Can a L.I. set transform into a L.D. set after Linear Transformation?

Question

Let $X$ be a real vector space. Suppose $\{v_1,v_2,v_3\} \subset X$ is a linearly independent set, and suppose $\{w_1,w_2,w_3\} \subset X$ is a linearly dependent set. Define $V= \operatorname{span}\{v_1,v_2,v_3\}$ and $W= \operatorname{span}\{w_1, w_2, w_3\}$.

(a) Is there a linear transformation $P : V \to W$ such that $P(v_i) = w_i$ for $i=1,2,3$?

(b) Is there a linear transformation $Q : W \to V$ such that $Q(w_i) = v_i$ for $i=1,2,3$?

What I am thinking is that is this question equivalent to "Can a L.I. set transform into a L.D. set after Linear Transformation ?"(I am a little bit confused about "$P(v_i)=w_i$" is this mean "for every set $V$ and $W$, there exist a linear transformation s.t. ..." ?) If not, then what should be the right solution to this question? Can anyone give me some hints on how to solve this problem？ Thanks!

Answer 1

Answer to a) is yes. You have to use the fact that $\{v_1,v_2,v_3\}$ cab be extended to a basis for $X$. Once you have a basis you can give arbitrary values to the basis elements and extend it to a linear function. Answer to b): any dependence relation among $\{w_1,w_2,w_3\}$ leads to the same relation among $\{v_1,v_2,v_3\}$ so the answer is no.

Answer 2

$P(v_i)=w_i$ means the image of $v_i$ under linear transformation $P$ is $w_i$.

Hint for part b:

Suppose $w_1=c_2w_2+c_3w_3$, then we have $Q(w_1)=c_2Q(w_2)+c_3Q(w_3)$, try to see a contradiction using definition of linearly independence.