Maths 1 2013/2014

Reduction , projector,endomorphism of rank 1 ...Hadamard’s inequality and some applications

CCP Maths 2〉〈

Solution

1. 1.1 A is a symmetric matrix so diagonalizable. 1.2 Diagonalization of A : A characteristic polynomial of A is χA =−(X +4)(X−1)(X−6), so Sp(A) =

−4,1,6.

A = PDP−1, where P =

3 −4 3−5 0 54 3 4

and D = diag(−4,1,6).

1.3 With wxMaxima :

An = PDnP−1 =

950(−4)n + 9

50 6n + 1625 − 3

10(−4)n + 310 6n 6

25(−4)n + 625 6n− 12

25− 3

10(−4)n + 310 6n 1

2(−4)n + 12 6n −2

5(−4)n + 25 6n + 16

25625(−4)n + 6

25 6n− 1225 −2

5(−4)n + 25 6n 8

25(−4)n + 825 6n + 9

25

2. Expression of sequences un, vn and wn : With wxMaxima

For n ∈ N, we put Un =

un

vn

wn

, the recurrence relation between the sequences, (un)n , (vn)n and (wn)n can

be written matricially ; Un+1 = AUn, so for each n ∈ N ; Un = AnU0, we get :un = 21

50(−4)n + 2150 6n + 4

25vn = − 7

10(−4)n + 710 6n

un = 1425(−4)n + 14

25 6n− 325

Exercise 1.

1. 1.1 p is a projector of E , so the polynomial X2−X = X(X − 1) is an annihilator of p, since the two

polynomials X and X − 1 are coprime , by the primary decomposition theorem , we have ; E = ker p⊕ker(p− IdE) ; other hand x ∈ Im p if and only if x = p(x). This show that ker(p− IdE) = Im p.Whence E = ker p⊕ Im p. 1.2 We denote r = rg p = dim(Im p), (v1, . . . ,vr) a basis of a subspace Im p and (vr+1, . . . ,vn) a basis ofa subspace ker p ; it is clear that (v1, . . . ,vn) is a basis of E , the matrix of p in this basis has the form :(

Ir 00 0

), it follows that tr p = r = rg p.

1.3 Let u be the endomorphism of R2 canonically associated to the matrix A =

(1 10 1

), we have rgu =

2 = tru , but u is not a projector, since A2 6= A.

Exercise 2.

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2. We consider a matrices A =

1 0 00 0 00 0 0

and B =

0 0 10 0 00 0 0

we see that rgA = rgB = 1.

A matrix A is diagonal , so , it is diagonalizable.A matrix B is not diagonalizable : B is nilpotent , and it’s minimal polynomial is X2 , it is a split polynomialbut it has a double root. 3. u an endomorphism of E of rank 1. 3.1 By the rank formula dimkeru = n− rgu = n− 1 , so keru is an hyperplane of E . Now let

(e1, . . . ,en−1) be a basis of keru there exist en ∈ E such that (e1, . . . ,en−1,en) be a basis of E ,obviously en /∈ keru, we can write u(en) = a1e1 + . . . ,anen , for all i ∈ 1, . . . ,n− 1, we haveu(ei) = 0. Thus the matrix of u with respect to B = (e1, . . . ,en) has desired form. 3.2 With the notations of previous question , we have tru = an, and the characteristic polynomial of u isχu = Xn−1(X−an).If tru 6= 0 i.e an 6= 0, in this case Spu = 0,an , the eigenspace associated to the eigenvalue 0 is keruand dimkeru = n− 1 , other an = tru is an eigenvalue of u then dimker(u− IdE) ≥ 1 as keru∩ker(u−an IdE) = 0 it result that dimker(u−an IdE) = 1, that is n = dimkeru+dimker(u−an IdE), we deduce that u est diagonalizable.If tru = 0 i.e an = 0, then 0 is the unique eigenvalue of u, hence u is a nilpotent endomorphism , but theunique nilpotent endomorphism which is diagonalizable is the zero endomorphism , since u is not a zeroendomorphism, it follows that u is not diagonalizable.Remark : If f is a nilpotent endomorphism of vector space of dimension n that is diagonalizable , thenf = 0 ; In fact, let B be a basis consisting of eigenvectors of the endomorphism f , the matrix of f withrespect to B is a zero matrix, hence f is a zero endomorphism. 3.3 rgu = 1 et tru = 1 6= 0, by the result of previous question , u is diagonalizable (dimkeru = n− 1and dimker(u− IdE) = 1) and there exist a basis B consisting of eigenvectors of u such that A :=

MB(u) =

0 . . . 0...

