Jordan chain Aim lecture: We introduce the notions of Jordan chains & Jordan form tableaux which are the key notions to proving the Jordan canonical form theorem. Throughout this lecture we fix the following notn. Let T : V −! V be linear & λ 2 F be an e-value & N = T − λI . Defn n n−1 Let v 2 Eλ(n) − Eλ(n − 1) so N v = 0 but N v 6= 0.A Jordan chain of length n is a row matrix of the form C = (Nn−1v Nn−2v ::: v) 2 V n. We say v is a seed for the Jordan chain & Ni v are vectors of the chain. The Jordan chain n space associated to C is im (C : F −! V ) ≤ V . E.g. T = J3(λ) Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 1 / 10 T -invariance of Jordan chain spaces Prop Let C = (vn−1 vn−2 ::: v0) be a Jordan chain. The Jordan chain space im C is T -invariant. In fact 1 T vi = vi+1 + λvi for i = 0;:::; n − 2. 2 T vn−1 = λvn−1. Proof. By our criterion for T -invariance lect. 20, suffice prove 1) & 2). But i vi = N v0 so i T vi = (N + λI )vi = Nvi + λvi = NN v0 + λvi = vi+1 + λvi n if we define vn = N v0 = 0. The propn follows. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 2 / 10 Linear independence of Jordan chain Prop Let C = (Nn−1v Nn−2v ::: v) be a Jordan chain of length n. i 1 N v 2 Eλ(n − i) − Eλ(n − i − 1). n 2 C : F −! V is injective so the vectors of the Jordan chain form a basis for the Jordan chain space im C. n−1 n−2 n−i 3 im C \ Eλ(i) = Span(N v; N v :::; N v) Proof. 1) Recall v 2 Eλ(n) − Eλ(n − 1) so in particular is lin indep modulo i Eλ(n − 1). Key lemma lecture 28 shows that N v 2= Eλ(n − i − 1). Also n−i i n i N [N v] = N v = 0 so N v 2 Eλ(n − i) & 1) is proved. n−1 n−2 i We prove 2) by showing inductively that Ci = fN v; N v;:::; N vg is lin indep. Suppose that Ci is lin indep. Note that 1) =) Span(Ci ) ⊆ Eλ(n − i). i−1 Hence 1) also shows N v 2= Span(Ci ) so Ci−1 is lin indep too. P j For 3) just note w = j βj N v 2 im C \ Eλ(i) iff n−i−1 i X i+j 0 = N w = βj N v: j=0 Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 3 / 10 Jordan chains & blocks Prop n Let C = (vn−1 ::: v0): F −! V be a co-ord system. Then C is a Jordan chain iff the matrix representing T wrt C is Jn(λ). In particular, I = (e1 ::: en) is a Jordan chain for Jn(λ). Proof. This is an easy calculation. Recall from our propn on T -invariance of Jordan chain spaces that T vi = vi+1 + λvi for i = 0;:::; n − 2 & T vn−1 = λvn−1 Hence the first column of C −1 ◦ T ◦ C is −1 −1 T C (T (Ce1)) = C λvn−1 = (λ, 0;:::; 0) whilst the (n − i)-th column (for i < n − 1) of C −1 ◦ T ◦ C is −1 T C (vi+1 + λvi ) = (0;:::; 0; 1; λ, 0;:::; 0) −1 where the λ is the (n − i)-th entry. Hence C ◦ T ◦ C = Jn(λ). The computation reverses to give the converse. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 4 / 10 Role in existence of Jordan canonical forms The Jordan canonical form thm states in particular, that if dim V < 1 & F is alg closed, then wrt some co-ord system, T is a direct sum of Jordan blocks. By the thm on decomposition into gen e-spaces, in proving this, we may as well assume that V = Eλ(1) is a single gen e-space. Upshot The existence of Jordan canonical forms thus reduces to showing that each gen e-space Eλ(1) is a direct sum of Jordan chain spaces. