Question

Find the dual basis for the basis {1,1+x,x²-1} of the vector space P3={a0+a1x+a2x²:a0,a1,a2 belongs to R}

Answer 1

The given vector space is ,

$P_3=\{ a_0+a_1x+a_2x^2:a_0,a_1,a_2\in \R\}$ and the given basis is

$\{ 1,1+x,x^2-1\}$ .

We known that ,if $B=\{ v_1,v_2,......,v_n\}$ is a basis of $V$ over $K$ and

$\phi_1,\phi_2,......,\phi_n \in V^*$ be the linear functionals as defined by

$$\phi_i(v_j)=\begin{cases} 1 \ \ if \ i=j \ \\ 0 \ \ if \ i\neq j \end{cases}$$

Then $\{ \phi_1,.......\phi_n\}$ is a basis of $V^*$ called the dual basis of $B$ .

According to question,

Let $B=\{v_1=1,v_2=1+x,v_3=x^2-1\}$ be a basis of $P_3$ .

Now ,any elements of $P_2$ is written as linear combination of elements of $B$ .

If $a_0+a_1x+a_2x^2 \in P_2$ be any arbitrary element then

$a_0+a_1x+a_2x^2=(a_0-a_1+a_2).1+a_1(1+x)+a_2(x^2-1)$

$\implies \ a_0+a_1x+a_2x^2=(a_0-a_1+a_2)v_1+a_1v_2+a_2v_3$

Again, we known that if $U \ and \ V$ are two vector space over a field $K$ and $\{ v_1,.........,v_n \}$ be a basis of $V \ and \ let \ u_1,.......,u_n$ be any vector in $U$ .Then there exist a unique linear mapping $F:V \rightarrow U \ such \ that \ F(v_1)=u_1,..................,F(v_n)=u_n$

Then ,we can define linear functional as follows,

$\phi_1:P_2\rightarrow \R$ defined by $\phi_1(v_i)=\begin{cases} 1 \ if \ i=1 \ \\ 0 \ if \ i\neq 1 \\ \end{cases}$

Since,$a_0+a_1x+a_2x^2=(a_0-a_1+a_2)v_1+a_1v_2+a_2v_2$

$\phi_1(a_0+a_1x+a_2x^2)=\phi_1\{(a_0-a_1+a_2)v_1+a_1v_2+a_2v _3\}$

$=(a_0-a_1+a_2)\phi_1(v_1)+a_1\phi_1(v_2)+a_2\phi_1(v_3)$

( Since $\phi_1$ is a linear mapping)

i,e, $\phi_1(a_0+a_1x+a_2x^2)=a_0-a_1+a_2$

similarly, $\phi_2:P_2\rightarrow\R$ defined by

$\phi_2(v_i)=:\begin{cases} 1 \ if \ i=2 \\ 0 \ if \ i\neq 2 \\ \end{cases}$

I,e $\phi_2(a_0+a_1x+a_2x^2)=a_1$

And $\phi_3:P_2:\rightarrow\R$ defined by $\phi_3(v_i)=\begin{cases} 1 \ if \ i=3 \\ 0 \ if \ i\neq3 \\ \end{cases}$

I,e $\phi_3(a_0+a_1x+a_3x^3)=a_3$ .

Therefore $\{\phi_1,\phi_2,\phi_3\}$ is the required dual of $Basis \ B$ .