Testing for a Vector Space In Exercises 13–36, determine whether the set, together with the standard operations, is a vector space. If it is not, identify at least one of the ten vector space axioms that fails. 13. M, 4.6 14. M, 15. The a set of all third-degree polynomials 16. The set of all fifth-degree polynomials 17. The set of all first-degree polynomial functions ax, a t 0, whose graphs pass through the origin 18. The set of all first-degree polynomial functions ax + b a, b 0, whose graphs do not pass through the origin 19. The set of all polynomials of degree four or less

Step-by-step explanation:

13. The set M4,6 of all 4x6 matrices is a vector space as it is closed under vector addition as also scalar multiplication. Also the 4x6 zero matrix is in M4,6.

14. The set M1,1 is a singleton, i.e. a 1x1 matrix. It is a vector space for the same reasons as in 13. above.

15. The degree of the zero polynomial is undefined and it is usually treated as a constant (of degree 0). If we treat 0 as a polynomial of degree 0, then the set of all 3rd degree polynomials is not a vector space despite being closed under vector addition and scalar multiplication as it does not contain the zero polynomial.

16. The set of all 5th degree polynomials is not a vector space despite being closed under vector addition and scalar multiplication as it does not contain the zero polynomial.

17. The set of all first degree polynomials ax (a≠0) is not a vector space. If p(x) = ax and q(x) = -ax , then p(x)+q(x) = 0*x which is not in the given set. Hence the set is not closed under vector addition and, therefore, it is not a vector space.

18. The set of all first degree polynomials ax +b (a,b ≠0) is not a vector space. If p(x) = ax +b and q(x) = -ax -b , then p(x)+q(x) = 0*x +0 which is not in the given set. Hence the set is not closed under vector addition and, therefore, it is not a vector space.

19. The set P4 of all polynomials of degree 4 or less isa vector space. If p(x) = a1x4+a2 x3+a3x2+a4x+a5 and q(x) = b1x4+b2 x3+b3x2+b4x+b5 are 2 arbitrary elements of P4, then p(x)+q(x)is in P4. Similarly, αp(x) is in P4 for any arbitrat scalar α. Hence, P4 is closed under vector addition and scalar multiplication. Also, the 0 polynomial is in P4. Hence P4 is a vector space.

the difference between a number and 5 is -2.

step-by-step explanation:

notice that the equation you are given has -2 (negative 2) on the right side, not 2. when you subtract 5 from a number n, you are finding the difference between the number and 5.

the difference between a number and 5 is -2.

answer: true

step-by-step explanation: