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Given $V,W$ vector spaces from finite dimension and given $T:V\to W$ linear transformation. decide if the following statment is true/false: (according to the book the answer is true)

If $T$ transfrom basis to an linear independant group, than $T$ is injective.

I say that this statement is fasle since I can take $V=\mathbb{R^2}$ and $W=\mathbb{R}$

than $$T(1,0)=1$$ $$T(0,1)=1$$ so the group $1$ is linearly independant and obviously that $T$ is not injective.

The answer to this statement is that is true and I dont understand why.

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  • $\begingroup$ 1 and 1 is not linear independent. $\endgroup$
    – onRiv
    Feb 28, 2022 at 8:00
  • $\begingroup$ I think that he’s saying \{T(e_i)\} has to be a set of independent vectors. In your case is not so, 1 is not independent if 1 $\endgroup$ Feb 28, 2022 at 8:00

2 Answers 2

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I think the problem is that they mean that the set of images is linearly independent, counting multiplicity. In your counterexample, the image set is {1, 1} which is not a linearly independent set.

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  • $\begingroup$ Ok I understood my mistake,thank you! $\endgroup$
    – Sagigever
    Feb 28, 2022 at 8:03
  • $\begingroup$ Well, actually, $\{1, 1\}$ actually is linearly independent, since it is equal to $\{1\}$. This is one of the pitfalls in using sets to describe linear independence/bases. As a list/collection (ordered, with repetitions), $(1, 1)$ is not linear independent, whereas $(1)$ is. $\endgroup$ Feb 28, 2022 at 8:04
  • $\begingroup$ Sure, I guess I am abusing words and notation $\endgroup$
    – Aphyd
    Feb 28, 2022 at 8:06
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The hypothesis requires $T(1,0)$ and $T(0,1)$ to be independent . This is not true in your case. [$(1)1+(-1)1=0$].

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  • $\begingroup$ Ok I understood my mistake,thank you! $\endgroup$
    – Sagigever
    Feb 28, 2022 at 8:03

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