LECTURE NOTES, 11/10/04

18.700 LECTURE NOTES, 11/10/04 Cotets 1. Direct sum decompositios 1 2. Geeralized eigespaces 3 3. The Chiese remaider theorem 5 4. Liear idepedece of geeralized eigespaces 8 1. Direct sum decompositios Defiitio 1.1. Let V be a F vector space. Let W = (W 1,..., W s ) be a ordered s tuple of vector subspaces of V. (i) A ordered s tuple of vectors i W, (w 1,..., w s ), is a ordered s tuple of vectors i V such that for every i = 1,..., s, w i W i. (ii) The ordered s tuple W is liearly idepedet if the oly ordered s tuple of vectors i W, (w 1,..., w s ), satisfyig w 1 + + w s = 0 is (0,..., 0). (iii) The ordered s tuple W spas V, or is spaig, if for every vector v V, there exists a ordered s tuple of vectors i W, (w 1,..., w s ), satisfyig w 1 + + w s = v. (iv) The ordered s tuple W is a direct sum decompositio of V if it is liearly idepedet ad spas V. Remar 1.2. Please do ot cofuse the otio of liearly idepedet (resp. spaig) collectio of subspaces of V with the otio of liearly idepedet (resp. spaig) collectio of vectors i V. These are closely related, but differet. Propositio 1.3. A ordered s tuple of subspaces of V, W = (W 1,..., W s ), is liearly idepedet iff for every ordered s tuple (B 1,..., B s ) of liearly idepedet ordered subsets B i W i, the cocateatio B = B 1 B s is a liearly idepedet ordered subset of V. Proof. ( ) Assume W is liearly idepedet. For every i = 1,..., s, defie e i = #B i, which is 0 Bif i is empty. Defie e = e 1 + + e s. For every i = 1,..., s such that e i > 0, deote B i = (w (i,1),..., w (i,ei)). If e = 0, the B =. Otherwise, by defiitio, B is the uique ordered set of vectors B = (v 1,..., v e ) such that for every i = 1,..., s with e i > 0 ad every 1 j e i, v e1+ +e i e i+j = w (i,j). If B =, it is vacuously liearly idepedet. Thus assume e > 0. Let (c 1,..., c e ) be a liear relatio amog B. For every i = 1,..., s with e i > 0 ad every 1 j e i, defie c (i,j) = c e 1+ +e i e i+j. If e i = 0, defie w i = 0. If e i > 0, defie w i = c (i,1) w (i,1) + + c (i,ei)w (i,ei). The, e e i s 0 = c w = ( c e1+ +e i e i+jw e1+ +e i e i+j) = w i. =1 1 i s,e i>0 j=1 i=1 1

For every i = 1,..., s, because W i V is a vector subspace, w i W i, i.e., (w 1,..., w s ) is a ordered s tuple of vectors i W. Because W is liearly idepedet ad because w 1 + + w s = 0, w 1 = = w s = 0. So for every i = 1,..., s with e i > 0, (c (i,1),..., c (i,ei)) is a liear relatio amog B i. By hypothesis, B i is liearly idepedet. Therefore c (i,1) = = c (i,ei) = 0. So for every i = 1,..., s such that e i > 0, ad for every 1 j e i, c (i,j) = 0, i.e., the liear relatio (c 1,..., c e ) is (0,...,0). Sice the oly liear relatio amog B is the trivial liear relatio, it is liearly idepedet. ( ) Assume that for every ordered s tuple (B 1,..., B s ) of liearly idepedet ordered subset B i W i, the cocateatio B = B 1 B s is a liearly idepedet ordered subset of V. Let (w 1,..., w s ) be a ordered s tuple of elemets i W. For every i = 1,..., s, if w i = 0, defie B i =, otherwise defie B i = (w i ). Each B i W i is a liearly idepedet subset. By costructio, B = (w i 1 i s, w i = 0). By hypothesis, this is liearly idepedet. B If is oempty, the, s 0 = w i = w i. i=1 w i B Therefore (1,...,1) is a otrivial liear relatio amog B. This cotradictio proves that B is empty, i.e., w 