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•i; ' ~. ............•.. ~',.. I·'.; · .':~' .'Y .' i~. ·1.·. ,~·tI I ,I AN ELEMENTARY DERIVATION Of THE JORDAN NORMAL FORM I-lITH AN APPENDIX ON LINEAR SPACES H. R. van der Vaart Didactical Report Work supported by Public Health Service Grant GM-618 from the National Institute of' General Medical Sciences to the Biomathematics Training Program, Institute of Statistics Raleigh Section Institute of Statistics Mimeograph Series' ~o. 460 po ,ary, 1966 .·~;t ", ~ ~, ',' '.' ~.t-_" . ~." - ..: ,.. .'~~:.:~~_ lt~ .....-,." '~~. ",?.~. -;l!l"-'. ,"':10.;1 \"'- 0-1 INTRODUCTION '!he aim of this mimeograph is purely didacticel. a matrix is generally regarded as an advanced topic. !be Jordan normal form of It is hard to find in the literature a complete, somewhat leisurely expositionl which in all its phases is essentially based on nothing more than the concepts of linear space and sUbspace, basis and 'Clirect sum, dimension, and the fundamental idea of mapping. In this sense the present exposition is elementary. Most of the proofs assembled here are essentially known, but the present account constitutes a consistent effort to use none but the most elementary tools throughout the entire development. It is hoped that this report may help dispel whatever misunderstandings "users" of matrix theory may have about this topic. 0-2 CONTEm'S Page 1. 2. A. L. Preliminaries 1.1. Representation of a linear mapping by a square matrix 1.2. Change of basis 1.3. Special cases 1-1 1-1 1.4. 1.5. 1-6 1-8 Eigenvalues, eigenvectors, characteristic polynomial Cayley-Hamilton theorem Canonical representations 2.1. A singular mapping on ,.., V is nilpotent on a certain subspace of ,.., V 2.2. ,.., V is a direct sum of generalized eigenspaces 2.3. A diagonal block representation of linear mappings 2.4. A canonical representation of nilpotent linear mappings 2.5. The Jordan normal form of an n X n matrix 1-3 1-1'0, -" 2-1 2-2 2-4 2-7 2-10 2-16 A-l Appendix A.l. Fields A.2. Complex numbers A.3. Linear spaces A.4. Basis, dimension, direct sum A.5. Linear mappings A-2l Some literature on linear algebra L-l A-2 A-8 A-9 A-12 1-1 1. 1,1. Preliminaries Representation of a linear maRRing by a square matrix We will consider only such linear mappings (v, ~ Gay) into the same space .w._. (V), £ : V .... V. ~ ~ as map u certain linear space £ \-le could choose one 'b:Isi s in1 the domain of X and unother1:ssis in V as (containing) the range of t , '" us but WE; will discuss only the case in which these two bases are the same. --- ----- ---- Let now (e 1', e ', ••• , e'} ic of ,..., V. For each 1'1 be one such'!:as . 2 bdng a vector in :£, is a linear combination of basis vectors: j = 1, ••• ,1'1, £'e.' , J (j =1, ••• ,1'1) rn-. .L"u<::: ' run trJ.x ] i' :=1 ••• n A=[ a.. 1 . J.J J = ...... 1'1 completely determines the linear mapping X. fact, the vector is carried into, sny, which can be uniquely repres ented by c = r. j=l 1'1 1'1 d = ~ 0i e~ = .tc = r. I'!e ~ = E.E e'a 'Y ij i ij' j • i=l j=l j J e'1 1'1' constitutes a basis of "1 , it follows that (1.1,2 ) (i = 1, ••• , u) which is the familiar formula for the transformation of vectors (such as given by their co04d~nate61 i.e. in the form caE rjej, c) when or in matrix notation: j c = ••• I'ri -I Instead of the notation used in (1,1..2) we have the well known alternatives: r l6 all (1.112b) e rOll ');]2 -I "21 "22 "'J -I'" ... 2I an r all .'.. ••• I'll (X2: ••• ••• L ::tn1 0:1'12 ••• ann i I I i _I (1.l,2c) In 1'1 d = A c • 1'2 I ... I . i LI'n.J or 1-2 ~ The matrix A is said to be the matri~ 2! relative to the basis the number of columns of ... , e'l. n A equal n, & (or the matrix representing ~) Note that the number of rows of A and the dimension of v. """ e 1-3 ~ Change of basis. n H} l e2, ••• ,en be another basis of linear combination of the e "? • Let { en , H is a then clearly each e k 1, )J n = 2,=1 E Now .t eH h = .£ e /I = )J h where B is the matrix of , .t I: en nth = t I: e~ k I3kh = n:: 12 r. .E i k 'lt llk N (k = 1, • u, n) }(J , ei aU , 1t),h e' n I3 i ik kh but also , "} • ••• , en re the basis .I: e~ So obvious ly nth = Z ~f3kh' k z aU (1.2,2) t where rl , as or A n= nB , defined in eq. (1.2,1) is the matrix expressing the mew basis vectors in terms of the old ones. ~. As an alternative to eq. (1.2,1) it is also obVious that each linear combination of the e n g : , = ellXI Hence eIIk where 8gk =0 if g t k, g.2 Two matrices said to be similar. different bases. (t = 1, ... , n) e" p n k g g2 2 g = k. fl = "" e" uk I:t ~ g g g Thus, if (k 1 =, ••, ~ I = [8gk] , has an inverse, -n- P - n) P = [PgtJ l • by n- l (1.2,5) ~. = ~" "~ I , (1.2,2) II I: eg Pg.>:,1 g if (1.2,4) This allows us to replace is a A1-1= B. A and B satisfying an equation of the form They represent the same linear transformation .t : (1.2,5) are V ... V re ~ 1-4 ~ ~. The concept of similar matrices is easiest grasped by a discussion of a linear transformation £ bases. represented by different matrices relative to different Practical situations are usually less symmetrical in that in practice a linear transformation is often introduced by its matrix representation to begin with (re some unidentified basis). By the same token vectors are often introduced in matrix form (cf. eq. (1.l,2a», i.e., by their coordinates (re the same unidentified basis). Let L be such a matrix, and assume we have n linearly iadependent vectors ------ I-'Y 1'12 722 c2 = ••• • •• , , ••• c = n • •• 'Y nn 'Y2n scalars t>ij ln 'Y2n such that (i = 1, 2, ... , n) = I: c. t>"1 j J d ••• where etc. (C l c2 ••• c n ) denotes the matrix with first column cl ' writing C for this matrix we find from (1.2,6): (1.2,,6a) L (e l c ••• cn) = (c c ••• c ) l 2 n 2 ~ll ~2l t>12 1322 ••• • •• f3n2 t3 nl second column c Pln t32n ••• ••• • •• ••• • •• t3 nn hence (1.2,6b) LC=CB, or -1 C LC=B, In other words, under the assumption (1.2,6), the transforming matrix -1 or L = C B C k and ~ are similar matriges, and C is bUilt from the column matrices I'll ci = 1'21 ••• I'ni • 2 , e .l:..:l... 1-5 Special cuses. i) The identity mapping J" defined by J v =v for each v € 1 , is represented by one and the same matrix relative to all bases: 0 0 1 n ••• ••• ••• ••• ••• 0 ••• 1 0 = I 1 • J Prove: ii) ! £ carries each v re some basis is the null € V into the null vector Q iff the matrix representing fV ~~trix CJ o o • •• o = The null matrix then represents this mapping J: particular ! will be denoted by ~) o ... 0 o ••• 0 ••••••••• o ... 0 , II J re all bases. ProvG~ (This • Define £2 d~.f .t 0 !" 1. e." £2 v =J: (,t v) for each v e:L • Similarly ,gk ~ J: 0 .e k - l so "by induction" J:s is now defined for every natural s. Now lii) - , let the matrix L represent .t relative to some basis. Prove that the matrix L2 represents J'} re the same basis. Prove that the matrix LS represents J:s re the same basis, where Let <p be a (formal) polynomial" <p(~) r =.E s is any natural number. i O;i~ • Then prove that 1.=0 <p(1- )v = E 0i .£i v i for each v Note that e if ~ is a scalar" if ~ is a mapping" if ~ is a matrix, <p(~) cp(1)) cp«t) is a scalar" is a mapping, is a matrix • € V 1-6 1.4. Eigenvalues, eigenvectors, characteristic polynomial When a non-null vector v -------- € V and a scalar A exist such that ~ (1.4,1) J:v :: AV then A is called an ~!6~~Y~~~~ of .C, v an ~!6~~Y~£~~~ of J:. 