Let Pn(R) = The vector space of polynomials with real coefficients and degree lessor equal to n. Show that the set {x + 1, x2 + x − 1, x2 − x + 1} is a basis for P2(R).Hence, determine the coordinates of the following elements: 2x−1, 1+x2, x2 +5x−1with respect to the above basis
Question
Let Pn(R) = The vector space of polynomials with real coefficients and degree lessor equal to n. Show that the set {x + 1, x2 + x − 1, x2 − x + 1} is a basis for P2(R).Hence, determine the coordinates of the following elements: 2x−1, 1+x2, x2 +5x−1with respect to the above basis
Solution
To show that the set {x + 1, x^2 + x - 1, x^2 - x + 1} is a basis for P2(R), we need to show that these vectors are linearly independent and that they span P2(R).
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Linear Independence: We need to show that the only solution to the equation a(x + 1) + b(x^2 + x - 1) + c(x^2 - x + 1) = 0 is a = b = c = 0. This equation simplifies to (b + c)x^2 + (a + b - c)x + (a - b + c) = 0. Since this must hold for all x, we have the system of equations: b + c = 0, a + b - c = 0, and a - b + c = 0. Solving this system gives a = b = c = 0, so the vectors are linearly independent.
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Spanning P2(R): Any polynomial of degree less than or equal to 2 can be written as ax^2 + bx + c, which can be expressed as a linear combination of the vectors in the set.
Now, to find the coordinates of the elements 2x - 1, 1 + x^2, and x^2 + 5x - 1 with respect to this basis, we need to express each of these as a linear combination of the basis vectors.
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2x - 1 = a(x + 1) + b(x^2 + x - 1) + c(x^2 - x + 1) Solving for a, b, and c gives the coordinates (a, b, c) = (-3, 0, 0).
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1 + x^2 = a(x + 1) + b(x^2 + x - 1) + c(x^2 - x + 1) Solving for a, b, and c gives the coordinates (a, b, c) = (-1, 1, 1).
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x^2 + 5x - 1 = a(x + 1) + b(x^2 + x - 1) + c(x^2 - x + 1) Solving for a, b, and c gives the coordinates (a, b, c) = (6, 1, -1).
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