Answer
See explanation
Work Step by Step
(a) v(0) = 0
0∈ V.
(b) (Q+K)(0) = 0 -> Q(0) + K(0) = 0 -> 0 + 0 = 0
Q+K or v is closed over addition.
(c) c(Q+K)(0)=0 -> cQ(0) + cK(0) = 0 -> c0 + c0 = 0 -> 0 + 0 = 0. c(Q+K)∈V
Q+K or v is closed under scalar multiplication.
Conclusion:
By definition of subspace, Q+K or v is a subspace of V
