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

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.