Question

Let H={[xy​]:3x²+6y²≤1}, which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples-two vectors, and a vector and a scalar-to show that H is not a subspace of R². H is not a subspace of R² because the two vectors [3​1​0​],[06​1​​] show that H closed under (Use a comma to separate vectors as needed.) H is not a subspace of R² because the scalar 3 and the vector show that H closed under Determine if the given set is a subspace of P₃. Justify your answer. All polynomials of degree at most 3 , with integers as coefficients. Complete each statement below. The zero vector of P₃​ in the set because zero an integer.

Answer 1

To determine if H is a subspace of R², we need to check if it satisfies the three conditions for a set to be a subspace. However, H does not satisfy any of these conditions, so it is not a subspace of R².

Step-by-step explanation:

In order to determine whether H is a subspace of R², we need to check if it satisfies the three conditions that are required for a set to be a subspace:

1. The zero vector of R² must be in H.
2. H must be closed under vector addition.
3. H must be closed under scalar multiplication.

Let's analyze each condition:

1. The zero vector of R² is [0 0]. However, when we plug in the values of x and y into the equation 3x² + 6y² ≤ 1, we can see that it is not satisfied by [0 0]. Therefore, H does not contain the zero vector.
2. In order to check if H is closed under vector addition, we need to consider two vectors, let's say A = [3 1] and B = [0 6]. When we add A to B, we get [3 7]. However, when we plug in the values of x and y into the equation 3x² + 6y² ≤ 1, we can see that it is not satisfied by [3 7]. Therefore, H is not closed under vector addition.
3. In addition, let's consider the scalar 3 and the vector A = [3 1]. When we multiply the scalar 3 with A, we get [9 3]. However, when we plug in the values of x and y into the equation 3x² + 6y² ≤ 1, we can see that it is not satisfied by [9 3]. Therefore, H is not closed under scalar multiplication.

Since H does not satisfy any of the three conditions required for a set to be a subspace, H is not a subspace of R².