Find the dimension of the vector space V and give a basis for V
V = {p(x) in P2 : p(0) = 0}
*P2 denotes polynomials of degree less than or equal to 2
Sol:
standard basis of P2={1, x, x^2}
let p(x)=a+bx+cx^2
then p(0)=a+0+0 ====> a=0,
do we through away 1 here?
hence
I got Vdim=2 and basis {x, x^2}, but have a feeling that its not correct
Let v \in V. Then v is a polynomial of degree less than or equal to 2 with v(0)=0. Expressed this way v = a+bx+cx^2 where a,b,c are in the field over which V is a vector space. As you said this implies a = 0, so v = bx+cx^2. Therefore v (an arbitrary element of V) can be written as a linear combination of elements in {x,x^2} therefore {x,x^2} is a spanning set. Show this set is also linearly independent (by assuming ax + bx^2 = 0 and concluding that a = b = 0) and you've got yourself a basis.
Or show that {x, x^2} is the minimal spanning set; that is removing any element destroys the spanning property.
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