Determine if the given set is a subspace of ℙ8. Justify your answer:All polynomials of degree at most 8, with positive real numbers as coefficients.
To determine if a set is a subspace of another vector space, we need to check three conditions: closure under addition, closure under scalar multiplication, and contain the zero vector
To determine if a set is a subspace of another vector space, we need to check three conditions: closure under addition, closure under scalar multiplication, and contain the zero vector. Let’s analyze each condition.
1. Closure under addition: Suppose we have two polynomials, f(x) and g(x), in the set of all polynomials of degree at most 8, with positive real numbers as coefficients. We need to show that their sum, f(x) + g(x), is in the set as well.
Let f(x) = a0 + a1x + a2x^2 + … + anx^n, where ai > 0 for all i, and n ≤ 8.
Let g(x) = b0 + b1x + b2x^2 + … + bmx^m, where bi > 0 for all i, and m ≤ 8.
The sum of f(x) and g(x) is given by (a0 + b0) + (a1 + b1)x + (a2 + b2)x^2 + … + (an + bn)x^n.
Since the coefficients ai and bi are positive real numbers, the coefficients (a0 + b0), (a1 + b1), (a2 + b2), …, (an + bn) are also positive real numbers. Therefore, the sum f(x) + g(x) is a polynomial of degree at most 8 with positive real numbers as coefficients. Hence, the set is closed under addition.
2. Closure under scalar multiplication: Let c be a positive real number and f(x) be a polynomial in the set. We need to show that the scalar product c * f(x) is also in the set.
Consider f(x) = a0 + a1x + a2x^2 + … + anx^n, where ai > 0 for all i, and n ≤ 8.
The scalar product c * f(x) is given by c * (a0 + a1x + a2x^2 + … + anx^n) = (c * a0) + (c * a1)x + (c * a2)x^2 + … + (c * an)x^n.
Since c > 0 and ai > 0 for all i, the coefficients (c * a0), (c * a1), (c * a2), …, (c * an) are also positive real numbers. Therefore, c * f(x) is a polynomial of degree at most 8 with positive real numbers as coefficients. Hence, the set is closed under scalar multiplication.
3. Contain the zero vector: The zero vector in the vector space ℙ8 is the polynomial with all coefficients equal to zero. In this set, the zero vector would be the polynomial 0.
The zero polynomial can be written as 0 = 0 + 0x + 0x^2 + … + 0x^8.
Since all coefficients of the zero polynomial are zero, which is not a positive real number, the zero polynomial is not in the set.
Therefore, the given set is not a subspace of ℙ8 because it does not contain the zero vector.
In summary, the set of all polynomials of degree at most 8 with positive real numbers as coefficients is not a subspace of ℙ8 as it does not satisfy all three conditions: closure under addition, closure under scalar multiplication, and containing the zero vector.
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