. . ....

0 . . . 1

, it is clear that A2 = A, that is u2 = u hence u is a projector.

3.4 Let u be the endomorphism of R3 canonically associated to matrix A, it is easy to verify that trA = 1rgA = 1, so tru = rgu = 1 ; by the previous result , u is a projector.If we denote (e1,e2,e3) the canonical basis of R3 ; then u(e1) = u(e2) = −u(e3) = e1 + e2 + e3,hence Imu = Vect(e1 + e2 + e3).Let (x,y,z) ∈ R3, then ; u(x,y,z) = 0 if and only if x+ y− z = 0 if and only if z = x+ y, it followsthat keru = Vect((1,0,1),(0,1,1)).

Exercice 2. suite

PROBLEMFirst part :

Preliminary questions 1. 1.1 Any symmetric endomorphism is diagonalizable in an orthonormal basis.

Any symmetric matrix is orthogonally diagonalizable .

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Maths 1 2013/2014 1.2 The matrix S is symmetric, its characteristic polynomial is X2, since S is not a zero matrix, then its minimalpolynomial is X2 that is split an it has a double root, so S is not diagonalizable.Conclusion : A complex symmetric matrix is not in general diagonalizable . 2.

2.1 Let x ∈ E , by putting x =n

∑i=1

xiεi, we have s(x) =n

∑i=1

λixiεi, hence 〈s(x),x〉=n

∑i=1

λix2i .

2.2 Let x =n

∑i=1

xiεi ∈ S(0,1), then ‖x‖2 =n

∑i=1

x2i ( since β is an orthonormal basis of E ),

λ1 = λ1

n

∑i=1

x2i ≤ Rs(x) =

n

∑i=1

λix2i ≤ λn

n

∑i=1

x2i = λn

thus Rs(x) ∈ [λ1,λn]. 3. 3.1 Let λ be an eigenvalue of s, and let v be an associated eigenvector, that is s(v)= λv, we have 〈s(v),v〉=

λ‖v‖2, if s is positive then λ‖v‖2≥ 0 , and λ ≥ 0. If s is positive definite then λ‖v‖2 > 0, and λ > 0. 3.2 〈s(ei),e j〉= si, j, in particular si,i = 〈s(ei),ei〉= Rs(ei) ∈ [λ1,λn], thus λ1 ≤ si,i ≤ λn.

Remark : If x =n

∑i=1

xiεi and y =n

∑i=1

yiεi then 〈s(x),y〉= ∑1≤i, j≤n

xiy jsi, j. 4. The maps M 7→M and M 7→ tM are continuous, (linear in finite dimension), so M 7→ tMM continuous (productof continuous maps), thus M 7→ tMM− In is a continuous map. 5. Let A ∈ On(R),then these row vectors form an orthonormal basis of Rn, in particular each row is an unitary

vector . Now for all i ∈ 1, . . . ,n we haven

∑k=1

a2i,k = 1, thus for all j ∈ 1, . . . ,n, a2

i, j ≤n

∑k=1

a2i,k = 1, so

|ai, j| ≤ 1. 6. We endow Mn(R) with the norm N defined for any M = (mi, j) by N(M) = max1≤i, j≤n

|ai, j| , and we recall

that ; in finite dimension, all norms are equivalents .The map f : Mn(R)→Mn(R) defined by f (M) = tMM− In is continuous, and On(R) = f−10, soOn(R) closed subset of Mn(R), other hand, for any M ∈ On(R), we have N(M) ≤ 1, hence On(R) is abounded subset , since Mn(R) is finite dimensional vector space , the subset On(R) is a compact subset ofMn(R). 7. 7.1 Since S is symmetric , there exist orthogonal matrix P such that S= tP∆P, T (A)= tr(AS)= tr(AtP∆P)=

tr(PAtP∆), and B = PAtP ∈ On(R) is a suitable matrix . 7.2 The map M 7→ tr(MS) is a linear form of Mn(R) so it is a continuous map, thus its restriction T toOn(R) is a continuous map, since On(R) is a compact subset , then T is bounded and attain its upperbound its maximum . 7.3 Let A ∈ On(R), there exist B = (bi, j) ∈ On(R) such that T (A) = tr(B∆), for all i ∈ 1, . . . ,n,

we have (B∆)i,i = λibi,i, hence tr(B∆) =n

∑i=1

λibi,i, since B ∈ On(R), then |bi,i| ≤ 1. Now it is that

T (A) = tr(B∆)≤n

∑i=1

λi = tr(S) (the λi ≥ 0).

