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Example 1 is clear, but this line is confusing in example 3 (I don't understand why it shows that these are not an element of the set). I have pasted example 1 for reference below.

"Modify Example 1 to consider the set of polynomials of degree ≤ 3 with f(1) = 1. Is this a vector space? Suppose we add two of the polynomials; then the value of the sum at x = 1 is 2, so it is not an element of the set. Thus requirement 1 is not satisfied so this is not a vector space. Note that a subset of the vectors of a vector space is not necessarily a subspace."

For reference, here is Example 1:

Example 1. Consider the set of polynomials of the third degree or less, namely functions of the form f(x) = a0 +a1x+a2x2 +a3x3. Is this a vector space? If so, find a basis. What is the dimension of the space?

We go over the requirements listed above:

1. The sum of two polynomials of degree ≤ 3 is a polynomial of degree ≤ 3 and so is a member of the set.

2. Addition of algebraic expressions is commutative and associative.

3. The “zero vector” is the polynomial with all coefficients ai equal to 0, and adding it to any other polynomial just gives that other polynomial. The additive inverse of a function f(x) is just −f(x), and −f(x) + f(x) = 0 as required for a vector space.

4. All the listed familiar rules are just what we do every time we work with algebraic expressions.

So we have a vector space! Now let’s try to find a basis for it. Consider the set of functions: {1, x, x2, x3}. They span the space since any polynomial of degree ≤ 3 is a linear combination of them. You can easily show (Problem 1) by computing the Wronskian [equation (8.5)] that they are linearly independent. Therefore they are a basis, and since there are 4 basis vectors, the dimension of the space is 4.

Thanks,

Chris Maness