1! Y is th are roots of an E:: degree £-dimensional vector space, the eigenvalues of J: an ~- nomial equation. Proof. ----- Choose a basis of (re this basis) of v, basis. Y, Let 7.J. be . the coordinates v = re the same represent .t Let Then !v ::! ~ ej 7 j = J and .tv = AV ~... I e i O:i' 7 j j i J ~ LI , iff I:E e~ O:i. 7 . ij J. J J (1.4,la) = , E e~ A 7 i i J. i.e., iff (1.4,lb) \.. ~ (A~ij- a ij ) 7 j "'" 0 ) J where if i = j, solutions for 0ij"'" 0 if i:f j . for all r 1 = 1, ••• , n , Equation (1.4,lb) has non-null equation (1.4,1) holds for non-null vectors of v, iff A satisfies the equation :: I 0 f :: 0 Clearly this determinant is an det (AI - A) (1.4,2a) in which 1.A n nomial n-degree polynomial in A, \. (~) def X (A) A , is the highest degree term ("leading coefficient is 1 II). e~, ••• , e~} the eigenvalues The poly-. 1s culled the £~~~~£~~!:~2!!~ E~~~2~1l~~~ ~! the ~~~!:z ~. , Now choose another basis of ,... V, {ai, say XA(A) by equation (1.2,1). A of.t connected with Repeating the above argument, we find that are also to satisfy the equation 1 - ~(X) def det(XI - B) =0 7 , (n O} • One would not like the el , e0, ••• , en 2 roots of equation (1.4,2) to be different from those of eq. (l.4,2b). where B represents re the basis p ~ Indeed det(XI - B) det(XI - B) = det{TT -1 = det(XI =det(XI - A) for all scalars -1 - 1T A IT) = det(1T (AI1T .. AlT)} = det{TT -1 = det(TT ).det(XI - A)odet .Denote ~(x) = XA(X) by X(X). 1T -1 -1 for, by equation -1 (XI)'TT - TT" A IT) (XI - A)TTJ =det(XI X; - A), = for all A • The polynomial expression X(~) ------------~- --- = is called the that one can find it by computing the characteristic polynomial of the matrix representing £ ~. a produce of Equal. --- ----- n factors of form (X-X ) p • (X .. X.) , J. n roots of the equation Assume that n p relative to any basis. Since our scalars are complex numbers, X(X) complex valued) ------ p ~ where the Ai X(X) =O. n of them are different. can always be written as are the (possibly Some of these n roots may be nl n Then X(X) = (X-Xl) (X-X2 ) 2 ••• 1-8 tit 1.5. Cayle;¥' - Hamilton theorem. If X Proof. V -t ...., V ...., , polynomial of the linear maRRing X : churact~ristic '1 '1 , -t = f.!) • X(!) then is the and if A represents that basis. XC!) is According to sec. 1.3, point (iii), a linear mapping: re some basis then X(A) ,t represents According to sec. 1.3, point (ii), we have to show that the null matrix. Since = det X(A) (AI - A) xC~) re X(A) = 0 , one might think that a proof could be conducted as follows X(A) However, this is no good, since = det(AI - A) = det(D) det ( ••• ) (AI - A) , that is, :-cofactor of (AO . - a .. ) in j qi·(A) th ~ J~ J ,= (n-l) degree polynomial; ~ ( AI - A) -1 matrix (AI - A) = n-l i so Q(A) = Z Q A i i=o , Q(A) det(AI-A) = so Q(A).(AI - A) =det(7I.I - A) write XA(A) = n j E e. A , =1 en where J j=o the , Then eq. (1.5,1) can be written as n-1 i n ( E Q1 A ).(AI - A) = E 1=0 j =0 ej = e0 I e1 I -Q1A +~ = ~A + Ql =~ 2 for all scalar O A 1 • A 2 • A • I .............................n-,L, -Q n- 1A + Qn-2 = n + "'£n-1 en- 1 I =e n = ~(A).I. are functions of A • Hence -Q0 A • is scalar-valued, but X(A) is a matrix. Let the matrix Q(A) be the adjoint of Then for all A we have =0 • I. A An + A. 1-9 th Thus we have found an n-- degree polynomial, the characteristic polynomial .c : J. . . J. X of the linear mapping e? V .... ...... V ...... .c n-dimensional), which at There are other polynomials, of degree greater than n, value (:} at.£, X: ~ has the value which also have the but it is far more interesting that some linear mappings admit polynomials 'if of degree less than n for which V(.£) = & Calling a polynomial with at least one coefficient unequal to zero a non-trivial polynomial, we have the following definition: leading coefficient 1 value f) at ! mapping .£ ~ which is of least degree among all polynomials that assume the (or 0 A) at (or of the matrix ~. the non-trivial polynomial with is called the minim~ polynomial of the linear A) • Without proof we will cite the following statements: Every linear mapping .£ A) has a unique minimum polynomial • (or matrix .c Every polynomial which at (or A) (!J has the value (or 0) is divis ible by the minimum polynomial (which points up a way to find the minimum polynomial from the characteristic polynOmia~. Similar matrices have identical minimum polynomials. Exam.R.l~. A = 0 0 0 0 4 1 0 0 0 0 0 0 0 1 0 0 But already A2 =0. B = l! 2 2 o ,• and indeed det(AI - A) = X(A) = A ; ?Kample • 0 0 1 3]. vet) X. 4 0 0 A 0 A -1 0 A 0 0 L = O. =~ 2 • Find the characteristic polynomial X. 2 1-1 Using the second statement in the above is also A I~ 0 Hence the minimum polynomial is Compute, number by number, the matrix ~ AI - A = -1 X(B) and thus check the Cayley-Hamilton theorem. ~ote prove that the minimum polynomial of B 2-1 Canonical r~resel1tations*) ===========--============ y. Again let re this basis has as "simple" a form as possible. process will be played by singular at all. let (i n-dimensional vector space. Our aim will now be to find a basis of linear mapping. ;r be an Ai ••• , p) j vi € of ;r (i An important role in this be the eigenvalues of ;r; define Xi =;r - Ai = 1, AiJ ) vi = Xi vi = V• In the process of constructing a simplifying basis '" ~. is invariant under Q (1 , V is the direct sum of """' and non-singular on 1 ::r, Le., -::r!l1 i n 6) j=l jFi "', p) subspaces p 1 ~i such that ,) ~j i ~ 1, ••• , p • ~ ~i second property permits us to choose a basis in re this basis consists of diagonal blocks and zeros = 1, need not be j (::r - Since according to the first property £i ~i V such that the matrix of X.J. play an important role. is nilpotent on ~1~ be a V with We will first show that 2) ~i """ """ linear mappings, even though p.:s n)**) j is singular on the mappings : V~ V then, as a consequence of the definition of eigenvalue there exist non-null vectors i.e., j Singular mappings enter the picture through the following backdoor: (i = 1, ••• , p = 1, ~!96~1~~ Let V such that the matrix """' o~tside these blocks. is nilpotent on the subspace ••• , p) , the remaining problem is then to choose bases in ~i the corresponding matrices of the nilpotent mappings Xi are "simple". such that The solution of this problem will enable us to obtain "simple" forms for the above-mentioned diagonal blocks. *) Before etart1ng on this' chapter, the reader may want to at least glance through the Appendix, especially sections A.5, A.4, A.5,in order to get acquainted with or brush up on matters of definition and notation •. **) Some of the roots of the equation X(iI.) = 0 may be equal. number., say p , of different eigenvalues is less than n. *-------- Then the total ! A singular mapping on Let consider :It (£ )" ! is nilpotent on a certain subspace of 1 be an n-dimensional linear space. Let .c : ,..,y... ,..,V be singular; t~e following sequences of null-spaces and image spaces: :n (! i2 ) , :R(.c 3 ) , oJ • • ~ (X )" ~ (,£a ) , !R(X 3 ) ~" ••• WLe:1ever C holds as in !n(,es) c ~1(,eS+l}" 'D:l;' ':nsion of SUbspacs \) that fo:::, all integer k > (;: dim !R(,es) ·r.cllows jJ:. (SqC section A.4" under < dim ;((~,;2~'1.); ~.:n':le ~(,ek) these dimensior'J 2.'11.1.6-1.:, f:.~;a.' }~,c::;:; '~:)1all l; V ,.., n; or equal to hence the strict inequ21ity signs cannot go on 1ndef:i.r..:ltely, and m9.ny integers k must exist sue::. '·hat ~(J~k) == ~(£k+:t.). >!) EES::?£ ~ '~:;'nce tID+lv ... Q. !it(! m+l =~Cl: m) , ) Let ~ rn r'v'" G W!lenever it follows 't.'::.€- t Hence .,em+s v =! lli+l (.ts-1v) ----- rr. be the smallest such =g ~ ".' ';)..) ", • I\ n ,m+s \ c__ ;\ll(ro J \ . ')+s-l) 1"fi('~"+S) ..... \.~ , and so on down c- ;,("m+s-l) ,... .If" ,J, = ffi(pID+S-2) :J~ oe;.. C _ ••• t~1e ~.l ::1.:, C ro(pro+l) _ ;..r" W _. ~("m ",L I.. - cL.. • From the 1:eginning of this point (1) we alr~ady know that all these inequality (cf., end -P ,... ......... .v· O...- ." .5) . A 2 2 iii) 3l(.t ) = .e y ,..", S:milarly Since l;;;!