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

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?

• I think you are correct, and you may need to check if it is linearly dependent or independent. – Quang Hoang Nov 28 '18 at 17:19
• It is not true. Indeed, you need $T$ to be an injection for this to be true in general. – Thomas Andrews Nov 28 '18 at 17:21
• Or, if $T(u_1),\dots,T(u_n)$ are independent, then $u_1,\dots,u_n$ are independent. – Thomas Andrews Nov 28 '18 at 17:29

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 $$\mathbb{R}$$.

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.

• 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?) – Thomas Andrews Nov 28 '18 at 17:27

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.