6.8. The primary decomposition theorem Decompose into elementary parts using the minimal polynomials.

Theorem 12. T in L(V,V). V f.d.v.s. over F. p minimal polynomial. P=p 1 r_1 ….p k r_k.r i > 0. Let W i = null p i (T) r_i. –Then –(i) V = W 1  …  W k. –(ii) Each W i is T-invariant. –(iii) Let T i =T|W i :W i ->W i. Then minpolyT i = p i (T) r_i

Example: –Char.polyT=(x-1) 2 (x-2) 2 =min.polyT: Check this by any lower degree does not kill T by computations. –null(T-I) 2 = null = null = –Similarly null(T-2I) 2 =

Proof: idea is to get E 1,..,E k. –Let f i = p/p i r_i =p 1 r_1 …p i-1 r_i-1 p i+1 r_i+1 …p k r_k. –f 1,…,f k are relatively prime since there are no common factors. –That is, =F[x]. –There exists g 1,…,g k in F[x] s.t. g 1 f 1 +….+g k f k = 1. –p divides f i f j for i  j since f i f j contains all factors. –Let E i = h i (T)=f i (T)g i (T), h i =f i g i.

–Since h 1 +…+h k =1, E 1 +…+E k =I. –E i E j =0 for i  j. –E i =E i (E 1 +…+E k )=E i 2. Projections. –Let Im E i = W i. Then V = W 1  …  W k. –(i) is proved. –T E i = E i T. Thus Im E i = W i is T-invariant. –(ii) is proved. –We show that Im E i = null p i (T) r_i. (  ) p i (T) r_i E i a = p i (T) r_i f i (T)g i (T)a = p(T) g i (T)a =0.

(  ) a in null p i (T) r_i. If j  i, then f j (T)g j (T)a =0 since p i r_i divides f j and hence f j g j. E j a=0 for j  I. Since a=E 1 a+…+E k a, it follows that a=E i a. Hence a in Im E i. –(i),(ii) is completely proved. –(iii) T i = T|W i :W i ->W i. –P i (T i ) r_i =0 since W i is the null space of P i (T) r_i. –minpolyT i divides P i r_i. –Suppose g is s.t. g(T i )=0.

–g(T)f i (T)=0: f i = p 1 r_1 …p i-1 r_i-1 p i+1 r_i+1 …p k r_k. Im E i =null p i r_i. Thus Im f i (T) is in Im E i since V is a direct sum of Im E j s. –p divides gf i. –p= p i r_i f i by definition. –Thus p i r_i divides g. –Thus, minpoly T i = p i r_i.

Corollary: E 1,…,E k projections ass. with the primary decomposition of T. Then each E i is a polynomial in T. If a linear operator U commutes with T, then U commutes with each of E i and W i is invariant under U. Proof: E i = f i (T)g i (T). Polynomials in T. Hence commutes with U. –W i =Im E i. U(W i )= Im U E i = Im E i U in Im E i =W i.

Suppose that minpoly(T) is a product of linear polynomials. p=(x-c 1 ) r_1 …(x-c k ) r_k. (For example F=C). –Let D=c 1 E 1 +…+c k E k. Diagonalizable one. –T=TE 1 +…+TE k –N:=T-D=(T-c 1 I)E1+…+(T-c k I)E k –N 2 = (T-c 1 I) 2 E1+…+(T-c k I) 2 E k –N r = (T-c1I) r E1+…+(T-c k I) r E k

–If r  r i for each I, (T-ciI) r =0 on Im E i. –Therefore, N r = 0. N=T-D is nilpotent. Definition. N in L(V,V). N is nilpotent if there is some integer r s.t. N r = 0. Theorem 13. T in L(V,V). Minpoly T= prod.of 1st order polynomials. Then there exists a diagonalizable D and a nilpotent operator N s.t. –(i) T=D+N. –(ii) DN=ND. –D, N are uniquely determined by (i)(ii) and are polynomials of T.

Proof: T=D+N. E i =h i (T)=f i (T)g i (T). –D=c 1 E 1 +…+c k E k is a polynomial in T. –N=T-D a polynomial in T. –Hence, D,N commute. (Uniquenss) Suppose T=D’+N’, D’N’ commutes, D’ diagonalizable, N nilpotent. –D’ commutes T=D’+N’. D’ commutes with any polynomials of T. –D’ commutes with D and N. –D’+N’=D+N. –D-D’=N’-N. They commutes with each other. –Since D and D’ commutes, they are simultaneously diagonalizable. (Section. 6.5 Theorem 8.)

–N’-N is nilpotent: r is suff. large. (larger 2max of the degrees of N,N’) -> r-j or j is suff large. Thus the above is zero. –D-D’=N’-N is a nilpotent operator which has a diagonal matrix. Thus, D-D’=0 and N’- N=0. –D’=D and N’=N.

