Math Subspace in ℙn: Verifying Closure, Scalar Multiplication, and Zero Polynomials

Determine if the given set is a subspace of ℙn. Justify your answer.The set of all polynomials in ℙn such that p​(0)=0

To determine if the given set is a subspace of ℙn, we need to verify if it satisfies the three requirements for a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector

To determine if the given set is a subspace of ℙn, we need to verify if it satisfies the three requirements for a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector.

1. Closure under addition: Let’s take two polynomials, p1 and p2, that satisfy p1(0) = 0 and p2(0) = 0. Now let’s add these two polynomials, p1 + p2.

For any x ∈ ℝ, (p1 + p2)(x) = p1(x) + p2(x). Since p1(x) and p2(x) are polynomials, they are continuous functions for all real numbers. Since both p1(0) = 0 and p2(0) = 0, we can conclude that (p1 + p2)(0) = p1(0) + p2(0) = 0 + 0 = 0.

Thus, the set is closed under addition.

2. Closure under scalar multiplication: Let’s take a polynomial p(x) that satisfies p(0) = 0, and let’s multiply it by a scalar c.

For any x ∈ ℝ, (c * p)(x) = c * p(x), where c is a constant. Since p(x) is a polynomial and a continuous function for all real numbers, we can conclude that (c * p)(0) = c * p(0) = c * 0 = 0.

Thus, the set is closed under scalar multiplication.

3. Contains the zero vector: The zero vector in ℙn is the polynomial with all coefficients equal to zero, i.e., the zero polynomial. Since the zero polynomial evaluated at 0 is also equal to 0, it satisfies the condition p(0) = 0.

Therefore, the set of all polynomials in ℙn such that p(0) = 0 is indeed a subspace of ℙn, as it satisfies all three requirements for a subspace.

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