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 5 / 10 Jordan form tableau Suppose that V = Eλ(1)& T has a Jordan canonical form so V is a direct sum of Jordan chain spaces. Let C1;:::; Cr be the Jordan chains listed in weakly decreasing (= non-increasing) lengths m1 ≥ m2 :::; ≥ mr . Defn Assuming existence of the Jordan canonical form & V = Eλ(1), the Jordan form tableau of T is the diagram consisting of a column of m1 boxes on the left, another column of m2 boxes to the right of that with the same top, followed by another column of m3 boxes to the right of that, and so on. We may fill the tableau by putting the vectors of the Jordan chains in the boxes of the corresponding columns, in descending order so the seeds are at the bottom. E.g. T = J1(λ) ⊕ J3(λ) ⊕ J2(λ) ⊕ J3(λ) ⊕ J1(λ) A We first write down the Jordan chains. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 6 / 10 Example cont'd The Jordan form tableau here is Note the upside-down staircase shape. Prop 1 The vectors in the first i rows of the filled-in Jordan form tableau form a basis for Eλ(i). 2 dim Eλ(i) is equal to the number of boxes in the first i rows of the Jordan form tableau. 3 The i-th row of the Jordan form tableau is a basis for a vector space complement to Eλ(i − 1) in Eλ(i). E.g. above Proof. Note that 1) =) 2) & 3) so we only need prove 1). Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 7 / 10 Proof of proposition We have proved 1) in the case of a single Jordan chain. The case of a finite direct sum of Jordan chain spaces follows from the next lemma. Lemma (Kernel of direct sums) Suppose T = T1 ⊕ ::: ⊕ Tr : ⊕i Vi −! ⊕i Vi for some endomorphisms T1;:::; Tr of V1;:::; Vr ≤ V . Let f (x) 2 F[x]. Then 1 ker f (T ) = ker f (T1) ⊕ ::: ⊕ ker f (Tr ). 2 In particular, i i Eλ(i) = ker(T1 − λI ) ⊕ ::: ⊕ ker(Tr − λI ) : Proof. We saw in lecture 26 that f (T ) = f (T1) ⊕ ::: ⊕ f (Tr ) so T v = (v1;:::; vr ) 2 ker f (T ) iff Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 8 / 10 Uniqueness of Jordan canonical form Theorem Suppose T has two Jordan forms J(1) &J(2). The the Jordan blocks of J(1) & J(2) are the same up to permutation. Proof. Note that J(1) & J(2) are similar so it suffices to show that similarity invariants determine the Jordan blocks uniquely up to permutation. (1) Let Jn1 (λ);:::; Jnr (λ) be the Jordan blocks of J with e-value λ. It suffices to show these are determined by the similarity invariants dim Eλ(i) for i 2 N. But these invariants determine uniquely the Jordan form tableau by our previous propn. Furthermore, the columns of the tableau give the sizes of the Jordan blocks so we are done. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 9 / 10 The tableau shape from Eλ(n) Suppose now that dim V < 1. Our previous propn suggests that even if we don't know if T has a Jordan canonical form, we may form the Jordan form tableau of T wrt e-value λ by placing a row of dim Eλ(1) boxes at the top a row of dim Eλ(2) − dim Eλ(1) beneath that (with same left end) another row of dim Eλ(3) − dim Eλ(2) beneath that and so forth. The upside down staircase shape is guaranteed by the following Lemma dim Eλ(i + 1) − dim Eλ(i) ≤ dim Eλ(i) − dim Eλ(i − 1) Proof. Note that if fv1;:::; vr g ⊂ Eλ(i + 1) is a basis for a vector space complement to Eλ(i) in Eλ(i + 1) then dim Eλ(i + 1) − dim Eλ(i) = r. Our key lemma lecture 28 shows that fNv1;:::; Nvr g ⊂ Eλ(i) is lin indep modulo Eλ(i − 1). Hence dim Eλ(i) ≥ dim(Eλ(i − 1) ⊕ F Nv1 ⊕ ::: ⊕ F Nvr ) = dim Eλ(i − 1) + r: Re-arranging gives the inequality of the lemma. Daniel Chan (UNSW) Lecture 29: Jordan chains & tableaux Semester 2 2012 10 / 10.