Respuesta :
Answer:
1) It is not a vector space
2) It is a vector space
Step-by-step explanation:
1) This is not a vector space. There is no neuter element for the addition, since the null matrix is not invertible.
2) This is a vector space. Lets denote the set given by this item A. A is, In fact, this is a subspace of the vector space given by the real-line funcitons defined in all the real line. Note that if k is a real number and f is an element of A, then
k*A is an element of A, because k*f is defined everywere (kf(x) = k* (f(x)) )
and k*f(1) = k*0 = 0.
If f,g are elements of A, then f+g is an element of A: f+g(x) = f(x) + g(x) is defined everywhere and f+g(1) = f(1)+g(1) = 0+0 = 0.
Also, the zero function is definced everywhere and in 1 it takes the value 0. Since A is a subspace of the vector space given by the real line functions, then it s indeed a vector space.
