Question

How to show T is a linear operator from
P^2 to P^2

Answer 1

depends largely on \(T\)

Answer 2

Yea forgot (T_p)(x) = p(x+1)

Answer 3

what is \(P^2\) ?

Answer 4

Space of polynomials having no more than degree 2 and coefficients in R

Answer 5

Basis of P^2 is (1,x,x^2)

Answer 6

Just not sure how to show T(f+g) = Tf +Tg for

Answer 7

ohh ok

Answer 8

polynomial of degree two looks like
\[a_2x^2+a_1x+a_0\]

Answer 9

Yep

Answer 10

take two general ones, say \[a_2x^2+a_1x+a_0\] and \[b_2x^2+b_1x+b_0\]
add them, get \[(a_2+b_2)x^2+(a_1+b_1)x+(a_0+b_0)\] then take
\[T((a_2+b_2)x^2+(a_1+b_1)x+(a_0+b_0))\]

Answer 11

once you do that, hold on to that though, then compute
\[T(a_2x^2+a_1x+a_0)+T(b_2x^2+b_1x+b_0)\] and see if you get the same thing

Answer 12

Yea cool. Having trouble getting what the right side will look like using
(T_p)(x) = p(x+1).
What you said looks right but I think you have to use the T operator I mentioned

Answer 13

This should be enough though

Answer 14

i am a bit confused as to what \(p(x+1)\) actually means

Answer 15

it is the polynomial at \(x+1\) ?

Answer 16

It is p(x+1) where p is a polynomial but it is not (p)(x+1), ((p times (x+1)

Answer 17

if P = standard 2nd degree polynomial not sure how to write it in using p(x+1)

Answer 18

if so \[T(a_2x^2+a_1x+a_0)=a_2(x+1)^2+a_1(x+1)+a_0\]

Answer 19

wow

Answer 20

I was stuck or something but that is right, thanks a bunch!! saved me