Question

Determine whether or not the following transformation T :V + W is a linear transformation. If T is not a linear transformation, provide a counter example. If it is, then: (i) find the nullspace N(T) and nullity of T, (ii) find the range R(T) and rank of T, (iii) determine if T is one-to-one, (iv) determine if T is onto. : (a) T: R3 + R2 defined by T(x, y, z) = (2x, y, z) (b) T: R2 + R2 defined by T(x,y) = (x+1, y) (c) T: R2 + R2 defined by T(x,y) = (x2, x - y) : : -1 2 (d) T: M2x2(R) → M2x2(R) defined by T(A) = A 3 (e) T: P2(R) + R3 defined by T(p(x)) = (P(0), p(1), p(2)). (f) T: Mnxn (R) → R defined by T(A) = det(A). (g) T: P2(R) + P3(R) defined by T(f(x)) = xf(x) + f'(x) :

Answer 1

. If Given that T: VW is transformation (a) T: R3-Re defined by T(3,4,) = (2x, 4-Z). Note that, v and are rector spaces over the same field K; then function f:VW is said to be linear transfo@mation if VUKEV and Vcek, we have: flult se) = f(u) + flok) ; ☺ f(cu) = cflu). So, let all = (x3,99, 2.), k= (x2,82, 22) ER3 and DER te arbitrary. Then, T(4+1) T(*q + 22, & q +% , Zqt Zz) (26X4+x2), 44+ 42 - 24-72) (284 +222, Y - Zq + G-Z2) = (2x4,5,- Za) + (2x2, 42 – Z2) = T(X2, 72, 74) + T (22,42, 22) = T(4) + T(X) = = and Tla (x,y,z)) Tlax, 04,dZ) = (20%, dy-dZ) → Tla (8,4,2)) = d. (2%, -Z) = d.T(X,Y, Z). Hence T is a linear transformation, Thus: © Finding nullspace N(T) and nullity n (T):- :: T(2,4,7) = (0,0) = (0,0) = (2x, 4-Z) = (0,0) → 2X=0 and y-Z=0. > x=, y=Z. :: N(T) = 4(,Y,Z): T(318,2)=0,0)} = {(2,812): x=0,7=2} >> N(T) = {10, a, a); a=&= Z and QEIR } This is the nellspace of T. : Basis [n(+2] = {(0,1,2}, ; nullity (T) = 2(1) = 1. finding Range (t) and rank (T) :-

"Range (7) & T(%) : @14,3) E1R3} > Range (T) = {2x,4-7): Qeny, )ER}}. This is the range space. Finding basis of R(A) : first Fund the image of the standard basis of R3, viz. {(1,0,0), (0,1,0),(0,0,1)}"; thes T(2,0,0) = (2,0) ; T10,2,0) = (0,1); T(0,0,1)= (0,-1). "We know that, Result: Suppose Mg, 42, ---, kom spans va vector space V, and suppose TiVw is elinear transformation, then T(12), T(12), T(m) span Range (T). .. By this result, yectors (2,0), (0,2), (0,-1) will span note that (0, 2) = -100,-1),ue. (0,1) and (0 ,-1) are linearly dependent, so both camot stay in basis Simultaneously. i Range (T) = span {(2,0), (0,4)} and & (2.0), (011)} is a linearly independent set, Basis [Range (r)] {(2,0), (0,1) Rank (T) = Cardinality of Basis (range (+33. + Rank (T2 = 2. Range (T). And B then die Note that, If T(88,Z) = (0,0) = (x14,7) = (0,0) then T is one - one. That is it F (0,0,0)-is the only vector which is mapped to (0,0) say But . T(0, 1, 1) = (0,0) and (0,1,1) + (0,0,0). 7 T is not one-one. T is one cane. Note that Hence T is onto. Rank (t) = 2 => Range (t) = R?

T: R² R² by T(274) (2+1,8). : T (22,83) + (22,42)) = f(xq +22, +42) (22++1, 4+42) and T(24,42) + T (82742) = (84+1,71)+(2+1,42) = (Xg+x2 + 2, 4y + y2 42), .: T (623, %] +(22,42]) # t (22,82 ) + +(22,42). :: Tus not a linear transformation, Counter example: T(1,1) (2, 2) & T(0,1) = (1,1). → T (2,1) + T10,1) = (3,2) and T((1,1)+(2,1)) = T(3,2) = (4,2). and (3,2) # (4,2). 6) T:m2 R2 by T(x,y) (8%, x-7). T is not a linear transformation, because Tla (8,4))= T (ax,dy) (a??, d8-04) d. 7(8.4) = a (x?, 2-4). (ax², dx-dy) :: T (a ()q.T(3,4)., where DER, and Ex:- : T(2,2) = (1, -1) and T ( 2(1,2)) = T(2,4)= (4,-2) die. T(2(1,2)) = (4,-2) = 2 (2,-1) + 2.T (1,2). d) T: Max2 (R) → M2x2 (R) Mexa CIR) Jay TA = A by T(A) = A (15 %). Then let A,BE Max2 (AR) and dER. Then T(A+B) = 3) 1) +B11-4) (due to the right distributivity in matrices). (A+B) (34 A ( 23 24

T(+T(B) >>> T(A+B) = and T(A) = (01X*3%) = a.((1 2.))= d. T(A). IT is a linear transformation, Now it ® Nullspace "T(A) = 02x2 A Let A= at [32] =.:] [d][3-2)=[:] ]=[:] JE Maxo => C-att 2a-46 20-40 -C + 3d -Q + 36 = 0 ; 29-46 = 0; -C+ 30 = 0,2c-4d=0 a=0, p=0; C =0, 0 = 0. a :: N(T) = & A: T(A)=0} = {1: :]=0.x2} > N(T) = { 02x2] frere {here Oexe is the 242, 2X2} zero Matris. = din [N(r)] = 0 * Nullity 0. A Range space RCT) :- PR(T) = {TA): AE Max2 (P)} - RCT) = { 0 (0%): A E Mox2 ()} = R(T) : antic, d ER} This is the range space of T. { (0*3* 24-4* C+30 20-40

using Rank - Nullity Theorem, ue chave, Rank (T) + Nullity (T) = dim (M2x2 (R)) * Rank (T) to = 4 de Nullity (M)=0} >> Rank (I) =4. Since ene - one. T(A)= 02x2 → A=02X2 ď proved above a's ®} * T is • Rank (T) = 4, wil duino [Range (T)] = 4 = din Codomasis Space T . is onto. Hence