Let P2 denote the vector space of all polynomials ofdegree 2 or less, Show that {1+x+x^2, 1+2x-x^2, 1-2x-2x^2}is a basis of P2.
Solution
The 2 requirements for a set of vectors to be a basis are: 1) They should be linearly independent 2) They must span the entire vector space. Let\'s check the first condition: they must be linearly independent for the set of vectors to be a vector space, so, a(1+x+x^2) + b(1+2x-x^2) + c(1-2x-2x^2) = 0 must have only {a=b=c=0} as the solution for them to be linearly independent. And that\'s the only solution we get if we solve for a , b, c. So, they are linearly indepenent. Now, going to the 2nd point, They span the entire vector space p2 as we choose different values for a, b, c. So, {1+x+x^2, 1+2x-x^2, 1-2x-2x^2}is a basis of P2. For further reference: http://www.utdallas.edu/dept/abp/PDF_Files/LinearAlgebra_Folder/Basis.pdf
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Let P2 denote the vector space of all polynomials ofdegree 2 or less-.docx
1. Let P2 denote the vector space of all polynomials ofdegree 2 or less, Show that {1+x+x^2, 1+2x-
x^2, 1-2x-2x^2}is a basis of P2.
Solution
The 2 requirements for a set of vectors to be a basis are: 1) They should be linearly independent
2) They must span the entire vector space. Let's check the first condition: they must be linearly
independent for the set of vectors to be a vector space, so, a(1+x+x^2) + b(1+2x-x^2) + c(1-2x-
2x^2) = 0 must have only {a=b=c=0} as the solution for them to be linearly independent. And
that's the only solution we get if we solve for a , b, c. So, they are linearly indepenent. Now,
going to the 2nd point, They span the entire vector space p2 as we choose different values for a,
b, c. So, {1+x+x^2, 1+2x-x^2, 1-2x-2x^2}is a basis of P2. For further reference:
http://www.utdallas.edu/dept/abp/PDF_Files/LinearAlgebra_Folder/Basis.pdf
