Question

QUESTION 4 Let T R3-P2 be defined by T(a, b, c) - (a + b + e) +(a+b)a2 (4.1) Show that T is a linear transformation (4.2) Fınd the matrix representation [T]s, B, of T relative to the basıs in R3 and the basis in P2, ordered from left to right Determine the range R(T of T Is T onto? In other words, is it true that R(T)P2 Let x, y E R3 Show that x-y ker(T) f and only if x-y Is T one-to-one? Motivate your answer (4.3) (4 4) (4.5) [22]

Answer 1

(4.1). Let X= (a,b,c) and Y = (p,q,r) be 2 arbitrary vectors in R³ and let k be an arbitrary real scalar. Then X+Y = (a+p,b+q,c+r) and T(X+Y) = (a+p+b+q+c+r)+ (a+p+b+q)x +(a+p)x² = (a+b+c)+ (a+b)x +ax²+(p+q+r)+ (p+q)x +px² = T(X)+T(Y). Hence T preserves vector addition.

Also, T(kX) =(ka+kb+kc)+ (ka+kb)x +kax² =k[(a+b+c)+ (a+b)x +ax²] = kT(X). Hence T preserves scalar multiplication.

Therefore, T is a linear transformation.

(4.2). We have T(1,1,0) = 2+2x+x² = 2(1+x)+0(1-x)+ 1(x²), T(1,-1,0) = x² = 0(1+x)+0(1-x)+ 1(x²), and T(0,1,1) = 2+x= (3/2)(1+x)+(1/2)(1-x)+ 0(x²).

Therefore, the matrix of T relative to the bases B₁ ,B₂ is [T]_B2,B1 = A (say)=

2	0	3/2
0	0	1/2
1	1	0

It may be observed that the entries in the columns of A are the coefficients of (1=x),(1-x) and x² in T(1,1,0), T(1,-1,0) and T(0,1,1) respectively.

(4.3). The RREF of A is

1	0	0
0	1	0
0	0	1

The range of T, being equal to col(A) is whole of P₂. Further, since the columns of A span P_2, hence T is onto.

(4.4). We have X-Y = (a-p,b-q,c-r) and T(X-Y) = (a-p+ b-q+c-r) +( a-p+ b-q)x+ (a-p)x² . Now if X-Y is in ker(T), then T(X-Y) = 0 or, (a-p+ b-q+c-r) +( a-p+ b-q)x+ (a-p)x² = 0 so that a-p+ b-q+c-r = 0…(1), a-p+ b-q = 0 …(2) and a-p = 0…(3).

As per the 3^rd equation, a = p . On substituting a = p in the 2^nd equation, we get b-q = 0 so that b = q. On substituting a-p+ b-q = 0 in the 1^st equation, we get c-r = 0 or, c = r.

Hence X= Y.

Further, if X= Y, then a = p, b =q and c = r so that T(X) = T(Y) and, therefore, T(X_Y) = T(X)-T(Y) = 0. Hence X-Y is in ker(T).

(4.5). The columns of A are linearly independent . Hence T is one-to-one.