix A as above. (2) For each of these matrices A, compute the cardinality of the stabiliser {g GL 3 (K) gag 1 = A} of A, and deduce the cardinality of these orbits.