to each member ofB0and maps at least two members ofBto thesame member ofB0. Any linear transformation so defined must necessarily be onto without being one-to-one. Similarly, ifm>nthen we can map each member ofBto a unique member ofB0with at least one member ofB0not mapped to by anymember ofB. Any such transformation so defined will necessarily be one-to-one but not onto.Exercise 11:LetVbe a finite-dimensional vector space and letTbe a linear operator onV. Suppose that rank(T2)=rank(T).Prove that the range and null space ofTare disjoint, i.e., have only the zero vector in common.Solution:Let{α1, . . . , αn}be a basis forV. Then the rank ofTis the number of linearly independent vectors in the set{Tα1, . . . ,Tαn}. Suppose the rank ofTequalskand suppose WLOG that{Tα1, . . . ,Tαk}is a linearly independent set (it mightbe thatk=1, pardon the notation). Then{Tα1, . . . ,Tαk}give a basis for the range ofT. It follows that{T2α1, . . . ,T2αk}spanthe range ofT2and since the dimension of the range ofT2is also equal tok,{T2α1, . . . ,T2αk}must be a basis for the rangeofT2. Now supposevis in the range ofT. Thenv=c1Tα1+· · ·+ckTαk. Supposevis also in the null space ofT. Then0=T(v)=T(c1Tα1+· · ·+ckTαk)=c1T2α1+· · ·+ckT2αk. But{T2α1, . . . ,T2αk}is a basis, soT2α1, . . . ,T2αkare linearlyindependent, thus it must be thatc1=· · ·=ck=0, which impliesv=0. Thus we have shown that ifvis in both the range ofTand the null space ofTthenv=0, as required.Exercise 12:Letp,m, andnbe positive integers andFa field. LetVbe the space ofm×nmatrices overFandWthe spaceofp×nmatrices overF. LetBbe a fixedp×mmatrix and letTbe the linear transformation fromVintoWdefined byT(A)=BA. Prove thatTis invertible if and only ifp=mandBis an invertiblem×mmatrix.Solution:We showed in Exercise 2.3.12, page 49, that the dimension ofVismnand the dimension ofWispn. By Theorem9 page (iv) we know that an invertible linear transformation must take a basis to a basis. Thus if there’s an invertible lineartransformation betweenVandWit must be that both spaces have the same dimension. Thus ifTis inverible thenpn=mnwhich impliesp=m. The matrixBis then invertible because the assignmentB7→BXis one-to-one (Theorem 9 (ii), page81) and non-invertible matrices have non-trivial solutions toBX=0 (Theorem 13, page 23). Conversely, ifp=nandBisinvertible, then we can define the inverse transformationT-1byT-1(A)=B-1Aand it follows thatTis invertible.Section 3.3: IsomorphismExercise 1:LetVbe the set of complex numbers and letFbe the field of real numbers. With the usual operations,Vis avector space overF. Describe explicitly an isomorphism of this space ontoR2.