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.

  • $\begingroup$ 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. $\endgroup$ – DonAntonio Dec 20 '18 at 20:44
  • $\begingroup$ @DonAntonio Gotcha, fixed $\endgroup$ – Tegernako Dec 20 '18 at 20:48


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.

  • $\begingroup$ This doesn't seem correct: how do you know $\;v\in V\;$ can be written as linear combination of $\;v_1,..,v_n\;$ ? $\endgroup$ – DonAntonio Dec 20 '18 at 20:46
  • $\begingroup$ @greedoid I know they are linearly independent but I can't just assume they span V.. $\endgroup$ – Tegernako Dec 20 '18 at 20:47
  • $\begingroup$ @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 $v$. $\endgroup$ – Omnomnomnom Dec 20 '18 at 20:51
  • $\begingroup$ I took them from a basis. @DonAntonio $\endgroup$ – Aqua Dec 20 '18 at 20:52
  • $\begingroup$ @greedoid I know...that is what is wrong as the claim is true when $\;v_1,..,v_k\;$ is any lin. ind. set $\endgroup$ – DonAntonio Dec 20 '18 at 20:52

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