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Chapter 3 Exercise D


1. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22. It is almost the same.


2. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 25.


3. Solution: If there exists an invertible operator TL(V) such that Tu=Su for every uU, then S is injective since T is injective.

If S is injective. Assume u1, , um is a basis of U, we can extend it to a basis of V as u1, , um, vm+1, , vn. Since S is injective, by Problem 9 of Exercises 3B, we have Su1, , Sum is linearly independent in V. Hence we can extend it to a basis of V as Su1, , Sum, wm+1, , wn. Define TL(V) as below Tui=Sui,Tvj=wj,1im,m+1jn.The existence of T is guaranteed by 3.5(unique). Then for any u=a1u1++amum, aiF, we have Tu=T(a1u1++amum)=a1Tu1++amTum=a1Su1++amSum=S(a1u1++amum)=Su.Moreover, T is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

Compare this problem with Problem 11 of Exercises 3A.


4. Solution: If we assume nullT1=nullT2. Since W is finite-dimensional, so is range T2. Let w1, , wn be a basis of rangeT2, then there exist v1, , vnV such that T2vi=wi,i=1,,n.Now we will show that V=nullT2span(v1,,vn). For any vV, we T2v=a1w1++anwnfor some a1, , anF. Hence T2(va1v1anvn)=0,namely v=(va1v1anvn)+(a1v1++anvn),this implies V=nullT2+span(v1,,vn). Moreover, if a1v1++anvnnullT2, then we have T2(a1v1++anvn)=a1w1++anwn=0.Note that w1, , wn is linearly independent, it follows a1==an=0. Thus we haveV=nullT2span(v1,,vn). Similarly, T1v1, , T1vn is linearly independent. For if a1T1v1++anT1vn=0, we have T1(a1v1++anvn)=0.Note that nullT1=nullT2, it follows that 0=T2(a1v1++anvn)=a1w1++anwn.Thus a1==an=0. Now extend w1, , wn to a basis of W as w1, , wn, e1, , em and T1v1, , T1vn to a basis of W as T1v1, , T1vn, f1, , fm. Define SL(W) by Swi=T1vi,Sej=fj,i=1,,n;j=1,,m.Note that V=nullT2span(v1,,vn),any vV can be expressed as v=vnull+a1v1++anvn,where vnullnullT2=nullT1 and a1, , anF. Hence we have ST2(v)=ST2(vnull+a1v1++anvn)=ST2(a1v1++anvn)=S(a1w1++anwn)=a1T1v1++anT1vn=T1(a1v1++anvn)=T1(vnull+a1v1++anvn)=T1(v)namely ST2=T1. Moreover, S is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

If there exists an invertible operator SL(W) such that ST2=T1, then for any μnullT1, we have ST2μ=T1μ=0.As S is invertible, we have T2μ=0. Hence μnullT2, it follows that nullT1nullT2. Similarly, consider T2=S1T1, nullT2nullT1. Thus we conclude null T1= null T2.

Compare this problem with Problem 24 of Exercises 3B.


5. Solution: If we assume rangeT1=rangeT2. Let u1, , um be a basis of null T1, then we can extend it to a basis of V as u1, , um, w1, , wn. Then range T1 is span(T1w1,,T1wn) and T1w1,,T1wn is linearly independent. There exist v1, , vnV such that T1wi=T2vi for i=1,,n since rangeT1=rangeT2. As T1w1,,T1wn is linearly independent, it follows that v1, , vn is linearly independent by Problem 4 of Exercises 3A. Note that range T1= range T2 implies null T1 and null T2 have the same dimension. Let ζ1, , ζm be a basis of null T2, then ζ1, , ζm, v1, , vn is a basis of V by the proof of 3.22. Define SL(V) by Sui=ζi and Swj=vj, then we have T1wj=T2vj=T2Swj,j=1,,n,and T1ui=0=T2ζi=T2Sui,i=1,,m,hence T1=T2S by uniqueness in 3.5. Moreover, S is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

If there exists an invertible operator SL(V) such that T1=T2S, then for any μV, we have T1μ=T2SμrangeT2.Hence rangeT1rangeT2. As S is invertible, we have T2=T1S1. Similarly, we conclude rangeT2rangeT1. Thus range T1= range T2.

Compare this problem with Problem 25 of Exercises 3B.


6. Solution: If there exist invertible operators RL(V) and SL(W) such that T1=ST2R, then S1T1=T2R. Hence null T1= null T2R by Problem 4. Note that we have range T2R= range T2 by Problem 5, it follows that dimnullT2R=dimVdimrangeT2R=dimVdimrangeT2=dimnullT2.Hence dimnullT1=dimnullT2.

Conversely, if dim null T1=dim null T2. Let u1, , um be a basis of null T1, then we can extend it to a basis of V as u1, , um, w1, , wn. Let v1, , vm be a basis of null T2, then we can extend it to a basis of V as v1, , vm, ζ1, , ζn. Then T1w1, , T1wn is linearly independent in W, hence we can extend it to a basis of V as T1w1, , T1wn, α1, , αl. Similarly, T2ζ1, , T2ζn is linearly independent in W, hence we can extend it to a basis of V as T2ζ1, , T2ζn, β1, , βl. Define RL(V) by Rui=vi,Rwj=ζj,i=1,,m;j=1,,n.Define SL(W) by ST2ζj=T1wj,Sβk=αk,j=1,,n;k=1,,l.Since S and T map basis to basis, hence S and T are invertible(surjective). Moreover, it is easy to check T1ui=0=ST2Rui and T1wj=ST2ζj=ST2Rwj,hence T1=ST2R.

