The exercise I'm trying to prove is:
If $T:V\to W$ is a linear transformation with $\ker T=\{0_V\}$. If $v_1,v_2,\cdots, v_k \in V$ are linearly independent, then $T(v_1),T(v_2),\cdots, T(v_k) \in W$ are also linearly independent.
What I've tried to analyze from this, is that, if $\ker T=\{0_V\}$ and $v_1,v_2,\cdots, v_k\in V$ are linearly independent, then, $v=a_1v_1+a_2v_2+\cdots+a_kv_k=0_V=\ker T$ is a linear combination where $a_i \in \mathbb{R}$ and $a_i=0$ for $i=1,2,\cdots,k.$
Should that imply that $T(v)=0_W$ for all $v\in V$? And since $T(v)=0_W$, does that imply that $T(v)=0_W=b_1w_1+b_2w_2+\cdots+b_kw_k$ and hence $T(v_1),T(v_2),\cdots,T(v_k)$ are linearly independent elements of $W$?
Also, at this point I'm not really sure if it's really necessary to define $\ker T=\{0_V\}$ since we're already saying that $v_1,v_2,\cdots, v_k \in V$ are linearly independent. Thank you in advance!
Best Answer
Assume that $$b_1T(v_1)+\cdots+b_kT(v_k)=0$$ holds for some constants $b_1,..,b_k$. You can rewrite this as $$T(b_1v_1+\cdots b_kv_k)=0$$ Can you now use your assumption on the kernel of $T$ and finish the proof?