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: enter image description here

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.

  • 1
    $\begingroup$ If $T$ is injective then what is $ker(T)$? $\endgroup$ – Yadati Kiran Jan 27 at 2:21
  • $\begingroup$ is Ker(T) means Null(T)? is that ${0}$? $\endgroup$ – JOHN Jan 27 at 2:28
  • $\begingroup$ 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? $\endgroup$ – Yadati Kiran Jan 27 at 4:16
  • $\begingroup$ @YadatiKiran Injectivity? $\endgroup$ – JOHN Jan 27 at 4:39
  • $\begingroup$ Yes. And the rest follows. $\endgroup$ – 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 $\sum a_iv_i=0$.


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