Scalar Multiplication Condition.For a vectorv=(v1v2v3)∈R3, we haveλv=(λv1λv2λv3)∈R3, λ∈R, and hence2λ v¿¿¿5(λv1)−(λ v2)¿(¿(λ v2)¿¿)T(λ v)=¿T(λ v)=λ(v25v1−v2v2)T(λ v)=λT(v)Thus, the scalar multiplication condition is also satisfied and since it both conditions aresatisfied, It can be concluded that the functionT:R3→R2is defined by :T:(x1x2)=(x25x1−x2x1)for all(x1x2)∈R3is linear.Question 3.We are required to show that the functionS={x∈R4:3x1−x2+2x3+3x4=0}is a subspaceofR4.To show that the function is a subspace ofR4we can utilize the subspace theorem,firstcheck that the function satisfies the subspace theorem, which states that a function is asubspace if it satisfies the following conditions:i)It contains the zero element,0∈Sii)S is closed under vector additioniii) S is closed under scalar multiplicationi) Existence of Zero VectorSubstitute the vector(0000)into the function, so