-y '" = ~(,c). ' where the inequality follows from S:'V ~\!3) ~ ffi(!2)" etc. So: ~(,e) ~ ~(L2) ~ ~ ••• dim!R(,ck) + dim l; I""J m(.e 3 ) m(lk) : : n for all integer 'Theorem' )" it follows from point (1) that dim k>0 m(..r m) (cf. sec. :::: dim 9t(,\;m+B) J Y (<:=X: V'" V). ~ ,...., A.5, (of. sec. I'"W 2-3 hence ~(~m) A.4 'Dimension of Subspace') So J;(~(.tm» .. .tm+1V =~(!m+l) !R(.r: m) :II =~(~m+s) 211 integer s >0 • (cL sec. A.5, under 'Non- i.e. I for singular vs. singular'): Note the definition of ! iV) The intersection ffi(.t k ) in the above point (i) • n ~(.tk) though dim ~(.tk) + dim ~(.tk) !J1(.r) c !Jt(l2) , V € .tv ~(.t2), € ~(.t). Y. tv f. Q, v .c 2 v == E V such that Q, .tv m(r) , E: I if m (as in point (i» ~ !R(.~m) n m(.c m) == [Ql • m with .t v .. w; k. [~J even Let, for instance v·~ ~(£) 1 yet obviously is the smallest integer with ==.cmw:o:\: 2mv. ,£ mv = w; QED. ./4 ...... -.. - -... _-- .. I;: m+s m (since m(.c ) == !n(t) for all s > 0 • ) so Q -_ v) for all natural However, --.-._. m(,em) == ~(.tm+l) V E: =n then there exists i.e., - is not necessarily restricted to _-~. ._ _- V = ~(.cm) ED :J1(,tm), m defined as in points (1) and (iv) • '" ~~~~!. ~(.cm) is a linear subspace of ! , ~(.cm) is a linear subspace of ! dim 1)t(£ID) + dim !Jt(,tm) !R(.cm) n J1(Zm) = = dim! , , (Q} Claim follows by sec. A.4: 'Sufficient and necessary condition for An alternative way of delcribing the same result is, of course 2-4 they will be called 62~:~~!!~~~ :!S~~~~2~~~); different; m for is nilpotent on £i '1 write ~i m(£ii) =!Iti $ ffi non..singular on I i mi ; li!It = (Q) i j J:il)li = mi n. = dim ~i + dim mi • j i !It ; In addition we will show i) ;r.ni ~ !Iti; !ni Proof. v So all 3' .. images of vectors in ii) 3'ffi ~ l'Jl i i € !It i i.e. , to !Iti; ffi j (1 = 1 , is invariant under 3' i ~ m £i 1 m =Q ~ v 3'!It ~ !n i 3'oC l v = 3'Q i ~i belong i is invariant under 3' • ... --- m By pointm(iii) of section 2.1: i Proof. » =mCt i i ) J~i (m(£i V € mi ~ (Jv .. Ai v) and some vector of 1ii) mj 'cj !Tti=!Tt i for =m1 ~ oCf!R i € mi mi i~j; ~ ~ ~ ~ 3'v 3'v mj oC k oC j (3''' AiJ )!Ri =!Jti I i.e., is a linear combination of € mi !Tt i v (€ ~i) • = !Tti for i131 i/JA, k·/J. j 2-5 ~!:~2!. We know that :J!J'i ~ ~1' hence Xj~i r;;;; !ni , even if irJ • Hence if we £~~ ~r2!~ ~~~~ Xj is non-singular on !ni (1~j) m follow that l j!R i =!n i = 1,j j !R i == ~i • it will m ~~ !~£~, assume v £!R i mi (A j - A1 ) 1 = !R(X/) v = (X i - .&j) and also v mi &: m1 !R(X j). mi '!hen: k roi m1-k k v =.1: 1 v + k:l ( .. ) (k ) .£i .t j v 1 m where the first term is zero since v £ !n(t i 1 ) , and the second term is zero since v e !R(,c.) • Since A :} Ai' it follows that whenever J m J v e !n(1,/) n ~l(,c j) then v = Q. Hence no non-null vector in !)1i is annihilated by 1,j =:3' .. AjJ; i.e. Xj is non-singular on mi. (This rather ingenious proof from Nehring, p. 103, Th. 8.2). m. Now, obviously, since Z/!lli =:!Ri (i ~ j) it follows immediately that • ~ !k mj .t j ~ !Ri =.l: k m (irJ, jf:k, i/:k). :Il i ==!R i Generalization to more "factors" of the form .tjj is obvious • iV) The set of all vectors i n ! m. mj Jl 2 all vectors in .t. .t - Jl ._--' ._-- j2 . mi annihilated by 1,i is the same as the set of m jk m ••• .l: ! annihilated by .l: i t ; hence the latter jk --. set is still ~ (provided all Proof. Because of point (iii) in section 2.2 lJ1 Hence !n. ~ v) V ==!n G> !R ~ l 2 Proof. ----- i, jl' .)2' ••• , jk are different) r;;;; i == mj .c.J l.c l mj .)2 mj m. mj J2 1,j 1 x ••• .c k j 1j 2 k m @ ••• m. 2 Jk ••• t jk !ni 1. • ~ m @m@(1,P,cPl-l...,clV). p p ~ ~ According to point (v) of section 2.1 we have: if .c is singUlar on ,.., U then U is the direct sum of the set of all vectors in ,.... U ~ annihilated by S,m and the lID-image of U (where m is defined as in --- point (i) of section 2.1). ~ Thus eX for y I .&1 for l) : Now take !: V=J1IEJf,~V 1 1 r.J '1 r.J for ,!l and X2 for l . Using the reeult of point (iv) in this section we find ID2 ~ xmlV:!Jl 1"'" 2 (f) .c~.c~V 2 1.- :¥. for ;g and .t 3 for .t j etc. '!be claim made then l follows from borollary Associativity' at the end of sec. A.4. Now take .c vi) v =~l (jj 2 .t ~2 (£) ••• (f) !ll ;p E~22f. IJl. According to point (iii) of section 2.1 .t i ffi i =m i , i.e. m :¥. =Xi i :¥.. By repeated application of this result and of the Xi Xi 1 commutativity of the IIfactorsll in p ••• ,I, ••• ••• !. ~. • •• it follows that • •• n 2 n ···.t2 Xl 1 n ! ==] , n (3' - A2J) 2 (3' - X1J) 1 = X(3') = {fJ is the value of the characteristic polynomial at 3". .N = {el • == Thus 2-7 ~ ~. ! ~n i dl!S0nal b~ock matrix representation of linear mappin~s was defined to be (i = 1, ••• , p) defined as in the beginning of section 2.2. with mi ~i? But what is the dimension of We shall answer this question at the same time as we find a diagonal block ~ matrix representation of Denote dim!lli re a particular type of by (i = 1, ••• , p). Vi and sec. A.4, under 'Dimension of a direct sum' , ba~iD. Then because of sec. 2.2, pt. (Vi), P E i=l vi =n = dim V. ,.... Introducing the notation = 0 if i 1 ( j.L -., 1-1 i ) 1: ~ if i = 2, h=l b l let (Xj.Li+1 ' xj.Li~ I ••• , \ (i = 1, • tt, p), ••• , p) Xj.Li+"i} be a basis of ~i. Then because of sec. 2.2, pt. (Vi) : {XI' .••, Xv1 ' B = is a basis of V "'.I x +l'···'x + , ••• , x +1' 1i2 1i2 "2 lip ••• /X +v} lip p • that is P 3'~ with agh == 0 unless (g, h) = E ex b x g=l J.Li Hence the matrix representing (h g =1, ••• , n) belongs to anyone of the diagonal blocks described by ( 141 + 1 ::: g t g + 1 ~ :s l-1i .5 h ~ J.Li + Vi 1 (i = 1, ••• , p) + Vi ) re the basis B 1s of the form: 2-8 A 0 • •• 0 0 ~ • •• 0 ••• 0 ••• • •• • •• 0 • •• A ::; (v.'J. X Vi) is where Ai 1\ (i ::; 1, matrix , P ·.. , p) • ~ Consequently, according to section 1.4 the characteristic polynomial of is =iT det(AI - A) det(Ali - Ai)' where Ii is (ViX Vi) identity matrix. Note i=l is the matrix representing the restriction of ~ to ~i (cf. equation that Ai 2.3,1), hence det(AI.A.) is the J. J. :r to :noJ. • On the other hand, (Ao9 characteris~ -~) polynomial of this restriction of is singular on ;noJ. if and only if A ::; Ai (as is easily seen by means of the argument used in sec. 2.2, pt. (iii» so the characteristic polynomial of the restriction of V (A - Ai) i dim ~) = det(Ali - -Pr 1=1 Ai)' ~i 2, ••• , P Hence, the characteristic polynomial of IT det(Ali - Ai) eXRonent 2.! the power =1, to (degree of characteristic polynomial of a linear mapping det(AI - A) ::; i ~ = TIi=l (A -' Ai,vi. Therefore , must be ~~ ~ equals is Vi = dim ~li = the S2!. (" - Ai) 1!l the characteristic polynomial 2! L; for (In sec. 1.4 we used n for these exponents). 1 We shall now show that the next step in finding a basis re which IT has a tnice' matrix representation depends on the possibility of finding such a basis for a nilpotent linear mapping. ! to :n i , In fact, since relative to a basis of ~i' Ai represents the restriction of we know that (Ail i - Ai) represents the relative to the same basis. Now ("io9 - rr) is nilpotent mi def on ~i' because of (Aio9 -!) ~i = [Q}, so the matrix Aili - A.J. ---- - F.]. is m nilpotent, more precisely: Fi i 0i' SUppose we find a basis for ~i re which restriction of ("io9 - IT) = the restriction of (IT - "io9) We know that then a matrix that C i to mi has a nice matrix representation, say M1 • exists such that 2-9 , hence 1 \ C 2 A. I + M P P P The matrix in the first member is obviously similar to A, re a suitable basis. A.ili + Mi are too. I = • ~ iI A p • C I p that 1s, it represents If the M are in a 'nice' form, then certainly the i So the next section is about a canonical representation of nilpotent linear mappings. The section after that will combine all pieces, and demonstrate how a canonical form can be constructed for an arbitrary square matrix. ~ ~. A canonical representation of Let ] be an mapping on ] nil~otent n-dimensional vector space and X be a niltotent linear into ] , ~ > 0 exists such that i.e., an integer k !1,U = (o} '" for 1, lipear mappings ~~(X.t) = U .