# Thread: Dimension and basis of the Vector Space

1. ## Dimension and basis of the Vector Space

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

2. 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.

3. Why do you think that is not correct?
If $P(x)=ax+bx^2$ then $P(0)=0$. Correct?
What about $P(x)=c+ax+bx^2,\;c\not=0$ can $P(0)=0$?