Consider the space P2 of polynomials of degree less than or

Consider the space P2 of polynomials of degree less than or equal to 2, defined on the interval ?1 ? x ? 1. Define the inner product:

Prove that this definition satisfies the axioms of a vector space and that {1, x, x2} is a basis for P2.

Solution

The vector space Pn is the vector space of polynomials of degree less than or equal to n.

In particular, P2 is the vector space of polynomials of degree less than or equal to 2.

If p and q are vectors in P2 define

          <p, q>=p(1)q(1) +p(0)q(0) +p(1)q(1)

This rule will define an inner product on P2.

It can be verified that all four conditionsof an inner product are satisfied but we will only look at conditions 1 and 4, leaving the others as an exercise.

For condition 1:

           <p, q>=p(1)q(1) +(0)q(0) +p(1)q(1)=q(1)p(1) +q(0)p(0) +q(1)(1)

              =<q, p>

For condition 4:

           <p, p>= [p(1)]2+ [p(0)]2+ [p(1)]2

It is clear that this expression, being the sum of three squares, is always greater than orequal to 0,

so we have

<p, p> 0.

Next we want to show that <p, p>= 0

if and only if p= 0.

It’s easy to see that if p= 0 then <p, p>= 0.

On the other hand, suppose<p, p>= 0,

then we must have p(1) = 0, p(0) = 0, and p(1) = 0.

This means p has 3 roots, but since p has degree less than or equal to 2 the only way this is possible is if p= 0, that is, p is the zero polynomial.

and also P2 is at most degree is 2 so that

in our the axioms of a vector space and that {1, x, x2} is a basis for P2 is also have at most degree is 2.

   hence it was proved.