What is wrong in my counter example?

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.

• 1 and 1 is not linear independent. Feb 28, 2022 at 8:00
• 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 Feb 28, 2022 at 8:00

• 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. Feb 28, 2022 at 8:04
The hypothesis requires $$T(1,0)$$ and $$T(0,1)$$ to be independent . This is not true in your case. [$$(1)1+(-1)1=0$$].