We have In ∈ On(R) et T (In) = tr(S), by the previous question , for all A ∈ On(R), we get T (A) ≤tr(S) = T (In), hence t = T (In) = tr(S).

Second part :Hadamard’s inequality

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8. detS =n

∏i=1

λi et tr(S) =n

∑i=1

λi, by the arithmetic-geometric inequality , we getn

∏i=1

λi ≤

(1n

n

∑i=1

λi

)n

, i.e

detS≤ (1n tr(S))n. 9. tSα = t(tDSD) = tDtSD = tDSD = Sα , hence Sα is a symmetric matrix . For any column matrix X ,

tXSαX = t(DX)S(DX)≥ 0, then Sα ∈S +n (R).

By remarking that tD = D, tr(Sα) = tr(SD2), and for any i ∈ 1, . . . ,n, (SD2)i,i = si,iα2i , it follow that

tr(Sα) =n

∑i=1

si,iα2i . 10. Sα ∈S +

n (R), by the result of question 8., we get detSα ≤(1

n tr(Sα))n, but

Sα = detSdetD2 =n

∏i=1

λi

n

∏i=1

1si,i

=∏

ni=1 λi

∏ni=1 si,i

, and tr(Sα) =n

∑i=1

si,iα2i = n, so

detS =n

∏i=1

λi ≤n

∏i=1

si,i. 11. Sε is positive symmetric matrix. Moreover , for all i ∈ 1, . . . ,n, (Sε)i,i = si,i + ε > 0,by the result of the10. , it follows that detSε ≤∏

ni=1(Sε)i,i = ∏

ni=1(si,i + ε).

passing to the limit as ε tends to 0 in the previous inequality , and taking account the continuity of the mapsdet and tr, we obtain detS≤∏

ni=1 si,i, this show the result.

Third part :Application of Hadamard’s inequality : determining a minimum 12. The matrix B is symmetric, and for any column vector X we have tXBX = t(ΩX)A(ΩX)≥ 0. If tXAX = 0

then t(ΩX)A(ΩX) = 0 as A positive definite, then ΩX = 0, hence X = 0 (Ω is invertible matrix), other handdetB = detA = 1, thus B ∈U .tr(AS) = tr(AΩ∆tΩ) = tr(tΩAΩ∆) = tr(B∆). 13. If A ∈U , by putting B = tΩAΩ, we get B ∈U and tr(AS) = tr(B∆).If B ∈U , by putting A = ΩBtΩ, we get A ∈U et tr(B∆) = tr(AS).This show the equality between these sets .

If B = (bi, j) ∈ U , we have tr(B∆) =n

∑i=1

λibi,i, taking account the positivity of bi,i and λi , we obtain

tr(B∆) ≥ 0. It follows that the subset tr(B∆) , B ∈ U is non-empty set and it is bounded below, thus ithas a lower bounded. 14. Let B = (bi,i) ∈ U , we have tr(B∆) =

n

∑i=1

λibi,i, by arithmetic-geometric inequality , with ai = λibi,i, we

obtainn

∑i=1

λibi,i ≥ n

(n

∏i=1

λibi,i

) 1n

= n(λ1 . . .λn)1n (b1,1 . . .bn,n)

1n . this show the result. 15. B ∈U ⊂S +

n (R) , by applying the Hadamard’s inequality to the matrix B, we get b1,1 . . .bn,n ≥ detB = 1,hence tr(B∆)≥ n(λ1 . . .λn)

1n = n(detS)

1n . 16. Immediately we have : D ∈ U and tr(D∆) =

n

∑i=1

λiµi = n(detS)1n , so for all B ∈ U , we have tr(B∆) ≥

tr(D∆). It follows that m = tr(D∆) = n(detS)1n .

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Maths 1 2013/2014

END

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