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Chapter 3 Exercise B


1. Solution: Assume V is 5-dimensional vector space with a basis e1, , e5. Define TL(V,V) by Te1=e1,Te2=e2,Te3=Te4=Te5=0.Then nullT=span(e3,e4,e5), hence dimnullT=3. Similarly, rangeT=span(e1,e2), hence dimrangeT=2.


2. Solution: Since rangeSnullT, it follows that TSv=0 for any vV. Hence for any uV, (ST)2u=S[(TS)Tu]=S0=0,i.e. (ST)2=0.


3. Solution:

(a) Note that rangeT=span(v1,,vm), hence if v1, , vm spans V, it follows that T is surjective.

(b) Note that z1v1++zmvm=0 if and only if (z1,,zm)=(0,,0) since v1, , vm is linearly independent. Hence nullT={(0,,0)}, then T is injective.


4. Solution: Let e1, , e5 be a basis of R5 and f1, f2, f3, f4 be a basis of R4. Define S1 and S2 by S1ei=0,S1e4=f1,S1e5=f2,for~i=1,2,3; S2ei=0,S2e3=f3S2e5=f4,for~i=1,2,4.Then it is obvious that S1,S2{TL(R5,R4):dimnullT>2}. However, (S1+S2)e1=0,(S1+S2)e2=0and (S1+S2)e3=f3,(S1+S2)e4=f1,(S1+S2)e5=f2+f4.Then you can check that dimnull(S1+S2)=2. Hence {TL(R5,R4):dimnullT>2} is not closed under addition, this implies it is not a subspace of L(R5,R4).


5. Solution: Let e1, e2, e3, e4 be a basis of R4. Define TL(R4,R4) by Te1=e3,Te2=e4,Te3=Te4=0.Then nullT=span(e3,e4), and rangeT=span(e3,e4). Hence rangeT=nullT.


6. Solution: By 3.22, we know that dimrangeT+dimnullT=dim(R5)=5.If rangeT=nullT, we will get that rangeT=nullT=2.5. This is impossible since dimension is an integer.


7. Solution: Let v1, , vn be a basis of V and w1, , wm be a basis of W, then we have 2nm. Define T1,T2L(V,W) by T1v1=0,T1vi=wi,i=2,,nand T2v1=w1,T2v2=0,T2vi=wi,i=3,,n.Then T1,T2{TL(V,W):T~is not injective}. However, we have (T1+T2)v1=w1,(T1+T2)v2=w2,(T1+T2)vi=2wi,i=3,,n.Then by Problem 3 (b), it follows that T1+T2 is injective. Hence {TL(V,W):T~is not injective} is not closed under addition, which implies it is not a subspace of L(V,W).


8. Solution: Let v1, , vn be a basis of V and w1, , wm be a basis of W, then we have nm2. Define T1,T2L(V,W) by T1v1=0,T1vi=wi,T1vj=0,i=2,,m;j=m+1,,nand T2v1=w1,T2v2=0,T2vi=wi,T2vj=0,i=3,,m;j=m+1,,n.Then T1,T2{TL(V,W):T~is not surjective}. However, we have (T1+T2)v1=w1,(T1+T2)v2=w2,(T1+T2)vi=2wi,i=3,,m;Then by Problem 3 (a), it follows that T1+T2 is surjective. Hence {TL(V,W):T~is not surjective} is not closed under addition, which implies it is not a subspace of L(V,W).


9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 5.


10. Solution: Note that v1, , vn spans V, any vV can be written as a linear combination of v1, , vn. That is there are a1, , anF such that v=a1v1++anvn.Since TL(V,W), it follows that Tv=a1Tv1++anTnvn.Hence rangeTspan(Tv1,,Tvn). On the other hand Tv1, , Tvn are contained in rangeT. By the definition of span, we conclude that Tv1, , Tvn spans rangeT.

See a similar problem in Linear Algebra Done Right Solution Manual Chapter 3 Problem 7.


11. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 6.


12. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 8.


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


14. Solution: By 3.22, we have dimnullT+dimrangeT=dim(R8)=8.Note that nullT=U and dimU=3, it follows that dimrangeT=8dimnullT=83=5=dim(R5).Therefore T is surjective by Problem 1 of Exercises 2.C.


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


16. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 11.


17. Solution: By 3.22, it follows that for any injective TL(V,W), we have dimV=dimnullT+dimrangeT=dimrangeTdimW.Hence there exists an injective linear map from V to W, then dimVdimW.

If n=dimVdimW=m, then let v1, , vn and w1, , wm be the bases of V and W, respectively. Define TL(V,W) such that Tvi=wi,i=1,,n.Here we use nm. Similar to Problem 3(b), we can show that T is injective.


18. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 12. It is similar to Problem 17.


19. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 13.


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


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


22. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 16.


