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My Professor gives us a homework that contains this question:

Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U \to V$ be a linear transformation on $F$.

If $\{ u_1,...,u_n\}$ is a linearly independent of $U$, show that $\{ T(u_1), ..., T(u_n)\}$ is linearly independent.

Is that true? I took $T:\mathbb{R}^2 \to \mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) \in \mathbb{R}^2$

Then I found that $\{ (1,0),(0,1)\}$ is linearly independent in $ \mathbb{R}^2$, but $\{T(1,0)=0,T(0,1)=0\}$ is not linearly independent in $\mathbb{R}$ ?

What do you think?

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    $\begingroup$ I think you are correct, and you may need to check if it is linearly dependent or independent. $\endgroup$ – Quang Hoang Nov 28 '18 at 17:19
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    $\begingroup$ It is not true. Indeed, you need $T$ to be an injection for this to be true in general. $\endgroup$ – Thomas Andrews Nov 28 '18 at 17:21
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    $\begingroup$ Or, if $T(u_1),\dots,T(u_n)$ are independent, then $u_1,\dots,u_n$ are independent. $\endgroup$ – Thomas Andrews Nov 28 '18 at 17:29
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Your example is a valid counterexample to the claim. So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),\dots,T(u_n)$ are linearly independent, then $u_1,\dots,u_n$ are linearly independent.


Additionally, a small note on terminology: it is preferable to say that $\{ u_1,\dots,u_n \}$ is a linearly independent subset of $U$, and that $\{ T(1,0), T(0,1) \}$ is not linearly independent over $\mathbb{R}$.

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You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if $$\alpha_1 T(u_1) + \cdots + \alpha_n T(u_n) = 0$$ linearity tells us $$T(\alpha_1 u_1 + \cdots + \alpha_n u_n) = 0,$$ and hence if we apply $T^{-1}$ to both sides $$\alpha_1 u_1 + \cdots + \alpha_n u_n = 0.$$ Since $\{u_1,\ldots,u_n\}$ are linearly independent this implies that $\alpha_1=\cdots=\alpha_n=0$, and then by definition $\{T(u_1),\ldots,T(u_n)\}$ are linearly independent as well.

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    $\begingroup$ Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?) $\endgroup$ – Thomas Andrews Nov 28 '18 at 17:27
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In general, this isn't true as your counterexample shows.

If $T$ is invertible then the statement holds. Start from $$c_1T(u_1)+\cdots+c_nT(u_n)=0$$ Apply $T^{-1}$ on both sides and get $$c_1u_1+\cdots+c_nu_n=0$$ and $c_1=\cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.

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