-..; Note that a nilpotent X must be singular: k. if XU = "'" U then m be the smallest integer ,." ~ =Q for all t ~ contradiction. with ~(Xm) = !Jl(Xm+l ), so that X~] !Jlee) C !R(X 2 ) c ••• C !R(Xm- l ) Let, as before, C ~(J:m) =!R(,cm+s) \; .!!; Clearly m is also the smallest integer for which !R(X m) = ,... U defined integer k s >_0 • I so the above- equals m. Now we will construct a basis of ,... U so that the matrix of X re that basis is nice and simple. First observe that dim !R(Xm) Si (i = 1, ••• , m) f ( i (2.4,1) \ = dim ,... U =n • Next observe that all defined below are strictly positive: n .. dim !R(Xm- l ) ~ - s dim !Ree i ) - dim !R(.J~i-l) d"ef 0 s1 > 0 ~ dim 0t(,r) > m 8 1 (i =2, ••• , m-l} >0 Clearly m I: s i=l i = n • Define J Define J as the 1 X 1 matrix [0). l as the k X k matrix which has all entries equal to k in the first superdiagonal which consists of a ~~ing of = o o o o 1 0 0 1 0 0 0 0 1 0 0 0 0, (k - 1) 1'6. except those For example 2-11 e We !!.ll!. prove: i) 8 m .5 sm_1 .:s sm_2 .:s ••• .5 6 :s sl 2 ii) A basis of U exists such that the matrix of t ""J re this basis is of the following form: s m-1- smblocks smbl.o.cks " J-~.' .~"\r."' _-.d.. . . . . ~,.-.': ..........-;:",- / - -,...... m • r ~\ ! \\ I I j • I J \ m-1 I j t I I • I , i f . / J Whenever -_ _- Proof. •. m 1 / 1+1 = 0 it is understood that there are no blocks J i • To simplify notation we will discuss a special case. Assume s1- = 4, 8 63- s4= l , s2- s3= 1, 8 1 - 8 2 = 2, so n == E 8 i = 15 • i Then since ~(13) c ~(.c4) =,Q; and dim !1t(t 3 ) is s4= 2 less than dim U, we know 64= 2, (sec. A.4, 'Extension of a linearly independent set to a basis') that there is (even an infinite number of such sets) a set of s4= 2 vector~ ~ ~(.c3) , €)n(.t 3 ) , € ~(~4~ and which are mutually independent and no linear combination of which ; .. Let b 41 4 and b 42 be two such vectors. 1°) £ b 4i = g and £ k b 4i f Then k ~, £-04i € ~(! and 2°) i 4-k ), =1, 2 • The vectors in the first two columns of the following array are b 41 b 42 £b 41 Xb 42 ! 2 b 42 £ 2b 42 X3b 42 £ 3b 41 Indeed, assume b !b 31 31 b21 ,C~3l !b 21 3 l: 0:41i £ i=o 1 b 12 ·'11 3 i b 41 + .E 0:42i X b 42 1=0 =Q Apply X3 to both sides and find !3(Q:4l0b 41 + 0:420b 42) Hence 0:410b41 + Q:420b42 =~ :n(X 3 ); but b 41 and b 42 were explicitly assumed to have the property that no linear combination would belong to ~(£3); so 0:410 = 0 0:411 =0 =0:420' to Now apply !2 =0:421 ,etc. ! b m(x 2 ) , 41 € Claim under (2°) (2°) is proved. are linearly independent, they belong to m(t 3 ) , not 42 and no linear combination of them belongs to m(£2) (assume the and .£ b opposite, then ! 3b 4l and! 3b 42 point to both sides and similarly find that would be linearly dependent, which would contradict which was just proved). This proves that 2-13 In our caee say s4= 2, s3= 3. So there exists one other vector (again there is a huge choice), such that ,I; b 41 ' 5': b independent and no linear combination of them beloDfls to 42 ' b .,l(e 3 ), € 31 b 31 are linearly The reader will jl(Z2). prove 3°) £\31 40) € k A(J.: 3 - ); C\31~ Jl(S;2-k) for a < k < 2. ('.J1(!0) ~ (9) ) The vectors in the first three columns of the above array are independent. ~. ( ':2) , are linearly independent, €:.It linear combination belongs to ~(£l. Therefore s2 ~ s3· In our case no s2- s3 = 1. So there exists one other vector which shares these features with .~2b41' £~42 ' f q, and also 21 that the vectors in the first four columns of the above array are linearly independent. ~ b 3l ; call it b2l • The reader will prove that Finally the fact that 4It bl2 ' s - s 1 2 =2 '~21€ ~(!), !b entails that two more vectors, say b ll and have to be taken in, in order that the bottom line of this array may be a basis j1er.:) • for As a result we have 15 linearly independent vectors, which therefore form a 6 in the bottom line are a basis of Jt(£), the 4 in ihe next line complement this basis into a basis of ]L(~2); the 3 in the next line comple- basis of Q; the ment this basis into a basis of ~t(.f-3); the 2 at the top complement this basis into 4 A(C ) =Q a basis of Note that the number of vectors in each row and in each column depends on ••• , s ... . ~ m i.e., on the dimensions of the sets ; not in the least on the matrix representing £. This means that the lengths of the rows and columns of this array are the same for all matrices which are similar to each other. Now enumerate the vectors of the above array in the following order: column, bottom to top; second column, bottom to top, etc. be v , 12 b 21 21 v These vectors v (Note that the subscripts of the b are double l3 are single: the subscript of b21 is two-one; of v 12 would would be subscripts, of the twelve). For instance £b first v now satisfy the following equations: is 2-14 ! v4 ': v 3 = v3 = v2 x v7 ! V =v 2 1 ! v6 = v 5 .J: v 5 = = vll vlO :: v ! v lO 9 l" = v6 .c v = Q ~ 9 ,£ v =v 12 13 ,£ = v 12 Hence, in accordance with equation (1.1,1) the matrix '0 1 0 1 000 10 0 I 000 [a ij ] ,£ v1 4 =9 .e representing .1: v =e 15 with (v , v , ... , v } is as follows 1 2 15 respect to the basis r- Q 0 0 1 0 010 o 0 1 0 0 000 1 0 0 o 0 and 0 1 0 0 0 0 1 0 0 0 1 0 0 0 at all other entries. 0 0 The actual construction of such a basis is harder than it might seem because ~. of the condition that, e.g., no linear combination of b 41 and b 42 may belong to ~(!3). Although the existence of such vectors can be asserted without having an actual basis of ~(!3) available, this checking requires more doing (either one has first to choose bases for ~(.1:k), k = 1, 2, ••• , (cf. Nehring, p. 104) against which this type of conditions can then be checked; or one selects e.g. b 41 and b 42 on a trial-and-error basis checking if £3(1\ b 41 + t'2b42) :: t'l = 0 :: t'2' indeed entails e and if it does not, then chooses another b 41 and/or b 42 ) • ~. The Segre Characteristic of a nilpotent mapping gives the SUbsequent orders of the blocks J k ; in the above example (4, 4, 3, 2, 1, 1). 2-15 .4It The ~ characteristic of a nilpotent mapping is the above example [6, 4, 3, 2}. {Sl' 2 , 6 3, •.• , sm} , in Obviously the Weyr characteristic follows from 8 the row lengths of our above array of vectors, the Segre characteristic from its column lengths. 2-16 2.5. The Jordan normal form of an n X n matrix We shall now combine the results of sections 2.3 and 2.4. Let ~ be a linear transformation mapping an 8-dimensional linear space into itself, with characteristic polynomial The so-called Jordan normal matrix of d has then for its main diagonal -1 -1 -1 -1 -1 2 2 2 = 5 ; n2 = 3) l will then be completed by adding in keeping with the formula at the end of section 2.3. (n The first diagonal block (5 x 5) it, where (=:r + J M1 M1 to is a 'nice' representation of the restriction of ! - A1J in our case) to ~l' 5-dimensional linear space ~l Since ;r - A1J ~ We saw that to ~l Fl =A1- A1Il , n~ and on i.e., of the • Suppose in our case ~= 5 the reader will prove that M1 will consist of zeros except on the super diagonal, depending on the value of the dimensions of the null-spaces of the powers of restriction of d - A1J is nilpotent on the (WhiCh has the role of ~ in section 2.4), can be found by applying sec. 2.4. possibly for some 1 's (= 0' + J) (~+ J)k ~l (i.e., = (Q} for 2 ~ 5). dim ~( (~ + J)1) = i, si= i for 1 S i ~ 5. . 0 ~ = 1 0 1 0 1 0 1 0 So Then 2-17 Suppose ~ = 2, and dim ~'l( (3' + J) ) = 3. Show that now 1 o = 0 o 1 o 0 o Note that in the first case the scheme connecting Segre and Weyr and in the second case: . . . . ) characteristics is 6 w VI 2 = 2 s = 3 1 • vi .. . · L . I 2 2 I 5 1 s s As to the second diagonal block (3 adding ~ to it, where l' - ""2.J (= l' - 2J) dim !ll2 = 3 = 8 1 , X 3), it will have to be completed by i6 a nice representation of the restriction of ~ to !Jt2. Suppose m2 = 1 , i.e., ('T - 2J) !Jl2 = [Q}. and the above pattern is: and A transformation with • • • for • which combines ~ ; A ..