Maths /2014. CCP Maths 2. Reduction, projector,endomorphism of rank 1... Hadamard s inequality and some applications. Solution.

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1 CCP Maths 2 Reductio, projector,edomorphism of rak 1... Hadamard s iequality ad some applicatios Solutio Exercise 1. 1 A is a symmetric matrix so diagoalizable. 2 Diagoalizatio of A : A characteristic polyomial of A is χ A = (X +4)(X 1)(X 6), so Sp(A) = { 4,1,6} A = PDP 1, where P = ad D = diag( 4,1,6) With wxmaxima : 9 A = PD P 1 50 ( 4) ( 4) ( 4) = 3 10 ( 4) ( 4) ( 4) ( 4) ( 4) ( 4) Expressio of sequeces u, v ad w : With wxmaxima u For N, we put U = v, the recurrece relatio betwee the sequeces, (u ), (v ) ad (w ) ca w be writte matricially ; U +1 = AU, so for each N ; U = A U 0, we get : u = ( 4) v = 7 10 ( 4) u = 14 ( 4) Exercise 2. 1 p is a projector of E, so the polyomial X 2 X = X(X 1) is a aihilator of p, sice the two polyomials X ad X 1 are coprime, by the primary decompositio theorem, we have ; E = ker p ker(p Id E ) ; other had x Im p if ad oly if x = p(x). This show that ker(p Id E ) = Im p. Whece E = ker p Im p. 2 We deote r = rg p = dim(im p), (v 1,...,v r ) a basis of a subspace Im p ad (v r+1,...,v ) a basis of a ( subspace ) ker p ; it is clear that (v 1,...,v ) is a basis of E, the matrix of p i this basis has the form : Ir 0, it follows that tr p = r = rg p Let u be the edomorphism of R 2 caoically associated to the matrix A = 2 = tru, but u is ot a projector, sice A 2 A. ( ) 1 1, we have rgu = 0 1 MP 1 /5 Aqalmou Mohamed

2 Exercice 2. suite We cosider a matrices A = ad B = we see that rga = rgb = A matrix A is diagoal, so, it is diagoalizable. A matrix B is ot diagoalizable : B is ilpotet, ad it s miimal polyomial is X 2, it is a split polyomial but it has a double root. 3. u a edomorphism of E of rak By the rak formula dimkeru = rgu = 1, so keru is a hyperplae of E. Now let (e 1,...,e 1 ) be a basis of keru there exist e E such that (e 1,...,e 1,e ) be a basis of E, obviously e / keru, we ca write u(e ) = a 1 e ,a e, for all i {1,..., 1}, we have u(e i ) = 0. Thus the matrix of u with respect to B = (e 1,...,e ) has desired form. 3.2 With the otatios of previous questio, we have tru = a, ad the characteristic polyomial of u is χ u = X 1 (X a ). If tru 0 i.e a 0, i this case Spu = {0,a }, the eigespace associated to the eigevalue 0 is keru ad dimkeru = 1, other a = tru is a eigevalue of u the dimker(u Id E ) 1 as keru ker(u a Id E ) = {0} it result that dimker(u a Id E ) = 1, that is = dimkeru + dimker(u a Id E ), we deduce that u est diagoalizable. If tru = 0 i.e a = 0, the 0 is the uique eigevalue of u, hece u is a ilpotet edomorphism, but the uique ilpotet edomorphism which is diagoalizable is the zero edomorphism, sice u is ot a zero edomorphism, it follows that u is ot diagoalizable. Remark : If f is a ilpotet edomorphism of vector space of dimesio that is diagoalizable, the f = 0 ; I fact, let B be a basis cosistig of eigevectors of the edomorphism f, the matrix of f with respect to B is a zero matrix, hece f is a zero edomorphism. 3.3 rgu = 1 et tru = 1 0, by the result of previous questio, u is diagoalizable (dimkeru = 1 ad dimker(u Id E ) = 1) ad there exist a basis B cosistig of eigevectors of u such that A := M B (u) =....., it is clear that A 2 = A, that is u 2 = u hece u is a projector Let u be the edomorphism of R 3 caoically associated to matrix A, it is easy to verify that tra = 1 rga = 1, so tru = rgu = 1 ; by the previous result, u is a projector. If we deote (e 1,e 2,e 3 ) the caoical basis of R 3 ; the u(e 1 ) = u(e 2 ) = u(e 3 ) = e 1 + e 2 + e 3, hece Imu = Vect(e 1 + e 2 + e 3 ). Let (x,y,z) R 3, the ; u(x,y,z) = 0 if ad oly if x + y z = 0 if ad oly if z = x + y, it follows that keru = Vect((1,0,1),(0,1,1)). PROBLEM First part : Prelimiary questios 1 Ay symmetric edomorphism is diagoalizable i a orthoormal basis. Ay symmetric matrix is orthogoally diagoalizable. MP 2 /5 Aqalmou Mohamed

