Jordan Normal FormApril 26, 2007Definition: A Jordan block is a square matrix B whose diagonal entriesconsist of a single scalar λ, whose superdiagonal entires are all 1, and all ofwhose other entries vanish. For example:λ 1 0 0 · · · 00 λ 1 0 · · · 00 0 λ 1 · · · 0· · · · · · · · · · · · · · · · · ·0 0 0 · · · 0 λTheorem: Let T be a linear op erator on a finite dimensional vector spaceV . Suppose that the characteristic polynomial of V splits. Then there existsa basis for T such that [T ]βis a direct sum of Jordan blocks.The first step in the proof of this theorem is to us e the direct sum decom-position of V into generalized eigenspaces Kλ. Then it suffices to prove thetheorem for the restriction of T to each Kλ. On Kλ, let Sλ:= T − λI. If wecan find a basis β of Kλwith respect to which Sλis a sum of Jordan blocks,then the same will be true for T . On Kλ, there exists an r such that Srλ= 0.Thus it suffices to consider the special case of operators with this property.Let V be a finite dimensional vector space over a fie ld F . A linear operatorN: V → V is said to be nilpotent if Nr= 0 for some positive integer r. LetN be a nilpotent operator on a finite dimensional vector space V . For eachi, let Ribe the image of Ni. Each Riis a linear subspace of V and isN-invariant, and 0 = Rr⊆ Rr−1· · · ⊆ R1⊆ V . Since N is nilpotent itis not injective (unless V = 0). Thus the kernel K of N is not zero anddim R1= dim V − dim K < dim V .Let (v1, v2, · · · vs) be a basis for V Then [N]βis a Jordan block if and onlyif N (v1) = 0, N(v2) = v1, and N (vi) = vi−1for all i > 1. This motivates the1following definition.Definition: An N-cycle is a sequence (v1, v2, · · · , vs) of nonzero vectorssuch that N(vi) = vi−1for all i > 1 and N(v1) = 0.If (v1, · · · , vs) is an N-cycle, then v1= Ns−1(vs), so v1∈ Rs−1. Conversely,if v ∈ Rs−1, say v = Rs−1(x), then (Rs−1(x), Rs−2(x), · · · x) is an N-cyclewhose initial vector is v. If v belongs to Rs−1but not to Rs, then s is thelength of the longest N-c ycle starting with v.Definition: An N-cycle (v1, · · · , vs) is maximal if v16∈ Rs.It is clear that every nonzero element of the kernel K of N is containedin some maximal N-cycle.Lemma: Let (γ1, γ2, · · · γp) be a sequence of N-cycles. Then if the cor-responding sequence of initial vectors is linearly independent, so is the con-catenated sequence γ1∪ γ2∪ · · · ∪ γp.Proof: Say γi= (vi,1, vi,2, · · · vi,ni). Our assumption is that the sequence(v1,1, v2,1, · · · , vp,1) is linearly independent, and we want to prove that theentire (multi-indexed) sequece (vi,j) is linearly indep endent. We prove thisby induction on the maximum of the ni’s. If all the ni’s are 1, there isnothing to prove, since we assumed that the sequence of initial vectors islinearly independent. For the induction step, for each i let γ0ibe the (possiblyempty) Jordan cycle obtained by omitting the last term. The inductionassumption says that the union of these is linearly independent. SupposethatPai,jvi,j= 0. Applying N, we deduce thatPai,jNvi,j= 0, i.e., thatPi,jai,jvi−1,j= 0, where here for each j, i ranges between 2 and ni. This is thesum over the corresponding truncated cycles γ0i. The induction assumptionsays that ∪γ0iis linearly independent, so ai,j= 0 for i ≥ 2. Thus the originalsum reduces to a linear combination of the inital vectors, which we assumedto be linearly independent. Hence each a1,j= 0 as well.Recall that we have linear subspaces 0 ⊆ Rr⊆ Rr−1⊆ · · · V . Considerthe corresponding sequence of subspaces of K.0 = Rr∩ K ⊆ Rr−1∩ K ⊆ · · · ⊆ R1∩ K ⊆ K.We shall say that a basis α of K is adapted to N if for each i, α ∩ Riis abasis of Ri∩ K. It is clear that such bases always exist: start with a basisfor Rr−1, extend it to a basis for Rr−2, and continue.Definition: A sequence of maximal N -cycles (γ1, · · · γq) is full if thecorresponding sequence of inital vectors (v1, · · · , vq) is a basis of K which isadapted to N .2It is clear that full sequences of N-cycles exist: start with a basis for Kwhich is adapted to N, and for each vector v in the basis, find a a maximalcycle starting with v.Theorem: Every full sequence of maximal N-cycles forms a basis for V .Proof: Let (γ1, γ2, · · · γp) be a full sequence of maximal N-cycles. By as-sumption, the corresponding sequence of initial vectors is linearly indepen-dent, and hence by the lemma, the concatenation of γi’s is linearly indepen-dent. It suffices to show that it also spans V . We do this by induction on thesmallest r such that Nr= 0. If r = 1, then V = K and there is nothing toprove, since we assumed that the initial vectors span K. Let V0:= Im(N)and for each i, let γ0ibe γiwith the last element omitted. In fact, γ0i= N(γi),with zero omitted. Let N0be the restriction of N to V0. Each γ0iis containedin V0and is a maximal Jordan cycle for N0. Furthermore, γ0iis empty only ifγihas length one, which is true only if its initial (and only) vector does notbelong to V0. Thus the sequence of initial ve ctors of γ0icontains all the initialvectors of the original sequence which belong to V0. Let p0be the numberof nonempty γ0i’s. It follows that the sequence (γ01, · · · γ0p0) is maximal andfull for N0. By the induction assumption, it spans V0. Now let W be thespan of the all the γi’s. Note that by construction, W contains all of K.Furthermore, the image of W under N contains all the γ0i’s and hence all ofV0= Im(N). But then dim W = dim K + dim Im(N) = dim V , and henceW = V .Remark: For each i, let didenote the dimension of Riand let hi:=di−1− di. If α is any basis for K adapted to N, then diis the number ofelements of α which lie in Riand so hiis the number of elements of α whichlie in Ri−1but not in Ri. Corresponding to each such element there will bea maximal N-cycle of length i. Thus if β is the basis obtained as above, thecorresponding matrix [N ]βwill have exactly hiJordan blocks of length i.Let V and V0be two finite dimensional vector spaces over F , and letT be an operator on V and T0an operator on V0. Then T and T0aresometimes said to be similar if there exists an isomorphism Q: V → V0suchthat T0◦ Q = Q ◦ T , i.e., T0= Q ◦ …