- 2 -1 =2 1 -1 • has Jordan normal representation: ; --I o -1 1 -1 0 -1 o 2 0 2 0 2 ...J Then A transformation • • • for • ::r which combines : for (3' - 1 -1 1 -1 1 -1 1 -1 0 2 0 2 L Exercise. Ai = -1; • has Jordan normal representation: h,2= 2 ; -1 Ai"'; 0 2 What is the Jordan normal form of the matrix lJ -1 1 2 1 1 0 0 2 0 1 0 1 :; ] and of the matrix :; 0 0 0 0 ·3 o 1-1 0 3 -1 0 0 0 0 0 0 0 o o 2 0 02 ? with A-l A. = ===== ~endix 1 The principal purpose of this appendix is to make this report to a certain degree self-contained. Tb this end we have listed axioms, definitions, and standard results in so far as they have been used in the main part of this report. Sometimes a proof is provided, sometimes a plausibility argument or a heuristic motivation. For complete proofs see the textbooks listed in the section on literature. Since in the definition of linear spaces the algebraic concept of a field plays an essential role, section A.l discusses the axioms of fields. needed for the derivation of the Jordan normal form consists of the The field S~~~!~~ ~~~~E~. These have many additional properties, of which a few (viz., those needed in section A.3 and A.4) are given in section A.2. ~~, Then section A.3 gives the axioms for linear spaces, section A.4 discusses relevant properties of linear spaces, section A.5 of linear mappings. A-2 Fields • Denote the set of complex numbers by The system of complex numbers has ~. the following properties. I.l. A mapping: ~ ~ ~ X ~, Notation a + O = (3. l 2 1.2. called addition is defined. For any a l € ~, a2 € ~ their sum also For any a E ~, l Associative law of addition. a2 E ° (al + 2 ) + a:3=- 0:1+(0:2 + 0: 3 ) 1.3. Existence of solution of certain equations: there exists !,ileast ~ and at least one - E € 11 E ~ ~ with 0: + with ~ a: E I 3 .. • ~ • for any 0: ~ ~ € (3 E ~, (3 E ~: } • 11 + ex = (3 Corollaries. -----------.- !.: existence of ~~ !~2!~ 2~~ i.e., for any 0: -Note. ~ E right additive identity element, OJ' E cp; there is at least one OrE • with 0: + Or =-0 • Here the possibility is left open that there might be more than one right identity element for a given element of ~ 1 or different ones for different elements. Eo: existence of ~.~ !~~~~ ~~~ i.e., for any a e: ~. left additive identity element, CP there is at least one 0t E ~ 0,e with E 0t+ ex CP 1 =0 • However, because in 1.3 the Similar to the preceding note. existence of solutions to both equations is postulated: right left } --- .. additive identity element for ~ element - additive identity element for all elements Indeed, let let be a solution to T'j then 0:2 + 0r= 1} + 0:1 + Or .. Similar proof for 0t. T'j 0 1: = 11 + 0:1 0: 2 + 0:1 =0:2 ~ CP is a {right} left ~. E Or be a right identity for E (Xl + Or" (Xl 1 0:2 1a (Xl ; 0:2 + 0r= 0:2 • Now we shall prove: 1 A-3 ~: uniqueness of additive identitl element: identity, only ~ Indeed, let O~ there is only 2E£ right left identity, and they are the 0'r and 0"r ~, denoted by O. be two right identity elements, and some left identity element, then because of corollary £ (elaborate~ 0' r say. 0" r = 0 ~ , and r Similar proof holds for = 0~ + 0' = 0It + --- O.;(, = 01- + 0r = 0r = 0, called zero. say. 0" = r "" 0' -- 0"i- = Or. N It _0 hence O~, , say. 0' = 0" =0 r r Finally, r , is the _additive identity element, - -- such !.: g,: such uniguenessof additive inverse: additive inverse, only denoted by ~ ~ for any 0: E: ~ - _.---- there is only one right additive inverse, and the two are the ~, (~). In fact, let 0: + (~)' =0" r 0: + (-0:)" =0" (-o:L +0: =0" r "" then Similar proof for uniqueness of left additive inverse. Finally: ( -o:),~ = ("'Cl:)" +0 )J)J ~. = ("'Cl: )" ;(, + a; + (-0: ) r =0 + (-0: )r = (-0: ) r = (-0: ) The right and left inverse are the same, even without commutativity. Exercise. ~. Prove that f(o:+~» fotation: ex - f)ctq..fo: = (-~) + ("'Cl:). + (-f». Show that Prove that (-(-0:» 0: - 0: =0 • =0:. A-4 !!.: uniqueness 2! solution of 0 + uniqueness 2! solution of Indeed, given o £ } , f3 4> , £ let 0+£'=f3 == 0 + ( -a) + 0 + £ I == (-a) + f3 == (-a) + 0 + '!ben e So Similarly 1.4. = f3 +0=f3 T} ~ £ = ~" + f3 t" == ~ + (-a) == f3 T} == (-a) s" ~. Without commUl:ativity Note. 1he three postulates 1.1, 1.2, I.' are those of a{n additive) Commutative law of addition. £ need not be equal to For all 0 01 + 02 ~. 1 4>, £ =CX2 + CXl 1') • itggg. 02E 4> : • The four postulates 1.1, 1.2, I.', 1.4 are those of a commutative group, also called an Abelian group (~~~ ~~4!~!2~). ~ Complex numbers can also be multiplied: 1I.l. 4> x Ii>'" A mapping: Notation CX CX 1 2 For any a l E 11.2. == 4>, r ~, called multiplication is defined. (alternative cx ·a 'I ) l 2 € 4> their product also € :& O 2 Associative law of multiplication. ~ • For any alE 4>, cx 2 € t , C¥3£ • : (al a 2 )a, == a l (a20,) The addition and multiplication interact according to: III. D1stributive law. For any a l E ~, a 2 € 4> , CX, E 4> : °° 111.1. a l (a2 + 0,) = 1 2 + ala:, 111.2. (a2 + °,)°1 =cx20 1 + 0,01 Corollaries. -- .... _----- .!. a(f3 - r) III C$ - or for any a E 4>, f3 E 4>, r E ~ • Indeed, both members can be shown to satisfy (Use definition given of the notation Corollary ~ to point (a - f3) T} + ar III cxf3 • at the end of 1.3). {f3 - .,)a: = (3a - 'YO for any a ~ 4>, f3 E ~, ., E 4> • Indeed, both members can be shown to satisfy 1') + 'YO = I9a • b. ('1.0 =- 0 for any ex £ ~ • :wdeed, a.O. =- a(f) O.a a for any a 0 Indeed, c. O.a £ t» _C4'0 .. ~ 0 a t • = (f3 .. f3)ex .. th .. t=a = 0 • For any a E t, 13 (ooO).f3 .. -(a,,) , for both are the additive inverse of a. (-t3) "" - (a,,) , same argument (-0:).(-") ~. =o:t3 , £ t for both are the additive inverse of (a,,) (-0:)." "" -(at3) • ~. The postulates 1.1, 1.2, 1.3, 1.4, 11.1, 11.2, III are those of a 1! for a £ t t3 I E .---_. t, a" cannot be zero unless either a" 0 t3 .. 0 , or _.----- ---- --_._-_. the ring is said to be without zero divisors. l! ~~~!~!~~~~!2~ is also ~~~~~!!2 (postulate II.4 below), the ~!~ is said to be commutative. ------_.--- A ~£~~~2~~!~ !:ing !!~~2~~ !~!:£ ~!!!2£::!! is often called an !~~=6~!! ~2~!~: II.3. 111e set t \ to} .. t ' , say, is .E2l empty and. is a sroup under multiplication, that is: Existence of solution of certain equations: - - £ E ~' with ag there exists at least one and at that ~ E £ t, a - least one ~:a 0 0, ~ i.e., are solutions.. t3 = t3 t', "E ~ I : } rp "" t3 a If 0, then III, oorollary a .. 0, t3 /: 0, no g :2. shows E ~ or . respectively, can satisfy these equations (same coroll. !?) If ex" 0, t3 "" 0, every "1 0, neither s If E ~'with for any a E nor £ ~ E ~, .~ E t satisfies these equations. can be zero. --- Corollaries ......... • !: existence of!~ ~~!~~ 2!!= ri§ht multiplicative identity element, 1rE ~', i.e., for any ex E t' there is at least one 1 rE t' with ex.1 r "" a • - Note. :!?: - See note in I.3, a existence of !~ !~!2~ 2~~ ~ multiplicative identity element, 1.e E t' , i.e., for any ex E t' there is at least one 1 I.E t' with 1J..a = a • A-6 ~. See note in I.), -.. £. Now because in II.) the existence of - solutions to both equations is postulated: ~: a {~;~~t} {~;~t} multiplicative identity element for of -d: ~ multiplicative identity element for E::~~!. ~ elements E is a t/• +, and 1 instead instead of Replica of I.';" £." with €.' element O. uniqueness of multiplicative identity element: identity, only ~ - there is only one right left identity, and they are the~, denoted by 1 , called unity. +, and 1 instead of of ~: !: o. there is !~ !