Compare this problem with Problem 4 and Problem 5.


7. Solution: a) Let T,SE, then (T+S)v=Tv+Sv=0+0=0,hence T+SE, namely E is closed under addition. For every λF, (λT)v=λ(Tv)=λ0=0,hence λTE, namely E is closed under scalar multiplication. Therefore E is a subspace of L(V,W).

(b) Since v0, we can extend it to a basis of V, namely v, v2, , vn. Let w1, , wm be a basis of W. Under these bases, we have a isometric between L(V,W) and Fm,n by 3.60. Moreover, Tv=0 if and only if the first column vector of M(T) is zero. Hence dimE is exactly all matrices in Fm,n such the first column vector is zero. Now , it is easily seen thatdimE=m(n1)=dimW(dimV1).Or it is equivalent to the dimension all linear map from span(v2,,vn) to W. Then we can use 3.61.


8. Solution: Let w1, , wn be a basis of W. Since T is surjective, there exist v1, , vn such that Tvi=wi. Moreover, by Problem 4 of Exercises 3A, it follows that v1, , vn is linearly independent. Consider U=span(v1,,vn), then T|U maps a basis of U to a basis of W. Hence T|U is an isomorphism of U onto W.


9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22.


10. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 23.


11. Solution: By Problem 9, we have TU is invertible. Again by Problem 9, we have T is invertible. Multiply both side of STU=I by S1 on left, we get TU=S1. Multiply both side of TU=S1 by S on right, we have TUS=I. Multiply both side of TUS=I by T1 on left, we get US=T1.


12. Solution: Consider V=C. Define T, S, U by T(z1,z2,z3,)=(0,z1,z2,z3,)S(z1,z2,z3,)=(z2,z3,z4,)and U=I. Then STU=I. However T is not surjective.


13. Solution: Since V is finite-dimensional, the surjectivity of RST implies RST is invertible. Hence S is invertible by Problem 11.


14. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 20.


15. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 21.


16. Suppose V is finite-dimensional and TL(V). Prove that T is a scalar multiple of the identity if and only if ST=TS for every SL(V).

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 24.


17. Solution: Let e1, , en be a basis of V. Suppose E{0}, then there is nonzero TE. This implies there is some s{1,,n} such that Tes0. Let Tes=a1e1++anen,ajF,j=1,,n.Since Tes0, there is some t{1,,n} such that at0. Define EijL(V) by Eijek=δikej, where δik=0 if ik and δii=1 for every i.

By definition, for every i{1,,n}, we have EtiTEisej=atδijei,j=1,,n.By assumption, EtiTEisE. Note that E is a subspace of L(V). It follows that (Et1TE1s++EtnTEns)ej=atej,j=1,,n,and (Et1TE1s++EtnTEns)E. This implies ajIE, hence IE. Therefore for any SL(V), S=SIE, namely E=L(V).


18. Solution: For given vV, define φv:FV by φv(λ)=λv. Then φvL(F,V) (check it). Hence we can define φ:VL(F,V) by vφv. It suffices to show φ is an isomorphic from V to L(F,V). First, we should check φ is linear. For every v1,v2V and λ,aF, we have φv1+λv2(a)=a(v1+λv2)=av1+λav2=φv1(a)+(λφv2)(a)=(φv1+λφv2)(a).Hence φ is linear. Then φv0 imples av=0 for every aF, thus v=0. We conclude injectivity. For any fL(F,V), if f(1)=v then f(λ)=λv=φv(λ). Hence every fL(F,V) can be express as φf(1), namely φ is surjective.


19. Solution:

(a) Since degTpdegp for every nonzero polynomial pP(R), we have T|Pn(R):Pn(R)Pn(R) for any nN+. Note that T is injective, hence T|Pn(R):Pn(R)Pn(R) is injective. Note that Pn(R) is finite-dimensional, we conclude T|Pn(R):Pn(R)Pn(R) is surjective by 3.56. Hence T is surjective as any polynomial must be contained in some Pn(R).

(b) We argue it by induction on degp. It is true for deg=0. Suppose it is true for n. As T|Pn(R):Pn(R)Pn(R) is surjective, every polynomial p with degree n can be attained by Tq for some qPn(R). Moreover, T|Pn+1(R):Pn+1(R)Pn+1(R) is surjective, every polynomial p with degree n+1 can be attained by Tq for some qPn+1(R). If there exist pPn+1(R) with degree n+1 such that degTp<n+1. Then there exist some qPn(R) such that Tq=Tp since T|Pn(R):Pn(R)Pn(R) is surjective. Hence Tp=Tq. Note that T is injective and pq (different degrees), we get a contradiction. Hence degTp=degp for every nonzero pPn+1(R) with degree n+1. The proof is complete.


20. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 26.


Linearity

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