Effective procedure for computing the Jordan Normal Form of nilpotent matrix Given an n × n nilpotent matrix A, define the linear transformation n n f : R ! R by f(x) = Ax: Then A represents f with respect to the standard basis. Define the subspaces n 2 n V0 = R ;V1 = cs(A);V2 = cs(A );:::;Vn = cs(A ): Note that f maps Vk into Vk for each k: This is clear when k = 0. For k ≥ 1, k n let x 2 Vk be given. Then we have x = A y for some y 2 R , hence k+1 k k f(x) = Ax = A y = A (Ay) 2 cs(A ) = Vk: For each k define fk : Vk ! Vk by fk(x) = Ax: Then each fk is a nilpotent transformation. We will find a Jordan normal basis for each Vk, working backwards from the largest value of k such that Ak 6= 0. Find the largest value of k such that Ak 6= 0. We must have Ak+1 = 0. k k k+1 We also have Vk = cs(A ) 6= f0g and fk(x) = fk(A y) = A y = 0 for all x 2 Vk. Hence fk is represented by the zero matrix with respect to any basis for Vk. The size of this basis is determined by the dimension of Vk. Since the zero matrix is Jordan normal of type J0(1; 1;::: ), we have found a Jordan normal basis for Vk. Induction Hypothesis: we have found a Jordan normal basis for Vj where j ≥ 1. Assume it has the form n1−1 0 n2−1 0 np−1 0 fA v1;:::;A v1;A v2;:::;A v2;:::;A vk;:::;A vkg where n1 n2 nk A (v1) = A v2 = ··· = A vk = 0: 1 Now we find a Jordan normal basis for Vj−1. Find ui so that A(ui) = vi n1−1 nk−1 for each i ≤ k. Expand the vectors A (v1);:::;A (vk), which belong to n1−1 nk−1 the kernel of fj−1, to a basis A (v1);:::;A (vk); vk+1; : : : ; vp for ker(fj). Then by a previous theorem, n1 0 nk 0 0 0 A u1;:::;A u1;A uk;:::;A uk;A vk+1;:::;A vp is a Jordan normal basis for Vj−1. Example: 20 1 7 19 54 61 3 60 0 0 −7 −29 −417 6 7 60 0 0 1 3 −1 7 A = 6 7 60 0 0 0 1 6 7 6 7 40 0 0 0 0 0 5 0 0 0 0 0 0 20 0 0 0 11 66 3 60 0 0 0 −7 −427 6 7 2 60 0 0 0 1 6 7 A = 6 7 60 0 0 0 0 0 7 6 7 40 0 0 0 0 0 5 0 0 0 0 0 0 20 0 0 0 0 03 60 0 0 0 0 07 6 7 3 60 0 0 0 0 07 A = 6 7 : 60 0 0 0 0 07 6 7 40 0 0 0 0 05 0 0 0 0 0 0 A Jordan normal basis for V2 is 82 11 39 > > >6−77> >6 7> <6 1 7= 6 7 : >6 0 7> >6 7> >4 0 5> :> 0 ;> 2 This belongs to the kernel of f1. We must find other vectors we can add to this to create a basis for the kernel of f1. In order to do this efficiently we must first find a matrix representation of f1. A basis for V1 is 8213 2 19 3 2 54 39 > > >607 6−77 6−297> >6 7 6 7 6 7> <607 6 1 7 6 3 7= 6 7 ; 6 7 ; 6 7 : >607 6 0 7 6 1 7> >6 7 6 7 6 7> >405 4 0 5 4 0 5> :> 0 0 0 ;> The matrix representation of f1 with respect to this basis is 20 0 −83 40 0 1 5 : 0 0 0 The kernel of this matrix is 82 3 2 39 < 1 0 = span 405 ; 415 : : 0 0 ; This implies that a basis for