Respuesta :
Answer:
The set is a subspace
Step-by-step explanation:
We need to check 3 things: whether the 0 vector is in the set, whether the sum of 2 elements of the set is an element of the set and whether the product of an element of the set for a real scalar is an element of the set.
- 0 is in the set
Yes: the 0 vector (0, 0, ..., 0) satysfies the set property: 0-0+0-0........-0 = 0.
- Given 2 elements v = (v1, ..., vn), w = (w1, ..., wn), is the sum v+2 = (v1+w1, v2+w2, ..., vn+wn) an element of the set?
Yes: Note that (v1+w1)-(v2+w2)+(v3+w3)- ..... - (vn+wn) = v1-v1+v3 - ... - vn + w1 - w2 + w3 - ... - wn = 0+0 = 0.
- Given an element of the set v = (v1, ... ,vn), and a real number r, is rv = (rv1, ..., rvn) an element of the set?
Yes: By taking r as common factor, we have rv1 - rv2 + rv3 - ... - rvn = r * (v1-v2+v3 - ... - vn) = r*0 = 0.
Thus, the described set is effectively a subspace.
