Trouble with proofs: Let V be a vector space. If v∈V, define T_v : lR→V by T_v(x)=xv for all x in lR...?

First, notice that we're considering the domain R as a vector space over itself. So in this case, the "vectors" and scalars are both real numbers.

(a) Remember, showing T_v is linear means we need to show T_v(x+y) = T_v(x) + T_v(y) and T_v(λx) = λ*T_v(x) for any vectors x,y and scalar λ. Well, there's only one thing we can do, and that is to use the definition of T_v that we were given:

T_v(x+y) = (x+y)v = xv + yv = T_v(x) + T_v(y)

T_v(λx) = (λx)v = λ(xv) = λ*T_v(x)

Note: in the "vector addition" part above, we used the distributive property of scalar multiplication in V, since the "vectors" x,y are real numbers and so can be viewed as "scalars" when working in V.

(b) Let T:R→V be an arbitrary linear transformation. Let v = T(1), which is a vector in V uniquely determined by T. Then for any x in R, we have:

T(x) = T(x*1) = x*T(1) = xv = T_v(x)

Which shows that T = T_v.

merely write out the multiplication. all of us comprehend that for any vector v = (v1, v2, ...) and scalar ?, scalar multiplication is defined by ?v = (?v1, ?v2, ...). If v = Ø = (0, 0, ...), then ?v = (?*0, ?*0, ..) = (0, 0, ...) = Ø.