Advanced Calculus fi..

Hence all vectors x such that T(x) = (1 6,24, 8) are given byChapter I Vectors and Matrices 59where t is arbitrary.EXAMPLE 4T maps v3 into V7 andTo find the kernel, we must solve the equations .rl - 12 = 0, .r2- x3 = 0. Thesolutions are given by all (xl..r2, x3) such that .rl = .TI = .r3. hence by all vectorst(1, 1, 1 ), where t is an arbitrary scalar. Here T is not one-to-one. However, the rangeof T is all of v2, for the equationscan always be solved for . XI, s2. xi-for example, by taking .v3 = 0. xl_ = y2, and.XI = vl + vl. Therefore T maps V7 onto V' but is not one-to-one.EXAMPLE 5 T maps Vn into Vn' and T(x) = 0 forevery x. Thus T has the matrix0 = Om,. We call T the zero inupping and sometimes also denote this mapping by0. T is clearly linear and is neither onc-to-onc nor onto.EXAMPLE 6 T maps V" into V" and T(x) = x for every x. Thus T has the matrixI = I,,. We call T the identipmupping and sometimes also denote this mapping by I.T is clearly linear and is one-to-one and onto.EXAMPLE 7Let T map v3 into ~%nd have the matrixFind the range of TSolution. The range of T consists of all y = .rl ul + .r2u2 + ~ 3 ~ where 3 . ul , u:, uIare the column vectors of A:Thus the range consists of all linear combinations of ul, u2, u,. If ul, u,, u3 arelinearly independent, then they form a basis for v3. and the range is v3. However,we verify that 2ul - u2 - u3 = 0, so that these vectors are linearly dependent.Furthermore, we see that u,, u: are lincarly independent and that u3 is expressible asa linear combination of u, . u:. Hence the range is given by all linear combinationsof ul and u:, and this set is not all of v'. Thus T does not map V' onto v', andthe equation T(x) = yo has no solution for x, for some choices of yo; for example,T(x) = ( 1,O. 0) has no solution, as one can verify (the vector (l,O, 0) is not a linearcombination of u I and u?).