(Almost) Jordan Form

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1 (Almost) Jordan Form These notes will demonstrate most of the basic steps for getting to Jordan canonical form of a complex matrix They will also get to to an important diagonal-nilpotent decomposition, which we will require later [These notes owe a tremedous debt to the beautiful notes of Ed Nelson (Princeton) which are posted on a website of Andre Reznikov (Bar-Ilan) It is worth your while to do a little internet sleuthing to find these] The first result is not sexy, but actually does all of the hard work Theorem Let A M n (F) (F = C or R) and λ be an eigenvalue of A in F (i) There is a positive integer m (geometric multiplicity) for which ker(a λi) k ker(a λi) m for each positive integer k Each of the subspaces ker(a λi) m and ran(a λi) m are invariant for A, hence for any polynomial, p(a), in A (ii) We have F n = ker(a λi) m ran(a λi) m Hence there is g in GL n (F) for which [ ] λid + N 0 A = g g 1 0 R where d = dim ker(a λi) m and N M d (F) with N m = 0 Proof (i) To begin with, we simply observe that ker(a λi) ker(a λi) 2 By finite dimensionality of F n, this non-decreasing chain of subspaces must stabilise; let r be the smallest value for which ker(a λi) m = ker(a λi) k for each k m We observe that if x ker(a λi) m, then (A λi) m Ax = A(A λi) m x = 0 so A[ker(A λi) m ] ker(a λi) m x = (A λi) m y for some y Hence Finally if x ran(a λi) m, then Ax = A(A λi) m y = (A λi) m Ay ran(a λi) m so A[ran(A λi) m ] ran(a λi) m The same arguemet holds for p(a) (ii) If x ker(a λi) m ran(a λi) m, then on one hand 0 = (A λi) m x, while on the other, x = (A λi) m y for some vector y Thus 0 = (A λi) m x = 1

2 (A λi) 2m y, so y ker(a λi) 2m = ker(a λi) m, whence x = (A λi) m y = 0 By rank-nullity theorem, we find that n = dim ker(a λi) m +dim ran(a λi) m, so we find that F n is a direct sum of these subspaces Let B 1 = {ξ 1,, ξ d } be a basis for ker(a λi) m, and B 2 = {ξ d+1,, ξ n } a basis for ran(a λi) m Then the restricted operator (A λi) ker(a λi) m is nilpotent and admits matrix with respect to B 1 of the form N, with N m = 0 Let R be the matrix of A ran(a λi) m Then if g is the change of basis matrix from B 1 B 2 = {ξ 1,, ξ n } to the standard basis, we get the desired result The following is essentially a simple induction on the remainder block R from the theorem above The details are left to the reader We take it for granted that a complex matrix admits at least one eigenvalue and at least one complex eigenvector Corollary (Almost Jordan Decoposition) Let A M n (C) and λ 1,, λ s be a full list of distinct eigenvalues for A (s is the size of the spectrum) Let m i be so ker(a λ i I) m i ker(a λ i I) k for any positive integer k, and d i = dim ker(a λ i I) m i Then there are nilpotent matrices N i in M di (C) with N m i i = 0 and a g in GL n (C) for which λ 1 I d1 + N λ A = g 2 I d1 + N 2 0 g 1 ( ) 0 0 λ s I s + N s Furthermore, if all eigenvalues are in R, then we can arrange that g GL n (R), as well If one is willing to invest the extra effort to show that a d d nilpotent matrix N is similar to one of the form 0 η η η d where η 1, η d 1 {0, 1} then she has effectivley shown the usual Jordan form In fact if m is the smallest integer for which N m = 0, then there are 2

3 η i, η i+1,, η i+m which are all 1, and no consecutive chain of such η i = 1 may be longer than n Observe that if a matrix admits the form of a block decomposition A A A = 2 ( ) A s then for any polynomial p(z) we have p(a 1 ) p(a p(a) = 2 ) p(a s ) Corollary (Almost Cayley-Hamilton Theorem) Given A in M n (C), as above, the polynomial µ A (z) = s k=1 (z λ i) m i satisfies µ A (A) = 0 Proof Following ( ) and then ( ), we see that λ 1 I d1 + N λ µ A (A) = g µ 2 I d1 + N 2 A 0 g λ s I s + N s µ A (λ 1 I d1 + N 1 ) µ = g A (λ 2 I d1 + N 2 ) 0 g µ(λ s I ds + N s ) Each block contains a factor [(λ k I dk +N k ) λ k I dk ] m k = Nk m k = 0 and is thus 0 The polynomial µ A is the minimal polynomial of A We obtain a factorisation of the block decomposition given in ( ) A I d1 0 0 I d I A = d2 0 A I d I ds 0 0 I ds 0 0 A s 3