3 2 The matrix S is symmetric, its characteristic polyomial is X 2, sice S is ot a zero matrix, the its miimal polyomial is X 2 that is split a it has a double root, so S is ot diagoalizable. Coclusio : A complex symmetric matrix is ot i geeral diagoalizable Let x E, by puttig x = x i ε i, we have s(x) = λ i x i ε i, hece s(x),x = λ i xi Let x = x i ε i S(0,1), the x 2 = xi 2 ( sice β is a orthoormal basis of E), λ 1 = λ 1 i x 2 R s (x) = i xi λ 2 λ xi 2 = λ thus R s (x) [λ 1,λ ] Let λ be a eigevalue of s, ad let v be a associated eigevector, that is s(v) = λv, we have s(v),v = λ v 2, if s is positive the λ v 2 0, ad λ 0. If s is positive defiite the λ v 2 > 0, ad λ > s(e i ),e j = s i, j, i particular s i,i = s(e i ),e i = R s (e i ) [λ 1,λ ], thus λ 1 s i,i λ. Remark : If x = x i ε i ad y = y i ε i the s(x),y = x i y j s i, j. 1 i, j 4. The maps M M ad M t M are cotiuous, (liear i fiite dimesio), so M t MM cotiuous (product of cotiuous maps), thus M t MM I is a cotiuous map. 5. Let A O (R),the these row vectors form a orthoormal basis of R, i particular each row is a uitary vector. Now for all i {1,...,} we have k=1a 2 i,k = 1, thus for all j {1,...,}, a 2 i, j a 2 i,k = 1, so k=1 a i, j We edow M (R) with the orm N defied for ay M = (m i, j ) by N(M) = max a i, j, ad we recall 1 i, j that ; i fiite dimesio, all orms are equivalets. The map f : M (R) M (R) defied by f (M) = t MM I is cotiuous, ad O (R) = f 1 {0}, so O (R) closed subset of M (R), other had, for ay M O (R), we have N(M) 1, hece O (R) is a bouded subset, sice M (R) is fiite dimesioal vector space, the subset O (R) is a compact subset of M (R) Sice S is symmetric, there exist orthogoal matrix P such that S = t P P, T (A) = tr(as) = tr(a t P P) = tr(pa t P ), ad B = PA t P O (R) is a suitable matrix. 7.2 The map M tr(ms) is a liear form of M (R) so it is a cotiuous map, thus its restrictio T to O (R) is a cotiuous map, sice O (R) is a compact subset, the T is bouded ad attai its upper boud its maximum. 7.3 Let A O (R), there exist B = (b i, j ) O (R) such that T (A) = tr(b ), for all i {1,...,}, we have (B ) i,i = λ i b i,i, hece tr(b ) = λ i b i,i, sice B O (R), the b i,i 1. Now it is that T (A) = tr(b ) λ i = tr(s) (the λ i 0). We have I O (R) et T (I ) = tr(s), by the previous questio, for all A O (R), we get T (A) tr(s) = T (I ), hece t = T (I ) = tr(s). Secod part : Hadamard s iequality MP 3 /5 Aqalmou Mohamed

4 8. dets = λ i et tr(s) = λ i, by the arithmetic-geometric iequality, we get λ i ( 1 λ i ), i.e dets ( 1 tr(s)). 9. t S α = t ( t DSD) = t D t SD = t DSD = S α, hece S α is a symmetric matrix. For ay colum matrix X, t XS α X = t (DX)S(DX) 0, the S α S + (R). By remarkig that t D = D, tr(s α ) = tr(sd 2 ), ad for ay i {1,...,}, (SD 2 ) i,i = s i,i αi 2, it follow that tr(s α ) = s i,i αi S α S + (R), by the result of questio 8., we get dets α ( 1 tr(s α) ), but S α = detsdetd 2 1 = λ i = λ i s i,i s, ad tr(s α ) = s i,i αi 2 =, so i,i dets = λ i s i,i. 11. S ε is positive symmetric matrix. Moreover, for all i {1,...,}, (S ε ) i,i = s i,i + ε > 0,by the result of the 10., it follows that dets ε (S ε) i,i = (s i,i + ε). passig to the limit as ε teds to 0 i the previous iequality, ad takig accout the cotiuity of the maps det ad tr, we obtai dets s i,i, this show the result. Third part : Applicatio of Hadamard s iequality : determiig a miimum 12. The matrix B is symmetric, ad for ay colum vector X we have t XBX = t (ΩX)A(ΩX) 0. If t XAX = 0 the t (ΩX)A(ΩX) = 0 as A positive defiite, the ΩX = 0, hece X = 0 (Ω is ivertible matrix), other had detb = deta = 1, thus B U. tr(as) = tr(aω t Ω) = tr( t ΩAΩ ) = tr(b ). 13. If A U, by puttig B = t ΩAΩ, we get B U ad tr(as) = tr(b ). If B U, by puttig A = ΩB t Ω, we get A U et tr(b ) = tr(as). This show the equality betwee these sets. If B = (b i, j ) U, we have tr(b ) = λ i b i,i, takig accout the positivity of b i,i ad λ i, we obtai tr(b ) 0. It follows that the subset {tr(b ), B U } is o-empty set ad it is bouded below, thus it has a lower bouded. 14. Let B = (b i,i ) U, we have tr(b ) = λ i b i,i, by arithmetic-geometric iequality, with a i = λ i b i,i, we obtai 15. B U S + (R), by applyig the Hadamard s iequality to the matrix B, we get b 1,1...b, detb = 1, hece tr(b ) (λ 1...λ ) 1 = (dets) Immediately we have : D U ad tr(d ) = λ i µ i = (dets) 1, so for all B U, we have tr(b ) ) 1 λ i b i,i ( λ i b i,i = (λ 1...λ ) 1 (b1,1...b, ) 1. this show the result. tr(d ). It follows that m = tr(d ) = (dets) 1. MP 4 /5 Aqalmou Mohamed

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