the kernel of f1 is 8213 2 19 39 > > >607 6−77> >6 7 6 7> <607 6 1 7= 6 7 ; 6 7 : >607 6 0 7> >6 7 6 7> >405 4 0 5> :> 0 0 ;> We must find a vector v such that 82 11 3 9 > > >6−77 > >6 7 > <6 1 7 = 6 7 ; v >6 0 7 > >6 7 > >4 0 5 > :> 0 ;> 3 have the same span as 8213 2 19 39 > > >607 6−77> >6 7 6 7> <607 6 1 7= 6 7 ; 6 7 : >607 6 0 7> >6 7 6 7> >405 4 0 5> :> 0 0 ;> We can choose 213 607 6 7 607 v = 6 7 : 607 6 7 405 0 We must also find u so that 2 11 3 6−77 6 7 6 1 7 Au = 6 7 : 6 0 7 6 7 4 0 5 0 We choose 2 54 3 6−297 6 7 6 3 7 u = 6 7 : 6 1 7 6 7 4 0 5 0 The Jordan normal basis we have found for V1 is now 82 11 3 2 54 3 2139 > > >6−77 6−297 607> >6 7 6 7 6 7> <6 1 7 6 3 7 607= fAu; u; vg = 6 7 ; 6 7 ; 6 7 : >6 0 7 6 1 7 607> >6 7 6 7 6 7> >4 0 5 4 0 5 405> :> 0 0 0 ;> 4 The vectors Au and v belong to the kernel of f0. We will fill these out to a basis for the kernel of f0. We already have a matrix representation of f0: it is the matrix A with respect to the standard basis. The kernel of A has basis 8213 2 0 3 2 0 39 > > >607 6 98 7 6 0 7> >6 7 6 7 6 7> <607 6 0 7 6 14 7= 6 7 ; 6 7 ; 6 7 : >607 6−197 6−197> >6 7 6 7 6 7> >405 4 6 5 4 6 5> :> 0 −1 −1 ;> These vectors have the same span as the vectors 82 11 3 213 2 0 39 > > >6−77 607 6 0 7> >6 7 6 7 6 7> <6 1 7 607 6 14 7= 6 7 ; 6 7 ; 6 7 : >6 0 7 607 6−197> >6 7 6 7 6 7> >4 0 5 405 4 6 5> :> 0 0 −1 ;> So set 2 0 3 6 0 7 6 7 6 14 7 w = 6 7 : 6−197 6 7 4 6 5 −1 We must find also find vectors p and q so that Ap = u and Aq = v. We can use p = e5 and q = e2. So a Jordan normal basis for V0 should be 82 11 3 2 54 3 203 213 203 2 0 39 > > >6−77 6−297 607 607 617 6 0 7> >6 7 6 7 6 7 6 7 6 7 6 7> 2 <6 1 7 6 3 7 607 607 607 6 14 7= fA p; Ap; p; Aq; q; wg = 6 7 ; 6 7 ; 6 7 ; 6 7 ; 6 7 ; 6 7 : >6 0 7 6 1 7 607 607 607 6−197> >6 7 6 7 6 7 6 7 6 7 6 7> >4 0 5 4 0 5 415 405 405 4 6 5> :> 0 0 0 0 0 −1 ;> 5 One can check that the matrix representation of f with respect to this basis is 20 1 0 0 0 03 60 0 1 0 0 07 6 7 60 0 0 0 0 07 6 7 = J0(3; 2; 1): 60 0 0 0 1 07 6 7 40 0 0 0 0 05 0 0 0 0 0 0 Therefore 2 11 54 0 1 0 0 3−1 2 11 54 0 1 0 0 3 6−7 −29 0 0 1 0 7 6−7 −29 0 0 1 0 7 6 7 6 7 6 1 3 0 0 0 14 7 6 1 3 0 0 0 14 7 6 7 A 6 7 = J0(3; 2; 1): 6 0 1 0 0 0 −197 6 0 1 0 0 0 −197 6 7 6 7 4 0 0 1 0 0 6 5 4 0 0 1 0 0 6 5 0 0 0 0 0 −1 0 0 0 0 0 −1 As a corollary, we get 2 11 54 0 1 0 0 3−1 2λ 1 7 19 54 61 3 2 11 54 0 1 0 0 3 6−7 −29 0 0 1 0 7 60 λ 0 −7 −29 −417 6−7 −29 0 0 1 0 7 6 7 6 7 6 7 6 1 3 0 0 0 14 7 60 0 λ 1 3 −1 7 6 1 3 0 0 0 14 7 6 7 6 7 6 7 = 6 0 1 0 0 0 −197 60 0 0 λ 1 6 7 6 0 1 0 0 0 −197 6 7 6 7 6 7 4 0 0 1 0 0 6 5 40 0 0 0 λ 0 5 4 0 0 1 0 0 6 5 0 0 0 0 0 −1 0 0 0 0 0 λ 0 0 0 0 0 −1 Jλ(3; 2; 1): 6.