1 = = w s = 0. Therefore W is liearly idepedet. Propositio 1.4. A ordered s tuple of subspaces of V, W = (W 1,..., W s ), spas V iff for every ordered s tuple (B 1,..., B s ) of spaig subsets B i W i, the uio B = B 1 Bs is a spaig set for V. Proof. ( ) Assume W spas V. Let v V be ay vector. If v = 0, by covetio it is i spa( B) (eve if B = ). Assume v = 0. Because W is spaig, there exists a ordered s tuple of vectors i W, (w 1,..., w s ), such that v = w 1 + + w s. For every i = 1,..., s with w i = 0, because B i is a spaig set for W i, there exists e i > 0, vectors w (i,1),..., w (i,ei) B i ad scalars c (i,1),..., c (i,ei) F such that w i = c (i,1) w (i,1) + + c (i,ei)w (i,ei). The the followig collectio of vectors is a collectio i B, For the choice of coefficiets, (w (i,j) 1 i s, w i = 0, 1 j e i ). (c (i,j) 1 i s, w i = 0, 1 j e i ), the liear combiatio of these vectors B i is, 1 i s, w i = 0w i = v. So v spa(b). ( ) Assume that for every ordered s tuple (B 1,..., B s ) of spaig subsets B i W i, the uio B is a spaig set for V. For i = 1,..., s, defie B i = W i. This is certaily a spaig subset of W i. By hypothesis, B = W 1 Ws is a spaig set for V. Let v V be a vector. If v = 0, the (0,..., 0) is a ordered s tuple of vectors i W such that v = 0 + + 0. Therefore assume v = 0. Because B is a spaig 2

set for V, exists a iteger e > 0, ozero vectors v 1,..., v e B, ad scalars c 1,..., c e F such that, v = c 1 v 1 + + c e v e. For every j = 1,..., e, let 1 i(j) s be a iteger such that v j W i(j) ; because B = W 1 Ws, there is at least oe such i(j). For every i, if i(j) = i for every j = 1,..., e, defie w i = 0. Otherwise defie, w i = c j v j. 1 j e,i(j)=i For every i = 1,..., s, because W i V is a vector subspace, w i W i. (w 1,..., w s ) is a ordered s tuple of vectors i W. Ad, e s s v = c j v j = ( c j v j ) = w i. Therefore W spas V. j=1 i=1 j,i(j)=i i=1 Thus Propositio 1.5. A ordered s tuple of subspaces of V, W = (W 1,..., W s ), is a direct sum decompositio of V iff for every ordered s tuple (B 1,..., B s ) of bases B i for W i, the uio B = B 1 B s is a basis for V. Proof. This follows from Propositio 1.3 ad Propositio 1.4. 2. Geeralized eigespaces Defiitio 2.1. Let T : V V be a liear operator. For every iteger 0, the th iterate of T is the liear operator T : V V recursively defied by T 0 = Id V, ad T +1 = T T. For every polyomial f(x) = a x + + a 1 x + a 0 F[x], the associated liear operator, f(t ) : V V, is defied to be f(t ) = a T + + a 1 T + a 0 Id V. Propositio 2.2. (i) For every pair of polyomials f(x), g(x) F[x], f(t ) + g(t ) = (f + g)(t ). (ii) For every polyomial f(x) F[x] ad every scalar a F, a (f(t )) = (a f)(t ). (iii) For every pair of polyomials f(x), g(x) F[x], f(t ) g(t ) = (f g)(t ) = g(t ) f(t ). Proof. (i) Let f(x) = a x + +a 1 x+a 0 ad let g(x) = b x + +b 1 x+b 0. By defiitio of polyomial additio, f + g = (a + b )x + + (a 1 + b 1 )x + (a 0 + b 0 ). By defiitio of the associated liear operator, (f + g)(t ) = (a + b )T + + (a 1 + b 1 )T + (a 0 + b 0 )Id V. By associativity ad commutativity of additio of operators, ad by distributivity of scalar multiplicatio of operators ad additio of operators, this equals, (a T + + a 1 T + a 0 Id V ) + (b T + + b 1 T + b 0 Id V ). By defiitio of the associated liear operator, this f(t is )+g(t ), i.e., (f+g)(t ) = f(t ) + g(t ). (ii) Let f(x) = a x + +a 1 x+a 0. By defiitio of scalar product of polyomials, a f(x) = (aa )x + + (aa 1 )x + (aa 0 ). By defiitio of the associated liear operator, (a f)(t ) = (aa )T + + (aa 1 )T + (aa 0 )Id V. 3

