Show that T is a linear transformation and that it is biject

Show that T is a linear transformation and that it is bijective

Let V be a vector space and let a = {v_1, v_2,..., v_n} be a basis for V. Let T: V -> R^n defined by for every x e V.

Solution

T is a linear transformation as

T(v₁+v_2....+v_n)=T(v₁)+T(v₂)+....+T(v_n)

and T(av_n)=aT(v_n)

for showing it bijective we have to show that T is injective

let the equation is a₁T(v₁) + · · · + a_nT(v_n) = 0

This means that T(a₁v₁+· · ·+a_nv_n) = 0, implying that a1v1 +· · ·+a_nv_n = 0 by injectivity.

But this is a linear combination of vectors in S, giving a_i = 0 for all i.

hence T(S) is linearly independent.

Conversely if T maps spanning sets to spanning sets, then T(V ) = R(T) must span W. But since R(T) is a subspace of W, this means R(T) = W and T is onto.

hence T(x) is bijective