
Amer R. answered 12/27/21
Mathematician who enjoys teaching
P2 is the space of polynomials of degree less that or equal to 2. Then Ker(T) are the polynomials in the form (P(x) = ax^2+bx). Now we are looking for polynomails f(x) in P2 such that
<ax^2+bx, f(x) > = 0 for any a,b
First note that
<ax^2+bx , x^2 > = 2b/5
<ax^2+bx, x > = 2a/5
<ax^2 +bx , 1 > = 2b/3
We look for f(x)=rx^2 + sx + t which will satisfies <ax^2+bx , f(x)> = 0 for any a,b
<ax^2+bx, rx^2+sx + t > = 2rb/5 + 2sa/5 + 2tb/3 = 0 for any a,b (set b=0 we get s=0, set a=0 then -3r=5t)
So r=-5 , t = 3 will work. So the base is Span( -5x^2 + 3 )