Prove that $Tv_1,…,Tv_n$ is linearly independent in $W$

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.

• If $T$ is injective then what is $ker(T)$? – Yadati Kiran Jan 27 at 2:21
• is Ker(T) means Null(T)? is that ${0}$? – JOHN Jan 27 at 2:28
• Is $\ker(T)=Null(T)$? Yes. Too answer "is that $0$?", for a linear map, $T(0)=0$ always and in addition $T$ is given to be injective, so does there exist $x\in V,\;x\neq0$ such that $T(x)=0$. If so what assumption will be violated? – Yadati Kiran Jan 27 at 4:16
• @YadatiKiran Injectivity? – JOHN Jan 27 at 4:39
• Yes. And the rest follows. – Yadati Kiran Jan 27 at 4:41

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 $$\sum a_iv_i=0$$.