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**Definition**: A function $ f: A \to B $ is linear if $ f(x+y) = f(x)+f(y) $ and $ cf(x) = f(cx) $ for $ x,y \in A $ and $ c \in \mathbb{R} $ (or, rather, any field $ \mathbb{F} $). Furthermore, it is one-to-one if and only if $ f(x) = f(y) \Rightarrow x = y $.

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**Problem**: Let $ V $ and $ W $ be vector spaces, and let $ T: V \to W $ be linear. Then $ T $ is one-to-one if and only if $ T $ carries linearly independent subsets of $ V $ onto linearly independent subsets of $ W $.

**Solution**: First we will show that if $ T $ is one-to-one, then it carries linearly independent subsets of $ V $ onto linearly independent subsets of $ W $.

Let $ S $ be a linearly independent subset of $ V $. Suppose $ S = \{x_1, x_2, \ldots, x_k\} $. We want to show that $ R = \{T(x_1), T(x_2), \ldots, T(x_k)\} $ is also linearly independent. Suppose that $ R $ is not. Then there exist $ a_1, a_2, \ldots, a_k $ not all zero such that

$ a_1T(x_1)+a_2T(x_2)+\cdots+a_kT(x_k) = 0 $.

Since $ T $ is linear, we can rewrite this as

$ T(a_1x_1)+T(a_2x_2)+\cdots+T(a_kx_k) = T(a_1x_1+a_2x_2+\cdots+a_kx_k) = 0 $.

Since $ T(0) = 0 $ and $ T $ is one-to-one, we must have

$ a_1x_1+a_2x_2+\cdots+a_kx_k = 0 $.

But $ S $ is linearly independent, so this is only possible if $ a_1 = a_2 = \cdots = a_k = 0 $, a contradiction. So $ R $ is linearly independent.

Now, we will prove the converse - if $ T $ takes linearly independent subsets of $ V $ to linearly independent subsets of $ W $ then $ T $ is one-to-one. Let $ x, y \in V $ be nonzero such that $ T(x) = T(y) $. It suffices to show that $ x = y $.

If $ \{x, y\} $ is linearly dependent, then $ x = cy $ for some $ c \in \mathbb{R} $, so

$ T(x) = T(y) = T(cx) = cT(x) $.

If $ T(x) = T(y) \neq 0 $, we have $ c = 1 \Rightarrow x = y $. In the case that $ T(x) = T(y) = 0 $, the set $ \{x\} $ is linearly independent but $ T $ takes it onto $ \{0\} $, which is linearly dependent, a contradiction.

If $ \{x, y\} $ is linearly independent, then $ \{T(x), T(y)\} $ is also linearly independent. But since $ T(x) = T(y) $, the set is clearly linearly dependent, a contradiction.

Lastly, we have $ T(0) = 0 $. So $ T $ must be a one-to-one function, as desired. QED.

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Comment: Basically, we made extensive use of the properties of a linear transformation to prove both directions. A nice corollary to this proof is that if $ T $ is one-to-one, then $ S $ is linearly independent if and only if $ T(S) $ is linearly independent.

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Practice Problem: Let $ V $ and $ W $ be vector spaces, and let $ T: V \to W $ be linear. Then $ T $ is one-to-one if and only if $ N(T) = \{0\} $ (here, $ N(T) $ is the kernal of $ T $, defined by $ N(T) = \{x | T(x) = 0\} $).

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