Final answer:
a. {x−2x², −(x²+1), 2x²+x+4} is not a basis for P2. b. One possible basis for P2 is {1, x, x²}. c. A basis for the subspace HH is {x−2x²}.
Step-by-step explanation:
a. To determine if {x−2x², −(x²+1), 2x²+x+4} is a basis for P2, we need to check if these vectors are linearly independent and if they span the entire space. We can do this by setting up a system of equations and solving for the coefficients.
Let a, b, and c be the coefficients for x−2x², −(x²+1), and 2x²+x+4 respectively. We can set up the following system of equations:
a(x−2x²) + b(-(x²+1)) + c(2x²+x+4) = 0
For this system to have a non-trivial solution (other than all coefficients being zero), the determinant of the coefficient matrix must be zero. Solving the determinant:
-5ac + 4bc + 6ac + 6ab + c - b = 0
2ac + 6ab - 2bc + c - b = 0
From this, we can see that there are infinitely many solutions for a, b, and c. Therefore, {x−2x², −(x²+1), 2x²+x+4} is not a basis for P2.
b. Choosing a basis for P2 involves finding a set of vectors that are linearly independent and span P2. One possible basis for P2 is {1, x, x²} as these three polynomials are linearly independent and can span the space P2.
c. To find a basis for the subspace HH, we need to determine which vectors in {x−2x², −(x²+1), 2x²+x+4} are linearly independent. We can do this by setting up a system of equations and solving for the coefficients of the vectors. If a vector's coefficients can only be zero, then it is linearly independent. In this case, the vector x−2x² is linearly independent. Therefore, a basis for the subspace HH is {x−2x²}.