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
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}\n1 \\ if \\ i=1 \\ \\\\\n0 \\ if \\ i\\neq 1 \\\\\n\\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}\n1 \\ if \\ i=2 \\\\\n0 \\ if \\ i\\neq 2 \\\\\n\\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}\n1 \\ if \\ i=3 \\\\\n0 \\ if \\ i\\neq3 \\\\\n\\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" .
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