Application to differential equations. Primary decompostion theorem holds when V is infinite dimensional and when p is only that p(T)=0. Then (i),(ii) hold. This follows since the same argument will work. A positive integer n. V = {f| n times continuously differentiable complex valued functions which satisfy ODE } C n ={n times continuously differentiable complex valued functions}

Let p=x n +a n-1 x n-1 +…+a 1 x + a 0. Let D differential operator, Then V is a subspace of C n where p(D)f=0. V=null p(D). Factor p=(x-c 1 ) r_1 …(x-c k ) r_k. c 1,..,c k in the complex number field C. Define W j := null(D-c j I) r_j. Then Theorem 12 says that V = W 1  …  W k In other words, if f satisfies the given differential operator, then f is expressed as f =f 1 +…+f k, f i in W i.

What are W i s? Solve (D-cI) r f=0. Fact: (D-cI) r f=e ct D r (e -ct f): –(D-cI) f=e ct D(e -ct f). –(D-cI) 2 f= e ct D(e -ct e ct D(e -ct f))…. (D-cI) r f=0 D r (e -ct f)=0: –Solution: e -ct f is a polynomial of deg < r. –f= e ct (b 0 + b 1 t +…+ b r-1 t r-1 ). Here e ct,te ct,t 2 e ct,…, t r-1 e ct are linearly independent. Thus {t m e c_j t| m=0,…,r j -1, j=1,…,k} form a basis for V. Thus V is finite-dimensional and has dim equal to deg. p.

7.1. Rational forms Definition: T in L(V,V), a vector a. T-cyclic subspace generated by a is Z(a;T)={v=g(T)a|g in F[x]}. Z(a;T)= If Z(a:T)=V, then a is said to be a cyclic vector for T. Recall T-annihilator of a is the ideal M(a:T)= =p a F[x]. p a is the T-annihilator of a.

Theorem 1. a  0. p a T-annihilator of a. –(i) deg p a = dim Z(a;T). –(ii) If deg p a =k, a, Ta,…,T k-1 a is a basis of –(iii) Let U:=T|Z(a;T):Z(a:T)->Z(a;T). Minpoly U=p a. Proof: Let g in F[x]. g=p a q+r. deg(r ) < deg(p a ). g(T)a=r(T)a. –r(T)a is a linear combination of a, Ta,…,T k-1 a. –Thus, this k vectors span Z(a;T). –They are linearly independent. Otherwise, we get another g of lower than k degree s.t. g(T)a =0. –(i),(ii) are proved.

–U:=T|Z(a;T):Z(a:T)->Z(a;T). –g in F[x]. –p a (U)g(T)a= p a (T)g(T)a (since g(T) a is in Z(a;T).) = g(T)p a (T)a = g(T)0=0. –p a (U)=0 on Z(a;T) and p a is monic. –If h is a polynomial of lower-degree than p a, then h(U)  0. (since h(U)a=h(T)a  0). –Thus, p a is the minimal polynomial of U.

Suppose T:V->V has a cyclic vector a. deg minpolyU=dimZ(a;T)=dim V=n. minpoly U=minpoly T. Thus, minpoly T = char.poly T. We obtain: T has a cyclic vector minpoly T=char.polyT. Proof: (->) done above. (<-) Later, we show for any T, there is a vector v s.t. minpolyT=annihilator v. (p.237. Corollary). So if minpolyT=charpolyT. Then dimZ(v;T)=n and v is a cyclic vector.

Study T by cyclic vector. U on W with a cyclic vector v. (W=Z(v:T) for example and U the restriction of T.) v, Uv, U 2 v,…,U k-1 v is a basis of W. U-annihiltor of v = minpoly U by Theorem 1. Let v i =U i-1 v. i=1,…,k. Let B={v 1,…,v k }. Uv i =v i+1. i=1,…,k-1. Uv k =-c 0 v 1 -c 1 v 2 -…-c k-1 v k where minpolyU=c 0 +c 1 x+…+c k-1 x k-1 +x k. (c 0 v+c 1 Uv+…+c k-1 U k-1 v+U k v=0.)

This is called the companion matrix of pa. (defined for any monic polynomial.)

Theorem 2. If U is a linear operator on a f.d.v.s.W, then U has a cyclic vector iff there is some ordered basis where U is represented by a companion matrix. Proof: (->) Done above. (<-) If we have a basis {v 1,…,v k }, –then v 1 is the cyclic vector.

Corollary. If A is the companion matrix of a monic polynomial p, then p is both the minimal and the characteristic polynomial of A. Proof: Let a=(1,0,…0). Then a is a cyclic vector and Z(a;A)=V. –The annihilator of a is p. deg p=n also. –By Theorem 1(iii), the minimal poly for T is p. –Since p divides char.polyA. And p has degree n. p=char.polyA.