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.
Best Answer
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.