~~~~ ~~~ risgt multiplicative inverse, for any a e t' a -1 such that r for any a a.ar..1 = J E IP -1 such that at • 1 • there is 0: !~ ~~!2~ 2~~ ~ the two are the saIll;,e" denoted by a - Note. ~ multiplicative inverse, and • The reader will model the proof after I.), £. The right and left inverse are the same, even without commutativity. Exercise. Prove that (013(1 Notation. %~ .!!2. ~ -1 - I for any a e III there is only one £16ht multiplicative inverse" only 2E£ Proof. -1 multiplicative .!Everse" at = 1 • £: uniqueness of multiplicative inverse: ~: instead d,ivisors. a.~ ..1 j ~. _+ Let af3 .. O. our ring has zero divisors) = f3-1. 0-1 • Prove f3 def 0:.f3-1 Assume a f 0 , Then a -1. o~ .. f3 = 0 = 0: • Show that ~.. 1 • f3 j f 0 (i.e., assume contradiction. A ring satisfying postulate II.3 is called a division or a skew-field. ~ We saw that a skew-field cannot have zero divisors. e II.4. Cmmnutative law of multiplication. 0:1(X2 ~. For all 0: 1 E Q) , 0: 2 : = 0:2°1 A skew-field satisfying postulate II.4 is called a rational domain. E ~ fiel~ or a (Some authors use field instead of skew-field, and commutative field instead of field). A ring satisfying postulate II.;4 is a commutative ring. ~. '!he postulates I, II, III are shared by complex numbers, real numbers, rational numbers. Of the various properties which ~ bas in addition to I, II, III, we will discuss only those cited in the discussion of Linear spaces and those .used in deriving the Jordan normal form. e A-8 A.2. Complex numbers The field axioms listed in section A.l are satisfied by the systems of rational numbers, of real numbers, and of complex'numbers, to name a few examples. We have assumed that the reader is familiar with these various types of numbers. In the remaining sections of the appendix and in the main part of this report we need only the following two properties of complex numbers: !I The field of com~lex numbers is algebraically closed, i.e., for any th positive integer n, for any polynomial, P say, of the n-- degree, there n exist a complex number p (s n) '1 complex numbers (1' { P positive integers such that for all complex S2' ••• , n , n , ... , n l 2 p ~p with p I: k=l ~ = n z: = p ~ 1· (z - ~k) k=l . 1T (**) In fact, according to the 'Fundamental Theorem of Algebra' all that needs to be Pn , is that this implies the validity of postulated in order that this equality may hold for all polynomials the equation z2+ 1 = 0 has at least one root: equality (**) ~ A one-to-one correspondence exists between the set of complex numbers and the set of ordered pairs of real numbers. numbers z is: Z Let zl = The traditional notation for complex ~ + 1~, =x + iy , z2 = x2 + iy2' were h x ER , Ye R , then zl+ z2 = (xl + x2 ) + i(Yl+ Y2) zl z2 = (xl~ - Y1 Y2) + i(~Y2 + x2 Yl) 1·2 =-1 • e A-9 A.3. Linear spaces. - A linear space (or alternatively: V, vector space) is made up of a set ,." ~ whose elements (denoted by lower case Latin letters) are called vectors, a field, in our case 0, scalars, certain rules according to which the vectors may be related to each ~ whose elements (denoted by lower case Greek letters) are called other and to the scalars. single letter ! Following a common abuse of notation we will use the not only for the set of vectors, but also for the linear space. More precisely, we say that ! linear space ~!! ~ ~ -!.. field iff 1.1. A mapping: ,... V X ,... V~ ..... V, ·called (vector) addition is defined. V € 2 1, their sum also a common abuse of notation, we will use the same sign, of vectors as for the addition of scalars. € Notation Again following V • ,." for the addition +, No confusion will arise, and certain formulae will even look simpler for it. 1.2. Associative law of (vector) addition. For any Existence of solution of certain equations: there exists and ~ x € 1 at least one y € I"t<J ~ least _ _ _ V l €!, for any a V € €!, v V, b € 2 ,." 3 €!: V: ,." with a + x = b } V with y +a =b Corollaries _.. _-------- (proofs same as in sec. A.l, under 1.3; here we cite only the most important ones): !!.: unique ident~ty element: there is one and only one element denoted by Q, called the null vector, such that for any a e V, tV 6 ..... V : a +Q = a Q a: unique inverse: written (-a), for any a + a =a E!, there is one and only one element, such that a + (-a) Notation: -------- =Q (-a) + a = Q def a + (-b) • a· b == A-10 1.4c -Note. uniqueness of solution of a + x =b } uniqueness of solution of y + a =b • For any vl E! COIDmutativitl of (vector) addition. ' V2 E! : vl + v2 = v~/ vl • V to be a The postulates 1.1, 1.2, I~3:1 and 1.4 together obviously require ,... commutative group under addition. 11.1. A mapping: called scalar multiplication, is defined. - t X ,... V~ V:I ,... Notation: given a the value of the mapping is av E ~ :I (sometimes for convenience written as va ). a(x + y) =ax + ay for any a E ~ , yeV. ,... xe.V, "'" Corollaries: ---------- .. .!.: '~Q = Q, In fact, 1?: a( -x) for any a e x =x +Q = ~(ax), In fact, Q So a(-x) • ~ =ax = a(x + Q) for any a e ~, -._---- =aQ e:! . = ax + aQ x = Q • = aQ = a: (x + (-x» = ax + a(-x) • is the additive inverse of ax (a + ~)x = ax + ~x for any a ~ E = a(-x) X E , V ,... = -(ax) • • -------_ Corollaries: .. _.. .!: Ox = for any x e V • "'" In fact, 0: = a + 0 = C:~ Q , (-o:)x = -(ax), In fact, So (-a)x II~4. o(~x) = (a~)x, 11.5. 1x .!.: =X , Q for any a = Ox = (0: = (0 + O)x =c:~_:t_Q~ = Ox + (-a»x = ax + (-o:)x • is the additive inverse of ax for any x E = Q • E ~ , for any a e ~, ~ E ~, X = (-o:)x = -(ax) • E! . V• ,... V• (-1)x = -x for any x E ,.., Indeed, by corollary £. of 11.3 above: V ,... This mapping satisfies the conditions: n.2. E (-1)x = (-1 x) -. -x • A-ll called scalar Note. Point 11.1 above postulates that a mapping: x '" -multiplication, is defined, and 11.2 through IIQ5 proceed to postulate certain ~ properties of this mapping. ... V , '" This does not clarify yet how to define scalar multiplication in specific instances of vector spaces example: V V '" over fields ~. One let !be the space of ordered triples of complex numbers (and ~ be the complex number system). Now let a € Q, (£1' ~2' £3) E ag Sl 0: £2 £2 ~ !, then the definition l 0:£2 aS 3 is easily seen to satisfy all postulates 11.1 through 11.5. e A-12 !:2t. Basis, dimension, direct sum Linear combination belong to 2! ! vectors. Given vectors ~) is said to be a !!~2~~ £~~2!~~~!~~ of If the field ~ then (XlXl + ... + C%kxk If X.€ V (i • 1, ... , l(), 1. ~ ~,x2' ••• , ~ • consists of the rational numbers only, the coefficients 0i only be rational numbers). Subspace. k E Prove from definition of linear space: if XiE: can Z ! · V is a linear space, not every subset ~ See the two figures; in both figures U of V is a linear space. ~ ~ V is the set of all points ~ (x, y) in the plane, a linear space over the field of real numbers. or ~ -: .set 3 po i Y'lts 0 o ------11---------- l! J( is said to be a (linear) !~s~c:.e_of Z iff U ~ ,..,. V and ,.,; U is itself ~ linear space (over the same field as ,..,. (hence V) ~ U contains all linear combinations of its own vectors) ~ - - - - -......" - - - - - - - - - _ . % A-13 Let U be a subspace of V, Subspace spanned. vectors in ..... V. ~ ~._-_._- and let B be a finite set of ------ ~ Then one says; (1) each linear combination of vecto~ E B is a vector in ..... U, and U is spanned by B ..... U is spanned by the vectors in ..... each vector in U is equal to at least one linear.....c ombination of vectors E B • B spans Jl the vectors in B span -3 "I I I I ;' ,. ------- Example: ~, 'L ! /~ (..-> ._e .!! 1 V ,..., same as above; B consists of B = {Xl' x2 ' x }. flDw: 3 any point in the plane is equal to anyone of the 3 points indicated: a number of linear combinations of these -------. three vectors: l3i B spans '1. Thus (Xi and can be found such that (Xl ~ + O:2x2+ C¥3x3 = 131~ + 132x2 + 133x3 ' Whence Linear independence. ~~~2!