Let V and W be vector spaces and T:V \rightarrow

Let us take ${\displaystyle \{w_1,w_2,w_3\dots w_k\}}$ be the linear independent subset of R(t) instead of ${\displaystyle \{y_1\dots y_k\}}$ and ${\displaystyle S=\{v_1,v_2,\dots v_k\}}$ is subset of V such that ${\displaystyle T\left(v_i\right)=W_i}$...(1).
The transformation is linear, where V and W are two vector spaces.
To show ${\displaystyle S=\{v_1,v_2,\dots v_k\}}$ is LI.
For any scalars ${\displaystyle a_1,a_2,\dots a_k}$ consider ${\displaystyle a_1{v}_1+a_2{v}_1+\dots +a_k{v}_k=0}$...(2)
Since T is a function, the above equation can be written as follows,
${\displaystyle T(a_1{v}_1+a_2{v}_2+\dots +a_k{v}_k)=T\left(0\right)}$.
Since T is linear transformation,
${\displaystyle T\left(a_1{v}_1\right)+T\left(a_2{v}_2\right)+\dots +T\left(a_k{v}_k\right)=0}$
${\displaystyle a_{1}T\left(v_1\right)+a_{2}T\left(v_2\right)+\dots +a_{k}T\left(v_k\right)=0}$
${\displaystyle a_1{w}_1+a_2{w}_2+\dots +a_k{w}_k=0}$ (From (1))
Since, the set ${\displaystyle \{w_1,w_2,w_3\dots w_k\}}$ is LI, the only choice is ${\displaystyle a_1=a_2\dots =a_k=0}$...(3)
So, from equation (2) and (3), the set ${\displaystyle S=\{v_1,v_2,\dots v_k\}}$ is LI.
Hence, proved.