Let T P2 rightarrow P4 be the transformation that maps a pol

Let T: P_2 rightarrow P_4 be the transformation that maps a polynomial p(t) into the polynomial p(t) - t^2p(t). a. Find the image of p(t) = 6-t+t^2. b. Show that T is a linear transformation. c. Find the matrix for T relative to the bases {1, t, t^2) and {1, t, t^2, t^3, t^4}. The image of p(t) = 6-t + t^2 is Enter your answer in the answer box and then click Check Answer.

Solution

(a)The image of p(t) = 6-t+t² is p(t) –t²p(t) = 6-t+t²-t²(6-t+t²)= 6-t+t²-6t²+t³-t⁴ = 6-t-5t²+t³-t⁴.

(b) Let p(t) = a +bt+ct² and q(t)= d +et+ft² be 2 arbitrary elements (vectors) in P₂, where a,b,c,d,e, f are arbitrary real numbers , and let k be an arbitrary scalar. Then, we have T(p(t)+q(t)) = T(a +bt+ct²+ d+et+ft²) =T((a+d)+(b+e)t+(c+f)t²)=[(a+d)+(b+e)t+(c+f)t²]-t²[(a+d)+(b+e)t+(c+f)t²] = (a+d)+t(b+e) + t²(c+f)   -t²(a+d )- t³(b+e)– t⁴(c+f) =(a+d)+t(b+e)+t²(-a+c-d+f) -t³(b+e) –t⁴ (c+f). Also, T(p(t)) +T(q(t)) =(a+bt+ct²)        -t²(a+bt+ct² )+(d+et+ft²)- t (d+et+ft²)= a+bt+t²(-a+c)-bt³ –ct⁴+d+et+t²(-d+f)-et³ –ft⁴ = (a+d)+t(b+e)+ t²(-a+c-d+f)-t³(b+e) –t⁴(c+f) = Thus, T(p(t)+q(t))= T(p(t)+T(q(t) so that T preserves vector addition. Further, T(kp(t))=T(k(a +bt+ct²)=T(ka +kbt+kct²)= ( ka +kbt+kct²) –t²(ka +kbt+kct²)=ka+kbt+t²(-ka+kc) -kbt³ –kct⁴ = k[a+bt+t²(-a+c) -bt³ –ct⁴] = kT(p(t)). Thus, T also preserves scalar multiplication. Hence T is a linear transformation.

(c ) T(1)=1-t²=-t²+1,T(t)=t-t³= -t³+t,T(t²)=t²-t⁴ =-t⁴ +t. Then the matrix of T relative to the basis {1,t,t²} is

0

0

-1

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-1

0

-1

0

0

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1

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Since T: P₂P₄ and since t³ and t⁴ P₂, hence T(t³) and T(t⁴) are not defined/do not exist. Hence the matrix of T relative to the basis {1,t,t²,t³,t⁴} does not exist.

0	0	-1
0	-1	0
-1	0	0
0	1	1
1	0	0