Suppose $T \in \mathcal{L}(V,W)$ is injective and $v_1,…,v_n$ is linearly independent in $V$. Prove that $Tv_1,…,Tv_n$ is linearly independent in $W$.
I generally follow the solution below:
But I don't understand the line "Because $T$ is injective, this implies that $a_1v_1 + … +a_nv_n = 0$". How injectivity implies the formula?
And my trial to this question is something in reverse order.
I started with
$$0 = a_1v_1 + … + a_nv_n$$
$$T(0) = T(a_1v_1 + … + a_nv_n)$$
$$0 = a_1Tv_1 +…+a_nTv_n$$
Because $a_1=…=a_n = 0$, $Tv_1,…,Tv_n$ is linearly independent. And I don't know where I used the property of injectivity.
Best Answer
For any linear map $T$ one has $T(0)=0$. Further if $T$ is injective, (1-1 function) no other vector can be sent to zero by $T$. Hence $T(\sum a_ivi)=0$ implies $\sum a_iv_i=0$.