#### Question

# How to identify a basis for P2?

Which of the following is a basis for P2 and why?

a. {x^2, x+1, x^2+x+1}

b. {x, 1, 0}

c. {x^2+x+1, x^2+x, x+1}

d. {x^2, x, 1, 0}

I know a basis exists when two conditions are met: linear independence and span, but I am unsure how to apply these ideas to the problem.

Thanks for looking.

#### Answers

The answer is choice c because it represents both a spanning set and a linearly independent set. For a deeper explanation read below.

P2 = the set of all polynomials of degree 2 or lower.

Since every such polynomial can be written as the linear combination ax^2 + bx + c(1) where a,b, and c are constants, we know that the set { x^2, x, 1 } spans P2.

Now for linear independence, suppose that

a(x^2) + b(x) + c(1) = the zero polynomial. The only way that can happen is if each of a, b, and c are zero. Thus the set {x^2, x, 1} is linearly independent.

Since the set {x^2, x, 1} is both a spanning set and linearly independent, it is a basis for P2.

Since this basis has 3 elements, P2 is three-dimensional (has dimension 3). Since all bases of finite-dimensional vector spaces have the same number of elements, we can rule out choice d which has 4 elements in it.

Choice b has 3 elements, but there is no way to combine x, 1, and 0 using only constants to obtain x^2 which is an element of P2. Therefore choice b does not span P2 (and is therefore not a basis). You can also show that it is not linearly independent because it contains 0 (any set containing 0 is not linearly independent).

Choice a is not a basis because it is not linearly independent. We can combine the first two elements to get the third element. Thus 1(x^2) + 1(x + 1) = 1(x^2 + x + 1) which can be rearranged to show that 1(x^2) + 1(x + 1) - 1(x^2 + x + 1) = 0. In other words we can combine these quantities using non-zero multipliers to get zero.

Thus the only remaining candidate is choice c. It has three elements and can be shown to be both a spanning set and a linearly independent set (though since it has the right number of elements, a theorem in linear algebra states that you only have to show one of the two properties and the other will follow).

Linear independence.

Suppose

a(x^2 + x + 1) + b(x^2 + x) + c(x + 1) = 0.

Then after simplifying, we have

(a + b)x^2 + (a + b + c)x + (a + c) = 0.

This means that

(a + b) = 0,

(a + b + c) = 0,

(a + c) = 0.

Subtracting the first eq. from the second gives us c = 0. Plugging c into the last equation gives us a = 0 and plugging a into the first eq. gives us b = 0. Thus a = b = c = 0. The only way to combine these three polynomial vectors to give zero is if we multiply each by zero. Thus they are linearly independent.

Spanning set.

Suppose we have a polynomial Ax^2 + Bx + C. We want to show that we can obtain this polynomial by a suitable linear combination of {x^2+x+1, x^2+x, x+1}. Writing this out we wish to solve

Ax^2 + Bx + C = a(x^2 + x + 1) + b(x^2 + x) + c(x + 1) = (a + b)x^2 + (a + b + c)x + (a + c).

Equating coefficients, we have

(a + b) = A,

(a + b + c) = B,

(a + c) = C.

Subtracting the first eq. from the second gives us c = B - A. Plugging c into the last equation gives us a = C - B + A and plugging a into the first eq. gives us b = B - C. Thus a suitable choice of a, b, and c can give us any polynomial in P2 so that choice c represents a spanning set.