Prove that the collection of vectors of the form a 3a a b w

Prove that the collection of vectors of the form {[a 3a a + b]}, where a and b are any real numbers, form a vector space.

Solution

1.Consider two vectors of the above form V1 anV2 , where V1=(a1,3a1,a1+b1)^t, V2=(a2,3a2,a2+b2)^t. To show commutativity note that a1+a2=a2+a1, 3a1+3a2=3a2+3a1 and (a1+b1)+(a2+b2)=(a2+b2)+(a1+b1).

2.Similarly the addition is associative since addition of real no:s are associative.

3.The zero-vector 0=(0,0,0), i.e., a=0 and b=0. and 0+V=V+0.

4.If V=(a,3a,a+b)^t then -V=(-a,-3a,-a-b)^t, i.e. the vector given with values -a and -b.

5. s(r V)=s(ra,3ra,ra+rb)^t=s(ra,3ra,r(a+b))^t=rs(a,3a,a+b)^t. i.e. multiplication is associative, where r,s are scalars.

6. Also similarly rV+sV=(r+s)V (scalar sum is distributive).

7. Now r(V1+V2)=r(a1+a2,3(a1+a2),(a1+a2)+(b1+b2))^t=(r(a1+a2),3r(a1+a2),r(a1+a2)+r(b1+b2))^t

={(ra1,3ra1,ra1+rb1)^t+(ra2,3ra2,ra2+rb2)^t}= rV1`+rV2. (Distributive over vector sums).

8. And 1.V=V.(scalar multiplication identity).