Question

Suppose that T is a one-to-one transformation, so that an equation T(u)=T(v) always implies u=v. Show that if the set of images {T(v1)......T(vp)} is linearly dependent, then {v1......vp} is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Answer 1

Answer:

Step-by-step explanation:

The objective is to  Show that if the set of images {T(v1)......T(vp)} is linearly dependent, then {v1......vp} is linearly dependent.

Given that:

$\mathbf{[T(v_1) +T(v_2) ...T(v_p)]}$ is linearly dependent set

Thus; there exists scalars $\mathbf{k_1 , k_2 ... k_p}$ ; ( read as "such that")  $\mathbf{k_1 T(v_1) +k_2T(v_2) ...k_pT(v_p)=0}$

$\mathbf{= T(k_1 v_1 +k_2v_2 ...k_pv_p)=0}$

T = 0                                  (for the fact that T is linear transformation)

$\mathbf{k_1 v_1 +k_2v_2 ...k_pv_p=0}$     (due to T is one-one)

NOTE: Not all Ki's are zero;

Thus;

$\mathbf{[v_1,v_2 ...v_p] }$  is linearly dependent

It negation also illustrates that :

If $\mathbf{[v_1,v_2 ...v_p]}$ is also linearly independent then $\mathbf{[T(v_1),T(v_2) ...T(v_p)]}$ is also linearly independent.