Injective linear transformation and linearly independent vectors

Question

The claim is :

If for all $(v_1, \ldots, v_k) \in V$ the vector $(v_1, \ldots, v_k)$ is linearly independent and $(T(v_1),...(T(v_k))$ is also linearly independent, then $T$ is injective.

Is this claim true or false? I know that $ (T(v_1),...(T(v_k))$ is linearly independent, therefore for $$ c_1 T(v_1) + \ldots + c_k T(v_k) = 0 $$ we know all $ 1 \leq i \leq k,c_i = 0.$ From linearity we can show that $ T(c_1v_1 + ... + c_kv_k) = 0$, and we know $(v_1,...,v_k)$ is linearly independent, therefore $T(0) = 0$, and $\text{Ker} \ T = \{0\}$, therefore T is injective?

I feel like my proof is either flawed or missing something vital.

Answer 1

Highlights:

Suppose $\;0\neq v\in V\;$, then $\;\{v\}\;$ is lin. ind. $\;\implies Tv\;$ lin. ind. $\;\implies Tv\neq 0\implies \ker T=\{0\}\;$

Suppose now $\;T\;$ is injective (and this happens iff $\;\ker T=\{0\}\;$), and let $\;v_1,...,v_k\;$ is lin. ind. Suppose there are scalars $\;a_,...,a_k\;$ s.t.

$$\sum_{i=1}^k a_i Tv_i=0\implies 0=\sum_{i=1}^k a_i Tv_i=T\left(\sum_{i=1}^k a_i v_i\right)\implies \sum_{i=1}^k a_i v_i\in\ker T=\{0\}\implies$$

$$\sum_{i=1}^k a_i v_i=0\implies a_i=0\;\;\forall i\implies Tv_1,...,Tv_k\;\text{ lin. ind.}$$lin. ind.

Answer 2

Your writings are a little messy. You must prove $\text{Ker}(T)=0$, so take any $v\in \text{Ker}(T)$ and write $v= \sum a_iv_i$. So $T(v)=0$ which implies $$ T\left(\sum a_iv_i\right) =0 $$

since $T$ is linear we have $$\sum a_i(Tv_i)=0$$ Now since $v_i$ are linear independent so are $Tv_i$ and thus all $a_i=0$ which means $v=0$ and thus a conclusion.