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Anonymous asked in Science & MathematicsMathematics · 5 years ago

Spanning P3(R) ?!?!?

Determine if the following set spans P3(R).

{1+2x-4x^3, 2-x+x^2+2x^3, 3+6x-x^2+x^3}

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  • 5 years ago
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    This set spans P₃(R) if an arbitrary vector in P₃(R) can be written as a linear combination of the vectors in the given set.

    Let a₀ + a₁x + a₂x² + a₃x³ = p be an arbitrary polynomial in P₃(R). We want to find scalars b, c, d ∈ R such that

    b(1 + 2x - 4x³) + c(2 - x + x² + 2x³) + d(3 + 6x - x² + x³) = a₀ + a₁x + a₂x² + a₃x³

    Note, we must have

    b + 2c + 3d = a₀

    2b - c + 6d = a₁

    c - d = a₂

    -4b + 2c + d = a₃

    We see that c = a₂ + d. Thus, the first equation becomes

    b + 2(a₂ + d) + 3d = a₀ ==> b + 5d = a₀ - 2a₂

    The second equation becomes

    2b - (a₂ + d) + 6d = a₁ ==> 2b + 5d = a₂ + a₁

    Finally, the last equation becomes

    -4b + 2(a₂ + d) + d = a₃ ==> -4b + 3d = a₃ - 2a₂

    We now have a system of 3 equation with 2 unknowns, namely b and d. From the first equation, we have

    b = a₀ - 2a₂ - 5d

    Plugging into the second equation, we get

    2(a₀ - 2a₂ - 5d) + 5d = a₂ + a₁ ==> d = (-2/5)a₀ + (-1/5)a₁ - a₂

    From this, we get b = a₀ - 2a₂ - 5[(-2/5)a₀ + (-1/5)a₁ - a₂] = 3a₀ + a₁ + 3a₂. Finally, we know that c = a₂ + d. Thus, we have

    c = a₂ + [(-2/5)a₀ + (-1/5)a₁ - a₂] = (-2/5)a₀ + (-1/5)a₁

    Thus, what we have is that for any polynomial a₀ + a₁x + a₂x² + a₃x³ = p, there are coefficients b, c, d ∈ R such that b(1 + 2x - 4x³) + c(2 - x + x² + 2x³) + d(3 + 6x - x² + x³) = a₀ + a₁x + a₂x² + a₃x³. Namely, these coefficients are

    b = 3a₀ + a₁ + 3a₂, c = (-2/5)a₀ + (-1/5)a₁, and d = (-2/5)a₀ + (-1/5)a₁ - a₂.

    Since an arbitrary polynomial in P₃(R) can be written as linear combination of the vectors in the given set, it follows that the set spans P₃(R).

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