By distributivity of multiplicatio of scalars ad scalar multiplicatio of operators, this is, a (a T ) +... a (a 1 T ) + a (a 0 Id V ). By distributivity of scalar multiplicatio of operators ad operator additio, this is, a (a T + + a 1 T + a 0 Id V ). By defiitio of the associated liear operator, this a is (f(t )), i.e., (a f)(t ) = a (f(t )). (iii) The claim is that for every pair of itegers m, 0, T m T = T m+. This is proved by iductio o m. For m = 0, T 0 = Id V so that T 0 T = Id V T = T = T 0+. Thus assume m > 0 ad assume the result is true for smaller values of m. By the recursive defiitio, T m T = (T T m 1 ) T. By associativity of compositio of liear trasformatios, (T T m 1 ) T = T (T m 1 T ). By the iductio hypothesis, T m 1 T = T m 1+. Therefore T m T = T T m 1+. By the recursive defiitio, T m+ = T T m 1+, i.e., T m T = T m+. So the claim is proved by iductio o m. I particular, sice m + = + m, T m T = T m+ = T +m = T T m. Let f(x) = a m x m + + a 1 x + a 0 ad let g(x) = b x + + b 1 x + b 0. Let N = max(m, ). For i > m, defie a i = 0. For j >, defie b j = 0. By defiitio of polyomial multiplicatio, N (f g)(x) = ( a i b i )x. =0 i=0 By defiitio of the associated liear operator, (f g)(t ) = ( a i b i )T. =0 i=0 By the claim above, T = T i T i. Together with commuativity, associativity ad distributivity, the formula above is, ( a i b i T i T i ) = ( (a i T i ) (b i T i )). =0 i=0 =0 i=0 By distributivity of additio ad compositio of liear trasformatios, this is, m ( a i T i ) ( b j T j ). i=0 j=0 By the defiitio of the associated liear operator, this f(t is ) g(t ), i.e., (f g)(t ) = f(t ) g(t ). Defiitio 2.3. For every liear operator T : V V ad every polyomial f(x) F[x], the associated geeralized eigespace is E T,f = er(f(t )). For every λ F ad every iteger 0, the th geeralized λ eigespace is E () = E T,(x λ). Propositio 2.4. (i) For every pair of polyomials f(x), g(x) F[x], E T,f E T,g E T,f+g. (ii) For every polyomial f(x) F[x] ad every scalar a F, E T,f E T,a f. If a 0, the E T,f = E T,a f. 4

(iii) For every pair of polyomials f(x), g(x) F[x], E T,f + E T,g E T,f g. Proof. (i) For every v E T,f E T,g, f(t )(v) = g(t )(v) = 0. By Propositio 2.2(i), (f + g)(t )(v) = f(t )(v) + g(t )(v) = 0 + 0 = 0. Therefore v E T,f+g, i.e., E T,f E T,g E T,f+g. (ii) For every v E T,f, f(t )(v) = 0. By Propositio 2.2(ii), (a f)(t )(v) = a (f(t )(v)) = a 0 = 0. Therefore v E T,a f, i.e., E T,f E T,a f. If a = 0, the f = a 1 (a f) so that also E T,f E T,a f, i.e., E T,f = E T,a f. (iii) For every v E T,f, f(t )(v) = 0. By Propositio 2.2(iii), (f g)(t )(v) = (g f)(t )(v) = (g(t ) f(t ))(v) = g(t )(f(t )(v)) = g(t )(0) = 0. So v E T,f g, i.e., E T,f E T,f g. By the same argumet, E T,g E T,f g. Because E T,f g is a vector subspace, E T,f + E T,g E T,f g. Corollary 2.5. For every λ F, {0} = E(1) V. E (0) E(2) Proof. For every iteger, by Propositio 2.4(iii), E () = E T,(x λ) E T,(x λ) +1 = E (+1). Defiitio 2.6. For every λ F, the geeralized λ eigespace is E ( ) = 0E (). Propositio 2.7. For every λ F, the geeralized λ eigespace E ( ) is a vector subspace of V. Proof. Because {0} = E (0) E ( ), the subset E( ) is oempty. For every v, w E ( ), there exist itegers m, 0 such that v E(m) ad w E (). Let N = max(m, ). By Corollary 2.5, E (m) E(N). Sice E (N), E() is a vector subspace, v + w E (N) E ( ), i.e., E( ) is stable uder additio of elemets. By a similar argumet, E ( ) is stable uder scalar multiplicatio of elemets. So E ( ) is a vector subspace of V. 3. The Chiese remaider theorem Aciet Chiese geerals discovered a beautiful ad efficiet method for coutig large umbers of soldiers very quicly a tas of great importace i determiig the umber of losses after a battle. Let 1,..., s be itegers such that for every 1 i < j s, the pair of itegers i, ( j ) have o commo factor. For every i = 1,..., s have the N soldiers lie up i rows of size i. Do ot cout the umber of rows! Istead, cout the remaider i, i.e., the umer of soldiers who caot lie up i a row of size i. There exist itegers a 1,..., a s depedig oly o 1,..., s such that for every iteger N, N is cogruet to a 1 r 1 + + a s r s, modulo 1 s. If 1 = 10, 2 = 11, 3 = 13, the umbers are a 1 = 429, a 2 = 650, a 3 = 220, so that N = 429r 1 +650r 2 200r 3, modulo 1430. These may seem lie large umbers, but it is much faster for a mathematicia to compute the product o a abacus tha to cout the soldiers by brute force. Sice the geeral presumably ows the umber of soldiers to withi a rage ±715, of this method allows the geeral to compute precisely the umber of soldiers very quicly. We will use the same 5

method to prove that for distict λ 1,..., λ s F, the ordered s tuple of geeralized eigespaces, W = (E ( ),..., E ( ) 1 s ), is liearly idepedet. Defiitio 3.1. For a s tuple of polyomials i F[x], (f 1,..., f s ), ot all zero, a greatest commo factor is a polyomial f(x) of maximal degree such that f(x) divides f i (x) for every i = 1,..., s. If f 1 = = f s = 0, the greatest commo factor is defied to be 0. A s tuple of polyomials (f 1,..., f s ) is coprime if there exist polyomials (g 1,..., g s ) such that 1 = g 1 f 1 + + gs f s. Theorem 3.2 (The Chiese remaider theorem). A s tuple of polyomials (f 1,..., f s ) is coprime iff 1 is a greatest commo factor. Proof. ( ) Assume (f 1,..., f s ) is coprime, i.e., 1 = g 1 f 1 + + gs f s. Let f be ay commo factor of f 1 (,..., f s ). For every i, because f i is divisible by f, also g i f i is divisible by f. Because every g i f i is divisible by f, the sum g 1 f 1 +... g s f s is divisible by f, i.e., 1 is divisible by f. But the oly polyomials dividig 1 are ozero costats. Thus 1 is a greatest commo factor 1, of..(f., f s ). ( ) Assume 1 is a greatest commo factor fof 1,..., f s. The claim is that 1, (f..., f s ) is coprime. This will be proved by iductio s. o If s = 1, the f 1 is a greatest commo factor of f( 1 ). Sice also 1 is a greatest commo factor, fdeg( 1 ) = 0, i.e., f 1 is a scalar. By defiitio, the greatest commo factor of (0) is f 1 0, is so ozero. Sice F is a field, there exists a scalar g 1 such that 1 = f 1, i.e., (f 1 ) is coprime. Next assume s = 2. If ecessary, permute f 1 ad f 2 so that deg(f 1 ) deg(f 2 ). By hypothesis, 1 is a greatest commo factor of 1, (f f 2 ). If f 2 = 0, the 1 is a greatest commo factor of f( 1 ) ad by the last case there exists g 1 such that 1 = g 1 f 1. Puttig g 2 = 0, 