Final answer:
The set of all polynomials of the form p(t)a, where a is in ℝ, is a subspace of ℝⁿ because it satisfies the three criteria of being a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector.
Step-by-step explanation:
To determine if the given set is a subspace of ℝⁿ, we need to check if it satisfies the three criteria of being a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector.
Let's consider the set of all polynomials of the form p(t)a, where a is in ℝ. To check closure under addition, suppose we have two polynomials p₁(t)a₁ and p₂(t)a₂. Their sum is p₁(t)a₁ + p₂(t)a₂. Since both p₁(t) and p₂(t) are polynomials and a₁, a₂ are real numbers, the sum is also a polynomial of the same form, thus the set is closed under addition.
To check closure under scalar multiplication, let's consider a polynomial p(t)a. Multiplying it by a scalar c gives c(p(t)a), which is also a polynomial of the same form, thus the set is closed under scalar multiplication.
Lastly, to check if the zero vector is in the set, we can choose p(t) to be the zero polynomial. Then we have p(t)a = 0a = 0, which is in the set.
Therefore, the given set of all polynomials of the form p(t)a, where a is in ℝ, is a subspace of ℝⁿ.