Let T: P2 ? P2 be the transformation that maps a polynomial p(1) into the polyno
ID: 3138839 • Letter: L
Question
Let T: P2 ? P2 be the transformation that maps a polynomial p(1) into the polynomial tr(t) (that is T(p(t)) tp'(t)) and let E,, B-, I, t-P) be two bases for P2. Here p'(t) stands for the first derivative with respect to i of the polynomial p(t) and P2 is the space of polynomials of degree s2. i) [8 pts] Find the matrix M such that [p(t) lE M[p(t) la for any polynomial p) in P2 i) [6 pts] Compute T(q(t)) for q()-12t-3t2. ii) [8 pts] Show that T is a linear transformation. ???) [ 12 pts] Find the matrix A such that [T(p(t) )]E-A p(t) ]E for any polynom ial p@ ?n P2. b)Explanation / Answer
b) i). M is the change of basis matrix from B to E. Since E is the standard basis of P2, hence M =
1
1
0
0
1
1
0
0
-1
It may be observed that the entries in the columns of M are the scalar multiples of 1 and the coefficients of t,t2 in 1,1+t and t-t2 respectively.
ii). T(q(t)) = T(1+2t-3t2) =t(0+2-6t) = 2t-6t2.
iii). Let p(t) = a+bt+ct2 and q(t) = d+et+ft2 be 2 arbitrary vectors/polynomials in P2 and let k be an arbitrary scalar. ThenT(p(t)+q(t)) = T(a+d+(b+e)t+(c+f)t2) = t(0+(b+e)+2(c+f)t) = (b+e)t+2(c+f)t2 = bt+2ct2+ et+2ft2 = tp’(t)+tq’(t) = T(p(t)+T(q(t). Itimplies thatT preserves vector addition. Also, T(kp(t) = T(ka+kbt+kct2) = t(0+kb+2kc) = kbt+2kct2 = k(bt+2ct2)= ktp’(t). Hence T preseves scalar multiplication.Therefore, T is linear.
iv). We have T(1) = t*0 = 0, T(t) = t*1= t, T( t2) = t*2t = 2t2, T(1+t) = t*1 = t, T(t- t2) = t(1-2t) = t-2t2. Then [T]E =
0
0
0
0
1
0
0
0
2
It may be observed that the entries in the columns of [T]E are the scalar multiples of 1 and the coefficients of t,t2 in T(1),T(t) and T(t2) respectively.
Similarly, [T]B =
0
0
0
0
1
1
0
0
-2
It may be observed that the entries in the columns of[T]B are the scalar multiples of 1 and the coefficients of t,t2 in T(1),T(1+t) and T(t-t2) respectively.
Now, let N = [[T]E, [T]B ] =
0
0
0
0
0
0
0
1
0
0
1
1
0
0
2
0
0
-2
The RREF of N is
1
0
0
0
1
1
0
1
0
0
0
-1
0
0
1
0
0
0
Hence, A =
0
1
1
0
0
-1
0
0
0
1
1
0
0
1
1
0
0
-1
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