1 = g 1 f 1 + g 2 f 2, i.e., (f 1, f 2 ) is coprime. Therefore assume f 2 = 0, which implies f 1 = 0. The claim i this case will be proved by iductio o deg(f 2 ). If deg(f 2 ) = 0, i.e., if 2 is a ozero scalar, there exists a ozero scalar g 2 such that 1 = g 2 f 2. Settig 1 = 0, 1 = g 1 f 1 + g 2 f 2. Thus, by way of iductio, assume deg(f 2 ) > 0 ad assume the result is ow for smaller values of deg(f 2 ). By the divisio algorithm, there exist polyomials q(x) ad r(x) with deg(r) < deg(f 2 ) such that f 1 (x) = q(x)f 2 (x) + r(x). If h is a commo factor of 2, (fr), the h is also a factor of qf 2 ad so of the sum qf 2 + r, i.e., h is a commo factor of (f 1, f 2 ). Coversely, if h is a commo factor of 1, (ff 2 ), the h is a commo factor of qf 2 ad so of the sum f 1 + ( qf 2 ), i.e., h is a commo factor of 2, (fr). So the commo factors of 1, (ff 2 ) are precisely the commo factors of f 2, ( r). By hypothesis, 1 is a greatest commo factor of 2, r). (f Sice deg(r) < 0, by the iductio hypothesis there exist polyomials g 1, g 2 such that 1 = g 1 f 2 +g 1 r. Defie g 1 = g 2 ad g 2 = g 1 qg 2. The, g 1 f 1 + g 2 f 2 = g 2 (qf 2 + r) + (g 1 qg 2 )f 2 = g 1 f 2 + g g 1 2r = 1. So (f 1, f 2 ) is coprime. So the claim is proved by iductio o deg(f 2) if s = 2. By way of iductio, assume s > 2 ad the result is ow for smaller s. Let h be a greatest commo factor of 1, (f..., f s 1 ). The 1 is a greatest commo factor of (f 1 /h, dots, f s 1 /h). By the iductio hypothesis, there exist elemets g 1,..., g s 1 such that, 1 = g 1 (f 1 /h) + + g s 1 (f s 1 /h), 6

i.e., h = g 1 f 1 + + g s 1f s 1. The greatest commo factor of 1, (f..., f s 1, f s ) is the greatest commo factor of (h, f s ). By hypothesis, 1 is a greatest commo factor h, of f s ). ( By the case s = 2 above, there exist elemets g 1, g 2 such that 1 = g 1 h + g 2 f s. Defiig g 1 g 1 g 1 g 1 = g 1, g 2 = g 2,..., g s 1 = g s 1 ad defiig g s = g 2, g 1 f 1 + + g s 1 f s 1 + g s f s = g 1 (g s 1f s 1 ) + g 1 h + g 1 f 1 + + g 2 f s = g 2 f s = 1. So (f 1,..., f s ) is coprime. The theorem is proved by iductio s. o Corollary 3.3. Let s 2 ad let (f 1,..., f s ) be ozero polyomials. For every i = 1,..., s, defie, r i (x) = f j (x). 1 j s,j=i If for every 1 i < j s, 1 is a greatest commo factor (f of i, f j ), the (r 1,..., r s ) is coprime. Proof. By Theorem 3.2, it suffices to prove that 1 is a greatest commo factor of (r 1,..., r s ). This will be proved by iductio s. o For s = 2, (r 1, r 2 ) = (f 2, f 1 ). By hypothesis, 1 is a greatest commo factor. Thus, by way of iductio, assume s > 2 ad the result is ow for smaller values s. of Cosider the sequece (f 1,..., f s 1 ). The hypothesis holds for this collectio. For every i = 1,..., s 1, deote r i = f j. 1 j s 1,j=i By the iductio hypothesis, 1 is a commo factor of 1,..(r., r s 1 ). Therefore f s is a greatest commo factor of f s r( 1,..., f s r s 1) = (r 1,..., r s 1 ). So the commo factors of (r 1,..., r s 1, r s ) are the commo factors of s, (f r s ) = (f s, f 1 f s 1 ). Let h be a irreducible commo factor of s, (ff 1 f s 1 ). Because h is irreducible ad divides f 1 f s 1, there exists 1 i s 1 such that h divides f i, i.e., h is a commo factor of (f i, f s ). By hypothesis, 1 is a greatest commo factor i, of f s (f ). Therefore h is a ozero scalar. So for every commo factor h of (f s, f 1 f s 1 ), every irreducible factor of h is a ozero scalars, i.e., h is a ozero scalar. Therefore 1 is a greatest commo factor of 1, (r..., r s ). The corollary is proved by iductio o s. Corollary 3.4. For every s tuple of distict scalars λ 1,..., λ s F ad every iteger > 0, for every i = 1,..., s defie, r i (x) = (x λ j ). The (r 1,..., r s ) is coprime. 