Question

Determine whether the set, together with the indicated operations, is a vector space. If it is not, then identify one of the vector space axioms that fails. The set of all seventh-degree polynomials with the standard operations

Answer 1

The set of all seventh-degree polynomials with the standard operations forms a vector space.

Step-by-step explanation:

The set of all seventh-degree polynomials with the standard operations forms a vector space.

To verify this, we need to check if the vector space axioms are satisfied:

1. Closure under Addition: The sum of two seventh-degree polynomials is also a seventh-degree polynomial.
2. Closure under Scalar Multiplication: Multiplying a seventh-degree polynomial by a scalar results in a seventh-degree polynomial.
3. Associativity of Addition: The addition of polynomials is associative.
4. Commutativity of Addition: The addition of polynomials is commutative.
5. Identity Element: There exists a zero polynomial that acts as the identity element for addition.
6. Inverse Elements: Each polynomial has an additive inverse.
7. Distributivity: Scalar multiplication distributes over addition.

Since all of the vector space axioms are satisfied, the set of all seventh-degree polynomials with the standard operations is indeed a vector space.

Answer 2

Following are the solution to this question:

Step-by-step explanation:

Let X is a set of all polynomials of the third-degree with regular operations.

This set is not a space for a vector since it is not shut down

For instance:

`\to x^3+3x^2+2 \in X \ \ \ and \ \ \ -x^3+2x^2+2x \in X`

Calculating the sum:

`\to x^3+3x^2+2 + -x^3+2x^2+2x\\\\\to 5x^2+2x+2 \\otin X`

That's why X isn't a space for vectors because X isn't shut down.