# Injective linear transformation and linearly independent vectors

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.

• You should change, I believe, the first "then" to "and" . The claim is true, but again: you must state clearly that the condition must be fulfilled for all lin. indep set of vectors. – DonAntonio Dec 20 '18 at 20:44
• @DonAntonio Gotcha, fixed – Tegernako Dec 20 '18 at 20:48

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.

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.

• This doesn't seem correct: how do you know $\;v\in V\;$ can be written as linear combination of $\;v_1,..,v_n\;$ ? – DonAntonio Dec 20 '18 at 20:46
• @greedoid I know they are linearly independent but I can't just assume they span V.. – Tegernako Dec 20 '18 at 20:47
• @Tegernako If I understood the question correctly, the statement is supposed to hold for every linearly independent set $(v_1,\dots,v_k)$. So, the statement also applies to a linearly independent set that spans $v$. – Omnomnomnom Dec 20 '18 at 20:51
• I took them from a basis. @DonAntonio – Aqua Dec 20 '18 at 20:52
• @greedoid I know...that is what is wrong as the claim is true when $\;v_1,..,v_k\;$ is any lin. ind. set – DonAntonio Dec 20 '18 at 20:52