~~~!~~ of the In order that each vector in 1 equal exactly ~~~ ~!~~~t k vectors xl' x2 , ••• , xk in B, it is necessary (why not sufficient?) that rl~+ r 2x2+ ... + r k xk = Q implies r l = 0 = r = ... r k • '!his 2 occurrence has received a special name. Consider a set B of k vectors 0= The !!£~2t~ Ilbe 2~~ xl'~' ... , xk are said to be ~!~~~t~~ !~~~~~W~~~ } B is said to be ~!~~2;:~~ !W~E~~~~~! r 1Xl + ••• + r k"1t 9 ~ r1= 0 = r2 = ••• = r k ' i.e.,:Jff the only linear combination of these k vectors which equals trivial one (all coeff. zero). independent set (whyt) The null vector iff Q is the Q is never an element in a linearly A-14 Basis. --- ---B of-- vectors ------- is said to be a basis ----- of the linear space A finite set .. _--- U!7: V iff N "'J B spans U, and ""'-J _ B is linearly independent. So each vector of is a unique linear combination of the vectors in the basis. ~ The coefficients in that linear combination are called the coordinates relative to that basis. Let e , e , ••• , e be the vectors in some basis of l k 2 proved that some vector of Q equals: we can immediately conclude that a i = t3i (i = 1, ••• , k). ~. If we have No basis ever contains the null vector (why?) Variety of bases. E~~~! ~~~E~~. The linear space with ordered pairs of real B'= (3, 0), (0, -l)} , numbers as vectors, and real numbers as scalars has a basis but also a basis B" = {(l, 3), (-1, i)} • §~3~ ~~~!. '!he linear space with. ordered triples of real numbers as vectors, and real numbers as scalars has a basis BX= {(l,O,O), (0,1,0), (0,0,1)} and a basis ~ (1,1,1), (0,1,1), (0,0,1)} • However, once a basis has been chosen, the coordinates of any vector re that basis are uniquely determined. Dimension. (5, 4, 3) re BII V has one finite basis It can be proved that if a linear space ..... consisting of said to be Find the coordinates of the vector a vectors, then all its bases consist of n-dimensional, notation: Dimension depends on field. If n vectors. ! B , V is then dim ..... V= n • V is the set of vectors of a linear space, then "'J the dimension of the linear space may very well depend on the field Let ..... V be the set of ordered triples of complex numbers ~. ExaI!lPl e :- ------- A-15 ~ Show that in (1,0,0), (0,1,0), (0,0,1)} is a basis for the linear space with vectors 1 and scalars in the set of complex numbers. However [(1,0,0), (i,O,O), (0,1,0), (O,i,O), (0,0,1), (O,O,i)} is a basis for the linear space with the same set of vectors, but with real numbers for scalars. Extension of a linearly independent set to a basis. Let A be a set of k independent vectors in an n-dimensional vector space ...... V. additional vectors exist which together with the -------~-- ~------ V. ...... linearly - Ifk<n, then n-k k vectors in A form a basis of These additional vectors are by no means unique (think of examples~) It is clear that no linear combination of the n-k additional vectors can be equal to any linear combination of the Dimension of sUbspace. k vectors in A. Let ,..,. U be a sUbspace of a finite-dimensional linear space dim U < dim V and Then { dim ~: dim ~ iff ".., ,..." U =~ V """'" • If U C V then V must contain at least one vector not in U; ,..,.,,,.., ""'J ,..., thus we find that a basis of ,..,.V must have at least one more vector than any basis of g~~~!!~~l. Q. (For complete proof see Halmos, p. 18). Let] C! be a subspace of Then let A be a basis of ...... U. 1 I dim ,..,. V = . n I dimU=k<n. ...... Applying "extension of a linearly inde- pendent set to a basis" we find there are (n-k) linearly independent to -- U. ,..., Sum of two linear spaces. Let £1 and defined to be the set of all vectors itself a linear space. Ye be two linear spaces. £1 + J!e Xl + x2 wi th Xl E£1 and x2 E.!!e ; is it is A-16 4It (al , a 2 , ••• , a k } If of ~, then is ~~~!~ of ~l 5 basis, then 'Yl , ••• , 'Y k is a basis Of course, need ~2~ be a 2~~!~ (e.g., let ~l and ~ (a l , 52' ••• , ak , b l , b2, ••• , btl E~~ a spans ~l+ ~. (a l , 52' ••• , a k , b l , b2 , ••• , btl R3 ). two planes through the origin in (b l , b , ••• , b~} 2 and and _¥. be (a , a , ••• , a , b , b , ••• , b.e,} l l k 2 2 51' ••• , 51. k is exist with X, 1: 'Y i a i "" E 5. b j i=l j=l J such that at least one ~l n~ 'Y and one --- i contains non-null vectors. (a l , 52' ••• , ak , b l , b2 , ••• , btl to hold true is that all e 5j (Ei 'Y i a i E: ~l; OJ are ~~~2, and ~l motivates the distinction of the joint occurrence of ~l n lk = {Q} E: Ee). ¥.' however, (*) J. = ~l n Ee + ~ = This Q. and by a special term: D!rect sum of two linear spaces. J. "" l!l (notation: Ej 5j b j is ~ 2~~!~J then the only possibility for and ~!! 'Y i are not zero (why 'and'?). So in this case (f) J. ~!~~£~ is said to be the ~) ~! = ~l + Ee !:~<!. ~I nEe"" sum of ~l ~ and U~} • Uniqueness of vector decomposition relative to direct sums. Then in order that this sum be v ~!~:£~, it is necessary and sufficient that to each there corresponds 2E!l ~~~ vector E::t such that v = xl+ x2 ~ E: ~l and 9E!l 2~~ vector x2 E: Ee • such that v = xl+ X2 = x{ +~ belongs to both ~l • and ~. xl - x{ = Q "" x~ - ~, §~!!!~!:~£l. £1 n~ xl - x{ = x~ .. ¥2' So, if l!l n .2i2 a vector which clearly = ((~}, it follows that which proves uniqueness. (reductio ad absurdum). contains vectorc other than + (X .- u), 2 Then Q, If the sum is not direct, Le., if say u, which proves non-uniqueness, since then ~ + u E: v.::: ~ + x2 "" (xi+ u ) + Q and 1 x2 - u E: lk . A-17 Variety of direct sums. direct sum .1!1 $ ~ Although we just showed that each vector belonging to a is the sum of a uniquely determined vector Xl E:.Ql and a uniquely determined vector of subspaces l!l and Ee U' U' a... .-.;1 w..:;;e the reader should be cautioned that the choice ~E: ~ J itself is by no means unique. =•~l a-.w ~ • .,11 does .,11 t imply .-.;1 U' ~ .-1' =~l Not only: U' = ~ y.,ll and ..::e but even: " l' The situation is entirely similar to the one described under a variety of bases. In fact, saying that saying that Jl. = <~) space with basis (~, x } 2 $ (X2 ) (x }. 1 is a basis of a linear space , where (Xi) tL is tantamount to is the one-dimensional linear The reader may clarify the situation to himself by drawing some pictures. Dimension of a direct sum. V is a linear space f'J If .Ql ~ 1 is a linear space ~ .! is a linear space I:; dim ]1 = ~ } dim ~= 1 =.!!l $ then dim! = ~ + n2 = dim.Ql +. dim ~ • (finite) n2 ~ a set {xil , xi2 '···' Xini } of ni linearly U. (i = 1, 2). Given independent vectors exists, which is a basis of f'JJ, E!:22! • Since dim ~ V E: = ni ' ! , the preceding result shows that exactly one Xl E £1' and exactly one x E ~ exist such that v = ~ + x2 ' Now Xi can be represented 2 (uniquely) as a linear combination of basis vectors in .Qi (i = 1, 2). So v can be represented (uniquely) as a linear combination of the vectors ll ' ~, ... , Xl~' x21 I x22 " •• , x2n2 ' Hence these vectors do ~ They are also linearly independent, as follows along the lines of the X .! · A-18 argument around equation vectors (*) on page A-16 x ll ' • •• , Xl~' x21 ' ... , ~~. .. Consequently the form a basis of '1, n + n l 2 i.e., V=u Sufficient and necessary condition for ,..., ",1 @U ,;:;e • v ,..., 1 are finite-dimens1onal linear If u s;;;; J!l n~ ,;:;e V '" ,Jthen V =.!!l EE>.!!e. The tconverse is obviously true. = (Q} dim '1 = dim.!!l + dim Proof. Let of .!!l and a hasis .!!e dim.!!l = n l , dim ~ = n2 • Then a basis [x , x12 ' ll {X , x ' ••• , x } of U exist. The set 22 21 2n I'" 2 2 x12' ••• , ~nl' x2l ' x22 ' ••• , x2~1 of ~+ n2 vectors is linearly independent (see argument -around equation (*) on page A..16 spans .!!l+ ~ ~ '1, hence is a basis of .!!l+ ~ ), and clearly So = n l + n = dim J!l+ d1mt~ = dim JL. According to the previous 2 result on the dimension of subspaces it follows that J!l+ .!!e= JL. But we dim (!h +.!!e) y., n ~c; P"n know that Commutativity. '-.l.. = {Q}. J!l+ ~ =.!!e + J!l Commutativi t~. .!!l @ ~ Associativity. Associativity. @ (Property r.4 of section V= U ,...,1 '" @ U ..::e A.,) Ee= .!!e @ J!l J!l (f) ~ =.!!l + ~ Proof. Hence (definition of direct sum) .!!l = lIe +.!!l iff in addition J!l n ~ = {Q} • iff in addition ~ n.!!l = (Q J = (.!!l+~) + £3 (Property .!!l @(l1e (f) .2,) = (.!!l (f) l1e) @.2, in the l!l+ (Ee+ ~,) I.2 • of'$ection A.3) sense that i f the (f)-signs are valid in one member of the equality, they are valid in the other member as well. A-19 !:::S!S!!