Jordan Normal Form April 26 2007 Definition A Jordan block is a square matrix B whose diagonal entries consist of a single scalar whose superdiagonal entires are all 1 and all of whose other entries vanish For example 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 Theorem Let T be a linear operator on a finite dimensional vector space V Suppose that the characteristic polynomial of V splits Then there exists a basis for T such that T is a direct sum of Jordan blocks The first step in the proof of this theorem is to use the direct sum decomposition of V into generalized eigenspaces K Then it suffices to prove the theorem for the restriction of T to each K On K let S T I If we can find a basis of K with respect to which S is a sum of Jordan blocks then the same will be true for T On K there exists an r such that S r 0 Thus it suffices to consider the special case of operators with this property Let V be a finite dimensional vector space over a field F A linear operator N V V is said to be nilpotent if N r 0 for some positive integer r Let N be a nilpotent operator on a finite dimensional vector space V For each i let Ri be the image of N i Each Ri is a linear subspace of V and is N invariant and 0 Rr Rr 1 R1 V Since N is nilpotent it is not injective unless V 0 Thus the kernel K of N is not zero and dim R1 dim V dim K dim V Let v1 v2 vs be a basis for V Then N is a Jordan block if and only if N v1 0 N v2 v1 and N vi vi 1 for all i 1 This motivates the 1 following definition Definition An N cycle is a sequence v1 v2 vs of nonzero vectors such that N vi vi 1 for all i 1 and N v1 0 If v1 vs is an N cycle then v1 N s 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 cycle whose initial vector is v If v belongs to Rs 1 but not to Rs then s is the length of the longest N cycle starting with v Definition An N cycle v1 vs is maximal if v1 6 Rs It is clear that every nonzero element of the kernel K of N is contained in some maximal N cycle Lemma Let 1 2 p be a sequence of N cycles Then if the corresponding sequence of initial vectors is linearly independent so is the concatenated sequence 1 2 p Proof Say i 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 the entire multi indexed sequece vi j is linearly independent We prove this by induction on the maximum of the ni s If all the ni s are 1 there is nothing to prove since we assumed that the sequence of initial vectors is linearly independent For the induction step for each i let i0 be the possibly empty Jordan cycle obtained by omitting the last term The induction assumption says that the union of these is linearly independent Suppose P P that ai j vi j 0 Applying N we deduce that ai j N vi j 0 i e that P i j ai j vi 1 j 0 where here for each j i ranges between 2 and ni This is the sum over the corresponding truncated cycles i0 The induction assumption says that i0 is linearly independent so ai j 0 for i 2 Thus the original sum reduces to a linear combination of the inital vectors which we assumed to be linearly independent Hence each a1 j 0 as well Recall that we have linear subspaces 0 Rr Rr 1 V Consider the corresponding sequence of subspaces of K 0 Rr K Rr 1 K R1 K K We shall say that a basis of K is adapted to N if for each i Ri is a basis of Ri K It is clear that such bases always exist start with a basis for Rr 1 extend it to a basis for Rr 2 and continue Definition A sequence of maximal N cycles 1 q is full if the corresponding sequence of inital vectors v1 vq is a basis of K which is adapted to N 2 It is clear that full sequences of N cycles exist start with a basis for K which is adapted to N and for each vector v in the basis find a a maximal cycle starting with v Theorem Every full sequence of maximal N cycles forms a basis for V Proof Let 1 2 p be a full sequence of maximal N cycles By assumption the corresponding sequence of initial vectors is linearly independent and hence by the lemma the concatenation of i s is linearly independent It suffices to show that it also spans V We do this by induction on the smallest r such that N r 0 If r 1 then V K and there is nothing to prove since we assumed that the initial vectors span K Let V 0 Im N and for each i let i0 be i with the last element omitted In fact i0 N i with zero omitted Let N 0 be the restriction of N to V 0 Each i0 is contained in V 0 and is a maximal Jordan cycle for N 0 Furthermore i0 is empty only if i has length one which is true only if its initial and only vector does not belong to V 0 Thus the sequence of initial vectors of i0 contains all the initial vectors of the original sequence which belong to V 0 Let p0 be the number of nonempty i0 s It follows that the sequence 10 p0 0 is maximal and full for N 0 By the induction assumption it spans V 0 Now let W be the span of the all the i s Note that by construction W contains all of K Furthermore the image of W under N contains all the i0 s and hence all of V 0 Im N But then dim W dim K dim Im N dim V and hence W V Remark For each i let di denote the dimension of Ri and let hi di 1 di If is any basis for K adapted to N then di is the number of elements of which lie in Ri and so hi is the number of elements of which lie in Ri 1 but not in Ri Corresponding to each such element there will be a maximal N cycle of length i Thus if is the basis obtained as above the corresponding …