1 j s,j=i Proof. For every 1 i < j, 1 is a greatest commo factor of (x λ i ) ad (x λ j ). To see this, first observe, 1 = (x λ i )/(λ j λ i ) + (x λ j )/(λ i λ j ). Raisig both sides to the power 2 1 ad usig the biomial theorem gives, ( ( ) ) 1 = 1 1 2 1 = i=1 ( 1) i (x λ i ) 1 i (x λ j ) i /(λ j λ i ) 2 1 (x λ i ) ( i 2 1 ( ) ) + 2 1 i= ( 1) i (x λ i ) 2 1 i (x λ j ) i /(λ j λ i ) 2 1 (x λ j ) i. 7

I other words, ((x λ i ), (x λ j ) ) is coprime. By the easy part of Theorem 3.2, 1 is a greatest commo factor. 4. Liear idepedece of geeralized eigespaces Let s 2 be a iteger. Let λ 1,..., λ s F be distict, ad let > 0 be a iteger. Defie r 1,..., r s as i Corollary 3.4. By Corollary 3.4, there exist polyomials g 1,..., g s such that, 1 = g 1 (x)r 1 (x) + + g s (x)r s (x). Let T : V V be a liear operator. The idetity above gives a idetity of associated liear operators, Id V = 1(T ) = g 1 (T ) r 1 (T ) + + g s (T ) r s (T ). Lemma 4.1. For every 1 i s ad every w i E () i, for every 1 j s with j = i, r j (T )(w i ) = 0. Proof. Because j = i, by costructio r j (x) = q(x)(x λ i ) for some q(x) F[x], specifically, q(x) = (x λ ). 1 s,=i,j By Propositio 2.2(iii), r j (T ) = q(t ) (T λ i Id V ). By hypothesis, w i E () i := er((t λ i Id V ) ). Thus r j (T )(w i ) = q(t )((T λ i Id V ) (w i )) = q(t )(0) = 0. Lemma 4.2. For every i = 1,..., s ad every w i E () i, g i (T ) r i (T )(v i ) = v i. s Proof. By the idetity, w i = Id V (w i ) = j=1 g i(t ) r i (T )(w i ). By Lemma 4.1, for every j = i, g i (T ) r i (T )(w i ) = g i (T )(r i (T )(w i )) = g i (T )(0) = 0. Deote by W the ordered s tuple of vector subspaces of V, W = (E (),..., E () ) Propositio 4.3. For every ordered s tuple of vectors i W, (w 1,..., w s ), deotig v = w 1 + + w s, for every i = 1,..., s, w i = g i (T ) r i (T )(v). Proof. Because g i (T ) r i (T ) is a liear operator, 1 s s g i (T ) r i (T )(v) = g i (T ) r i (T )( w j ) = g i (T ) r i (T )(w j ). j=1 j=1 By Lemma 4.1, if j = i the g i (T ) r i (T )(w j ) = g i (T )(r i (T )(w j ) = g i (T )(0) = 0. Therefore, g i (T ) r i (T )(v) = g i (T ) r i (T )(w i ). By Lemma 4.2, g i (T ) r i (T )(w i ) = w i. Therefore g i (T ) r i (T )(v) = w i. Theorem 4.4. Let T : V V be a liear operator. For every iteger s 1 ad every ordered s tuple of distict scalars (λ 1,..., λ s ), the ordered s tuple of vector subspaces W = (E ( ),..., E ( ) 1 s ) is liearly idepedet. 8 s

Proof. If s = 1, this is trivial: for ay subspace W of V, (W) is liearly idepedet. Thus assume s 2. Let (w 1,..., w s ) be a ordered s tuple of vectors i W such that 0 = w 1 + +w s. By defiitio, for every i = 1,..., s, E ( ) = 0 E (). So for every i = 1,..., s, there exists a iteger i > 0 such that w i E (i) i. Defie = max( 1,..., s ). By Corollary 2.5, for every i = 1,..., s, w i E () i. By Propositio 4.3, for every i = 1,..., s, w i = g i (T ) r i (T )(0) = 0. Sice (w 1,..., w s ) = (0,..., 0), W is liearly idepedet. i i 9