6.6%20Direct%20sum%20decompositions - PowerPoint PPT Presentation

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6.6%20Direct%20sum%20decompositions

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6.6 Direct sum decompositions. The relation to projections. W1,...,Wk in V ... EjT= TEj. Theorem 11. T in L(V,V). T is diagonalizable. c1,...,ck distinct char. ... – PowerPoint PPT presentation

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Title: 6.6%20Direct%20sum%20decompositions


1
6.6 Direct sum decompositions
  • The relation to projections

2
  • W1,,Wk in V subspaces. W1,,Wk are linearly
    independent if a1ak0 implies ai0 for each i.
  • Lemma. TFAE
  • W1,,Wk are independent.
  • Bi basis for Wi -gt BB1,,Bk is a basis for W.
  • Proof omit
  • We write

3
  • Example
  • Example V Mnxn(F)
  • W1 all symmetric matrices AtA.
  • W2 all antisymmetric matrices At-A.
  • Then VW1?W2.
  • AA1A2, A1(AAt)/2, A2(A-At)/2.
  • Projections EV-gtV, E2E. E is called a
    projection.

4
  • Example E (x,y,z)-gt(x,y,0).
  • Properties Rrange of E. Nnull E.
  • b in R lt-gt Ebb.
  • (-gt) bEa, EbE(Ea)Eab.
  • (lt-) bEb. Done
  • I-E is a projection also
  • (I-E)(I-E)I-2EE2 1-2EE1-E.
  • Im(I-E) null E
  • Let a(I-E)b. ab-Eb. EaEb-E2bEb-Eb0. a in
    null E.
  • a in null E. Ea0. (I-E)aa-Eaa. a in Im(I-E).

5
  • a in V. aEa(a-Ea)Ea(I-E)a.
  • VIm E ? Im(I-E).
  • Ea(I-E)b. EaE2aEb-E2bEb-Eb0. (I-E)b0 also.
  • VR ?N.
  • Im ER. Im(I-E)N
  • Projection is diagonalizable
  • Let a1,,ar be a basis of R.
  • ar1,, an a basis for N.
  • B a1,,an a basis for V which diagonalizes E.

6
  • Cases where there are a number of projections
    (commuting)
  • Theorem 9. If V W1? ?Wk, then there exists k
    linear operators E1,,Ek on V s.t.
  • (i) Each Ei is a projection. Ei2Ei.
  • (ii) EiEj0 if i?j. (commuting)
  • (iii) IE1Ek.
  • (iv) Range Ei Wi.
  • Conversely, if E1,..,Ek are k linear operators
    satisfying (i)-(iii), then for Wirange Ei, V
    W1? ?Wk.

7
  • Proof
  • (-gt) V W1? ?Wk.
  • aa1ak, ai in Wi. Uniquely written.
  • Define Eja aj.
  • Then Ej2Ej.
  • N(Ej) W1??Wj-1?Wj1? ?Wk.
  • a E1aEka. IE1Ek.
  • Ei Ej0 if i?j. (Wj is in N(Ei)).
  • (lt-) Let E1,,Ek linear operators.
  • a E1aEka by (iii).
  • V W1 Wk.
  • The expression is unique.

8
  • a a1ak. ai in Wi Im Ei. ai Eibi.
  • Eja Ej(a1ak) Ej aj Ej Ejbj Ejbj aj .
  • By (b) of the Lemma, W1,,Wk are independent.
  • V W1? ?Wk
  • Note A finite sum of any collection of distinct
    Ei is a projection.
  • Check

9
6.7. Invariant direct sums
  • V W1? ?Wk, Wi T-invariant.
  • BB1,,Bk
  • Block form

10
  • Theorem 10 TV-gtV. V W1? ?Wk respective
    E1,,Ek.
  • Each Wi is T-invariant lt-gt TEiEiT, i1,,k.
  • Proof(lt-) Let a in Wj. aEja.
  • TaT(Eja) EjT(a). Ta in Wj.
  • Wj is T-invariant.
  • (-gt) aE1aEka.
  • TaT E1aT Eka.
  • Since T(Eja) in Wj , T(Eja)Ejbj for some bj.

11
  • EjTaEjTE1aEjTEka EjT Eja TEja.Thus,
    EjTTEj
  • Theorem 11. T in L(V,V). T is diagonalizable.
    c1,,ck distinct char. Value of T. Then there
    exists projections E1,,Ek s.t.
  • (i) Tc1E1ckEk
  • (ii) I E1Ek (iii) EiEj0 i?j. (iv) Ei2Ei.
  • (v) Range Eichar.v.s. of T ass. ci.
  • Conversely, given distinct c1,,ck, E1,,Ek with
    (i)-(iii). Then T is diagonalizable with char.
    values c1,,ck and (iv)(v) also hold.

12
  • Proof (-gt) T diagonalizable. c1,,ck dist. Char.
    Values.
  • V W1? ?Wk , Wi associated with ci.
  • aE1aEka.
  • TaT E1aT Ekac1 E1ackEka.
  • T c1 E1ckEk.
  • (lt-) We need to prove (iv)(v) and T is
    diagonalizable.
  • (iv) EiI Ei(E1Ek)Ei2.
  • T c1 E1ckEk by(i).
  • T Ei ciEi by (iii).
  • Since Ei is not zero, ci is a char.value.
  • T-cI (c1-c)E1(ck-c)Ek .

13
  • If c?ci for all I, then T-cI has the null-space
    0.
  • T has char.values c1,,ck.
  • T is diagonalizable since char.v.s. span V.
  • (v) null(T-ciI)Im Ei.
  • (?) (T-ciI)Eia ((c1-ci)E1(c i-1-ci)E i-1(c
    i1-c i)E i1(ck-ci)Ek ) Eia 0 by (iii).
  • (?) If Tacia, then (c1-ci)E1a(c i-1-ci)E
    i-1a(c i1-c i)E i1a(ck-ci)Eka 0.
  • (cj-ci)Eja0 for all j ?i.
  • Eja0 for all j ?i.
  • a in Im Ei.

14
  • Remark diagonalizable operator T is uniquely
    determined by c1,,ck and E1,,Ek.
  • T c1 E1ckEk.
  • g(T) g(c1) E1g(ck)Ek
  • Proof omitted.
  • Let
  • Thus, Ej is a polynomial of T and hence commutes
    with T.
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