• We shall assume the validity of y £ £1 take @ (lk ® 1!,). sums is unique" xl £ 1!1 y = condition that ~ ~, ~, j £(lk ($) 1!,) again x2 £ =~ + x2 + x, = (xl+ x2 ) + x:; (21 + Ee) write + 2:; is direct: = [Q} (!!l ($) we may 'Write follows from £1 .!k) $ £, associativity. Notat~£n and x, £.!:!, ~ hence j n y £:;; = {Q}, £ Ee I =" + (~+ x:;) = and x:; £ l!, are 2 (El + Ee) ~ ~ £.!!l' y = (~+ x ) + x:; J (Ee ($) £,) instead of are uniquely -~----. (" + ~) eEl+.!1e hence j uniquely determined by the condition that n Ee lk are ~g,~~~l g~~~~~~~~ by the condition that 1!, -------- .!:!l are uniquely determined by the x2 , = x2 + x, determined by the condition that x, £ Then Since vector decomposition relative to direct and + in the first member. ($) ------- so the sum tinally whence we are allowed to (£1 + !!e) $], • As is usual in cases where associativity holds we will omit parentheses altogether and 'Write ]1 ~ Ee $ 2, for ]1 $ (!!e $ £,) just proved the parentheses may be put up anyway we want to). (since it was The preceding proof suggests a handy definition for: Direc:t.....sum of 21, l:k.. !- linear spaces. J}2k (notation and in addition for any y 0 •• determines the vectors :t is said to be the :t = 21 e .!1e $ €! Xi e 2i g~::~~! ~~ of :t .!!k" iff =.!h + Et2 + ... + 2k the condition that y = ~ + x + •• " + x k uniquely 2 (i = 1, 2, ••• , k). ... ($) 21 e (!e e (£, $ 24» =1!1 e Ee e 2, e.!:!4 • (l!4 e (.!!, ~.P6 }») = 1!1 S Ee ~ 1!, e l!4 ~ ~ e.P6. Q2!~:sr associativity. 21 e (Ee $ (£, 19 Proof. etc • Similar to roughly the first half of the proof for three linear spaces summed directly. ~. For ]1 + sufficient that lIe + ... +.!!k 1!t n 2j = [QJ to be a direct sum of for all pairs (i, j) 21" .•• , with i ¥k I: j - it is not , see figure: A-20 Ii .\"~3 "- -- '" "'.. I tv " ~\ One (necessary and) sufficient condition in terms of intersections can be gathered from the above corollary: in case k =4 e A-21 &.2.. Linear mappings '!he first condition for a mapping 1, : be linear spaces. Definition. Let V... W to be linear is that ,.."" ,.,.., V and ,ow W ,...., For our purposes we need only consider the case that ,.,. W =,.,. V • 1 be a linear space over a field ~ • 'lhe ~~~!~ t :! ... ! -_.--- is said to be linear iff =~.t~ 1, (al ~ + a 2 x2 ) for any Xl E 1' ~E 1, (Xl E ~ + (X2.e~ , Ct E ~ • 2 - Note. Image of a set. e Let ..... U ~ U under .e is defined by .....V, then the image of ..... .eU = (y E V j there exists x c ,....U ~ucb that y. 1,x) ,....,.... Inverse image of a.point. under .c Let bE.!, then tbe (complete) inverse image of b is defined by .c.~ • (x E l . tx = b '!he range (or range space, or image set) of t The null-space (nucleus, kernel) of t ~(!,) By definition, m(.t ) is m(.c) def tv ...., • is ~(!,) def .t-li. are subsets of ,.... V. 'lhey are even sUbspaces: is a subspace of ,..,. V. ~:~~! . Let such that (X1Y + (X2 Y2 l y1 E ~ (.c ) and y2 Yl = .txl , E !!t(t). linear spaces. !Jl (.c ) and ~(.t) ) Y2 = !,x2 Xl E 1 and x2 ' hence ~Yl + (X2 Y2 = .c(<;.~+ (X2x2)' so E !!t (.t ) , then there exist This takes care of the properties I.l and II.l Q, hence The reader will check the other properties. is a subspace of ,..,. V. ~22f' Let Xl E ~(!,), .r. (a 1 Xl + a 2x2 ) x2 E ~(!,) , = al'1:xl + a2t~ =~ , then t~ = ~, so (O:l~ + Ct2x~ Same remark as at the end of the preceding proof. tX 2 E = ~(t) • of E 1 e A-22 Suppose dim m(.t) = k k ~ (for some mappings I k > 0, for others or rather 1). So we can choose a set of k independent vectors serving as a basis of 01(!). (n-k) Then we can extend this linearly independent set to a basis for J., adding vectors such that no linear combination of them is equal to any linear combination of the original additional vectorst dim k =0 ~(!) Theorem. = n-k t What about the !-image of the (n-k) (n-a) Evidently the images span .tV (whyt) (OW Is dim.tV = (OW The answer is yes: dim SR ex) + dim Proof. k vectors. m(! ) = n = dim.z .:. Still calling dim me!) = k, let (zl' ••• , Zk} be a basis of me!) • {zl' ••• , zk' xl' ••• , xn _k } of !. No linear combination of xl' ••• , x _ could then equal any linear combination h k of zl' ••• , zk (unless all coefficients be zero). For any V €J. I there Extend this basis to a basis exist (unique) and t3i Ct k n-k v = 1: t3 zi + 1: i=l i j=l n-k So Xv = 1: OJ .1':xj j=l such that j Ct j xj which proves the above argument that {,c~, Xx , ".J 2 !Xn _ } spans !l = ~(X). Now let r j be such that k n-k j:l r j !xj = !(1:j r j X j ) = Q, i.e., such that of the rj 1: zi' j xj € ~(!). rj xj but we know that this implies This means that the !x. J = n - dim !n(!) ~. Then 1:j I would equal a linear combination rj~ 0 for j = 1, ••• , n-k • are linearly independent, hence dim ~(!) = n - k which is what we wanted to prove. Section 2.1, point (iv), shows that not necessarily ~(!) hence not necessarily V = !Jl(.c) (OW e lJl(.c) • n !n(!) = {Q} , A-23 Non-singular vs. singular. iff m(l) =l-l Q • {Q}, is non-singular iff dim1,V ,.,..., :! ... ! A linear mapping l singular -------- iff - < n = dim V. Because = V, ,..., l ~ singular iff 1,V ,..,., subspace of 1, l. If ll:! C; C l:! ~ l : V -+ V is non-singular iff V (in which case l l:! is said to be 2!~!:!~~ on l:! In general, Nilpotent. such that l 1, l~ ,... = {Q}. ,-w maps many-to-one). is said to be invariant under iff !,l:! CQ I is said to be non-singular on is said to be ~ V, singular iff dim ~(1,) = '" maps ,...,,...,,, V into V, !V ,..,., ~ V, so that, because ,-w 1 ~ = dim 1,V = n = dim dim !R(l) '" ,."", ~2~:~!~!:!~E > 1. Equivalently X : V -+ V dim :n(l) of the result about dimensions of subspaces, !V is said to be -----------nilpotent on ------_.- U iff ~ {v E U; ~ Let U be a #"IItJ U, '" U iff lU =U • "'" ,...,... X • 1,v For such = Q} = (Q} • U iff some natural number k exists ,..." (For definition of lk see point (iii) of section 1.3.) L-l SOME LITERATURE ON LINEAR ALGEBRA *) Gel'fand, I.M. (1961), Lectures on linear algebra (transl.); New York, Interscience; ix + 185 pp. Halmos, Paul R. (1958), Finite-dimensional vector spaces; 2nd ed., Princeton, N.J., Van Nostrand;v1i + 200 pp. KowalsJg, Hans-Joachim (1963), Lineare Algebra, Berlin, W. de Gruyter and Co.; 340 pp. (GOschen' s Lehroucherei, 1. Gruppe, Reine und angewandte Mathematik, Bd. 27). Mal'cev, A.I. (1963), Foundations of linear algebra (transl.); San Francisco, Freeman; xi + 304 pp. Mostow, George D., Sampson, Joseph H., Meyer, Jean-Pierre (1963), Fundamental structures of algebra; New York, McGraw-Hill; xvi + 585 pp. Nerin~, Ever D. (1963), Linear algebra and matrix theory; 2 9 pp. New York, Wiley; xi + Thrall, Robert M. and Tbrnheim, Leonard (1957), Vector spaces and matrices; New York, Wiley; 318 pp. Waerden, B.L. van der (1959), Algebra II (4. Aufl. der ~dernen Algebra); Berlin Springer; x + 275 pp. (Section 111 has a sbort derivation of the Jordan normal form on tbe basis of more advanced conceptsj:- ---------me Zsmansky, Marc (1963), Introduction l'alg~bre et l'analyse modernes; 2 edition; Paris, Dunod; xvi + 435 pp. -.------ -------- a *)OnlY those titles are mentioned which have actually influenced our presentation.