4 from which it easy follows that det A = s k=1 det A s Hence if one is willing to show that det(λi d + N) = λ d, whenever N is a d d nilpotent matrix (this will follow form a form of Engel s Theorem, later in the course), then it is an easy step to show that the characteritic polynomial of A, above, is p A (z) = s k=1 (z λ i) d i and hence the Cayley-Hamilton Theorem holds Now for a different perspective on this result We say that A in M n (C) is diagonalisable if there is g in GL n (C) such that A = g diag(α 1,, α n ) g 1 for some α 1,, α n in C Notice that A is diagonalisable if and only if the minimal polynomial µ A (z), above, has multiplicity m k = 1 for each k Diagonal-Nilpotent Decomposition Theorem Let A M n (C) Then there is a unique decomposition A = A D + A N where A D is diagonalisable, A N is nilpotent, and [A D, A N ] = 0 Furthermore, there are polynomials p D (z) and p N (z) for which A D = p D (A) and p N (A) = A N Proof Let us exhibit, first, such a decomposition In the notation of ( ) let λ 1 I d1 0 0 N λ A D = g 2 I d1 0 N 0 g 1 and A N = g 2 0 g λ s I ds 0 0 N s Now define for each k = 1,, s, polynomials q k (z) = j=1,,s j k (z λ j ) and q k (z) = 1 q k (λ k ) q k(z) 4

5 As in the proof of the Almost Cayley-Hamilton Theorem we compute q k (A) = g qk (λ k I dk + N k ) 0 g Observe that q k (λ k +z) is simply a polynomial with constant constant coefficient 1, and hence q k (λ k I dk + N k ) = I + r k (N n ), where r is a polynomial with constant coefficient 0, Hence r k (N k ) is itself, nilpotent; in fact r k (N k ) m k = 0 Thus we have that [I + r k (N k )][I r k (N k ) + + ( 1) m k 1 r k (N k ) m k 1 ] = I Noting that r k (z) = q(λ k + z) 1 we find that the polynomial p k (z) = q(λ k + z)[1 (q(λ k + z) 1) + + ( 1) m k 1 (q(λ k + z) 1) m k 1 ] satisfies p k (A) = gp k g 1, where P k is block-diagonal with I dk in the kth block and zeros elsewhere Finally set s p D (z) = λ k p k (z) k=1 and we find that p D (A) = A D Hence p N (z) = p D (z) z Now we prove uniqueness Suppose A = D+N where D is diagonalisable, N is nilpotent and [D, N] = 0 Then [D, A] = [D, D +N] = 0, and, similarly, [N, A] = 0 Thus [D, A D ] = [D, p D (A)] = 0, and, similarly, [N, A N ] = 0 Hence the equation A D + A N = A = D + N implies A D D = N A N But then the binomial theorem implies that (N A N ) 2n = 0, so A D D = N A N is nilpotent Now let E k = gp k g 1 = p k (A), from above Then Ek 2 = E k and [D, E k ] = 0, so [g 1 Dg, P k ] = 0 It follows, just as in the proof of the first theorem, that we have block-diagonal form D g 1 0 D Dg = D s 5

6 But since the minimal polynomial µ D (z) has multiplicites m i = 1 by diagonalisability of D, and µ D (D k ) = 0 for eack k, it follows that each block D k is diagonalisable Hence there is a block-diagonal h in GL n (C) for which h 1 g 1 Dgh is diagonal Notice that h 1 g 1 A D gh = g 1 A D g remains diagonal Hence A D D is diagonalisable Thus A D D is both nilpotent and diagonalisable, so A D D = 0 We say two diagonalisable n n matrices A and B are simultaneously diagonalisable if there are complex numbers α 1,, α n, β 1,, β n and a g in GL n (C) such A = g diag(α 1,, α n ) g 1 and B = g diag(β 1,, β n ) g 1 In the course of proving the above result we showed the non-trivial direction of the following Corollary Two diagonalisable matrices are simultaneously diagonalisable if and only if they commute Written by Nico Spronk, for use by students of PMath 763 at University of Waterloo 6