Question

For the vector space P2,find the dual basis of {1+x,1+2x,1+x+x2}?

Answer 1

We are given a vector space of polynomials of degree less than or equal to 2. The given basis is, $\{1+x, 1+2x, 1+x+x^2\}$ . We have to find the dual basis for it.

So, we want to find $f_1, f_2, f_3$ such that, $f_i:P_2\rightarrow\mathbb{R}, i=1,2,3$ are linear functionals, and,

$$f_1(1+x)=1 , \\ f_2(1+2x)=1 ,\\ f_3(1+x+x^2)=1$$

Consider a general element of $P_2$ , say $P(x)=ax^2+bx+c, [a,b,c\in\mathbb{R}]$

As $\{1+x, 1+2x, 1+x+x^2\}$ is a basis set, $P(x)$ can be written uniquely as a linear combination of these 3 polynomials. Let the representation be,

$P(x)=\alpha(1+x)+\beta(1+2x)+\gamma(1+x+x^2)\\ =\gamma x^2+(\alpha+2\beta+\gamma)x+(\alpha+\beta+\gamma)$

Therefore, we must have,

$$\gamma=a,\\ \alpha+2\beta+\gamma=b,\\ \alpha+\beta+\gamma=c$$

Hence, $\gamma=a, \alpha=2c-b-a, \beta=b-c$

So, $P(x)$ can be written as,

$$P(x)=(2c-b-a)(1+x)+(b-c)(1+2x)+a(1+x+x^2)$$

So, we can define $f_1, f_2, f_3$ as follows:

$$f_1(ax^2+bx+c)=2c-b-a\\ f_2(ax^2+bx+c)=b-c\\ f_3(ax^2+bx+c)=a$$

$f_1, f_2, f_3$ are linear in coefficients of the polynimial. Hence, they are linear functionals. Hence $\{f_1, f_2, f_3\}$ is the dual basis for the given basis set.