23. Solution: It is obvious that rangeSTrangeS, hence dimrangeSTdimrangeS. Let u1,,um be a basis of rangeT, then by Problem 10, we have rangeST=span(Su1,,Sum).Hence dimrangeSTm=dimrangeT. Thus dimrangeSTmin{dimrangeS,dimrangeT}.


24. Solution: Suppose nullT1nullT2, we show that there exists SL(W,W) such that T2=ST1.

Since W is finite-dimensional, RangeT1 is finite-dimensional. Let T1v1,,T1vn be a basis of RangeT1, then v1,,vn is linearly independent. Let K=span(v1,,vn), then VKNullT1. Hence for any vV, we have v=i=1naivi+u, where aiF and uNullT1. Let SL(W,W) such that S(T1vi)=T2vi. Such an S exists since T1v1,,T1vn is linearly independent.

Since nullT1nullT2, we have T2v=i=1naiT2vi+T2(u)=i=1naiT2vi.Hence RangeT2 is spanned by T2v1,,T2vn.

On the other hand, we also haveS(T1v)=i=1nS(aiT1vi)+S(T1u)=i=1naiT2vi.

Hence T2v=ST1v for all vV, i.e. T2=ST1.

Conversely, if T2=ST1, it is  clear that nullT1nullT2.


25. Solution: If we assume rangeT1rangeT2. Let u1, , um be a basis of V, then we can find v1, , vmV such that T1ui=T2vi for i=1,,m since rangeT1rangeT2. Define SL(V,V) by Sui=vi, then we have T1ui=T2vi=T2Sui,i=1,,m,hence T1=T2S by uniqueness in 3.5.

If there exists SL(V,V) such that T1=T2S, then for any μV, we have T1μ=T2SμrangeT2.Hence rangeT1rangeT2.


26. Solution: Consider Problem 10 of Exercise 2C. Note that we have degDxn=n1,it follows that for any jN, there is a polynomial Dxj+1 with degree j. Note that rangeD is a subspace, hence span(Dx,Dx2,)rangeD.Moreover, by Problem 10 of Exercise 2C, we have span(Dx,Dx2,Dx3,)=span(1,x,x2,).Hence span(1,x,)rangeD, namely rangeD=P(R).


27. Solution: Denote the differentiation map by D, then 5D2+3DL(P(R),P(R)) is of the type in Problem 26. Hence 5D2+3D is surjective, thus there exists qP(R) such that (5D2+3D)q=p. Note that (5D2+3D)q=5q+3q, the proof completes.


28. Solution: For any vV, there is a uniquely linear combination Tv=a1w1++amwm,where a1,,amF are uniquely determined by v, as w1, , wm is a basis of range T. Denote ai by φi(v). Now we check that φiL(V,F), i=1,,m. For u,vV, we have Tu=φ1(u)w1++φm(u)wm and Tv=φ1(v)w1++φm(v)wm.Similarly, we also have T(u+v)=φ1(u+v)w1++φm(u+v)wm.Since TL(V,W), it follows that φ1(u+v)w1++φm(u+v)wm=T(u+v)=Tu+Tv=φ1(u)w1++φm(u)wm+φ1(v)w1++φm(v)wm=(φ1(u)+φ1(v))w1++(φm(u)+φm(v))wm. As w1, , wm is a basis of range T, it follows that φi(u+v)=φi(u)+φi(v),i=1,,m.Similarly, by consider T(λu)=λTu for λF, it follows that φi(λu)=λφi(u),i=1,,m.Hence φiL(V,F).


29. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 4.


30. Solution: If nullφ1=nullφ2=V, then φ1=φ2=0.

If nullφ1=nullφ2V, then there is a uV is not in null φ2, namely φ2(u)0. By problem 29, we haveV=nullφ1{au:aF}.Hence for any vV, v can be written as w+avu where wnullφ1. Then φ1(w)=φ2(w)=0, we have φ1(u)φ2(u)φ2(v)=φ1(u)φ2(u)φ2(w+avu)=φ1(u)φ2(u)avφ2(u)=avφ1(u)=φ1(w+avu)=φ1(v),i.e. φ1=φ1(u)φ2(u)φ2. Let c=φ1(u)φ2(u), we complete the proof.


31. Solution: Let e1, e2, e3, e4, e5 and f1, f2 be a basis of R5 and R2, respectively. Define T1L(R5,R2) by T1e1=f1,T1e2=f2,T1e3=T1e4=T1e5=0.Define T2L(R5,R2) by T2e1=f1,T2e2=2f2,T2e3=T2e4=T2e5=0.Then nullT1=nullT2=span(e3,e4,e5)and T1 is not a scalar multiple of T2. Otherwise, if T1=cT2, then f1=T1e1=cT2e1=cf1.We get c=1. However, from f2=T1e2=cT2e2=2cf2,we get c=1/2.


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