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Chapter 3 Exercise C


1. Solution: Suppose for some basis v1, , vn of V and some basis w1, , wm of W, the matrix of T has at most dimrangeT1 nonzero entries. Then there are at most dimrangeT1 nonzero vectors in Tv1, , Tvn. Note that rangeT=span(Tv1,,Tvn), it follows that dimrangeTdimrangeT1.We get a contradiction, hence completing the proof.


2. Solution: The basis of P(R3) is x3, x2, x, 1. The corresponding basis of P(R2) is 3x2, 2x, 1 (with the indicated order).(Check it!)


3. Solution: We use the notation and the proof of 3.22. Extend Tv1, , Tvn to a basis of W as Tv1, , Tvn, μ1, , μs. Then with respect to the basis v1, , vn, u1, , um of V and the basis Tv1, , Tvn, μ1, , μs of W, all entries of M(T) are 0 except that the entries in row j , column j , equal 1 for 1jdimrangeT.


4. Solution: If Tv1=0, then any basis w1, , wn of W will satisfy the desired conditions. If Tv10, then any basis w1, , wn of W such that w1=Tv1 will satisfy the desired conditions.


5. Solution: Let ν1, , νm be a basis of V, denote the first row of M(T) with respect to the bases ν1, , νm and w1, , wn by (a1,,am). If (a1,,am)=0, then we can choose vi=νi, i=1,,m. If (a1,,am)0, suppose ai0. Then let v1=νiai,vj=νj1aj1v1,vk=νkakv1for j=2,,i, k=i+1,,m. Then you can check that v1, , vm satisfies the desired conditions.


6. Solution: Suppose there exist a basis v1, , vm of V and a basis w1, , wn of W such that with respect to these bases, all entries of M(T) equal 1. Then Tvi=w1++wn,i=1,,m.Hence rangeT=span(w1++wn), it follows that dimrangeT=1.

Conversely, if dimrangeT=1, then dimnullT=dimV1. Let ν1, ν2, , νm be a basis of V such that ν2, , νmnullT. Note that Tν10, hence we can extend it to a basis of W as Tν1, w2, , wn. Let w1=Tν1w2wn and v1=ν1, vi=νi+ν1 for i=2,,m. T(v1)=T(vi)=w1+w2++wn,i=2,,m.It is obvious that v1, , vm is a basis of V and w1, , wn is a basis of W. Now we can directly check that all entries of M(T) with respect to these bases equal 1.


7. Solution: Given a basis v1, , vm of V and a basis w1, , wn of W, denote v and M(S) with respect to these bases by A and B, respectively. Then we have Tvj=k=1nAk,jwkand Svj=k=1nBk,jwk.Hence (T+S)vj=Tvj+Svj=k=1n(Ak,j+Bk,j)wk,it follows that the entries in row k , column j of M(T+S) with respect to these bases are Ak,j+Bk,j. By 3.35, we deduce that M(T+S)=M(T)+M(S).


8. Verify 3.38.

Solution: It is almost the same as the previous Exercise.


9. Prove 3.52.

Solution: It is almost the same as Problem 11. Just consider the entries.


10. Suppose A is an m-by-n matrix and C is an n-by-p matrix. Prove that (AC)j,=Aj,C for 1jm. In other words, show that row j of AC equals (row j of A) times C.

Solution: It is almost the same as Problem 11. Just consider the entries.

These exercises are tedious. I prefer solving other interesting exercises… If you have problems regarding to them, please make a comment.


11. Solution: By 3.41, we have (aC)1,k=i=1naiCi,k.It is obvious that (aiCi,)1,k=aiCi,k. Hence (aC)1,k=(a1C1,)1,k++(anCn,)1,k=(a1C1,++anCn,)1,kby 3,35. Thus we deduce that aC=a1C1,++anCn,.


12. Solution: Let A=(0110) and C=(1002), then we have AC=(0210)andCA=(0120).Hence ACCA.


13. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 17.


14. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 18.


15. Solution: Note that AAA=(AA)A, denote AA=B, then by definition (3.41) (1)Bj,r=p=1nAj,pAp,r.Similarly, the entry in row j , column k, of A3 is r=1nBj,rAr,k.Hence by (1), we have r=1nBj,rAr,k=r=1np=1nAj,pAp,rAr,k=p=1nr=1nAj,pAp,rAr,k.


Linearity

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