Let V be a vector space and \(S\subset V\) is said to be a subspace of V if the followings holds

\(0\in S\) for u,\(v\in S\) and for a scaler k \(u+kv\in S\).

(a)Here it is given \(\displaystyle{H}={\left\lbrace{\left({x},{y}\right)}:{y}={3}{x}−{1}\right\rbrace}.\) Since (0, 0) is not in H. Thus H is not a subspace of \(R^{2}\).

(b)Here it is given \(\displaystyle{H}={\left\lbrace{a}{t}+{b}:{b}={8}{a}\right\rbrace}.\) Clearly (0, 0) is in H. Let \(at+b,ct+d \in H\). Then \(b=8a\ and\ d=8c\). Now for a scaler k