# Prove Transformation is one-to-one iff it carries linearly independent subsets of $V$ onto Lin. Ind. subsets of $W$.

A textbook example asks you to prove that for the linear transformation $T\colon V \to W$ that it is one-to one if and only if $T$ carries linearly independent subsets of $V$ onto linearly independent subsets of $W$.

The answer key gives the following solution:

Suppose $T$ is one-to-one. Then let $S$ be a linearly independent subset of $V$, and let $S' = \{T(s)\mid s \in S\}$. (Notice that we’re not assuming that $S$ is finite.) Suppose there are some elements of $S’$ $T(s_1), T(s_2), \ldots , T(s_n)$ and scalars $a_1, \ldots, a_n$ such that $a_1T(s_1)+ \cdots + a_nT(s_n) = 0$. Then by the linearity of $T$,

$a_1T(s_1) + . . . + a_nT(s_n) = 0$

$T(a_1s_1) + . . . + T(a_ns_n) = 0$

$T(a_1s_1 + . . . + a_ns_n) = 0$

But $T$ is one-to-one, so $a_1s_1 + \cdots + a_ns_n = 0$, and thus $a_1 = \cdots = a_n = 0$ because $S$ is linearly independent. Thus $S’$ is linearly independent.

I don't understand this solution. Why does $(a_1s_1+...a_ns_n)$ need to equal 0 just because $T(\text{stuff}) = 0$? A one-to-one function doesn't mean that $T(1) = 1$ and $T(2) = 2$ etc... It simply means all numbers in the codomain must appear in the range.

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A one-to-one (or injective) function doesn't mean that the codomain must appear in the range. Those are called onto or surjective functions. One-to-one means that for any output, there is exactly one input that maps to it (one output - to - one input).

$$L(\mathbf{x}) = L(\mathbf{y}) \implies \mathbf{x} = \mathbf{y}$$

You may want to contrast the two types:

1. One-to-one functions have at most one input for each vector in the codomain.
2. Onto or surjective functions have at least one input for each vector in the codomain.

If you combine the two, you have exactly one input for each vector in the codomain. That's precisely the definition of a bijection mapping.

As for the above theorem, the more fundamental result is perhaps that a linear mapping is one-to-one if and only if it has a trivial kernel. The proof of this is rather simple.

If $L$ is a linear mapping with a trivial kernel, then $$L(\mathbf{x}) = L(\mathbf{y}) \implies L(\mathbf{x}-\mathbf{y}) = \mathbf{0}$$ since the kernel is trivial we must have $\mathbf{x}-\mathbf{y} = \mathbf{0}$ so that the mapping is injective.

If conversely $L$ is injective, then since $L(\mathbf{0}) = \mathbf{0}$ this automatically implies $$L(\mathbf{x}) = \mathbf{0} = L(\mathbf{0})\implies \mathbf{x} = \mathbf{0}$$

Applying the above result should yield what you need.

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 Thank you for clearing this up for me. – Imray Oct 5 '12 at 15:43

No, incorrect. Being one-to-one means that whenever $T(x) = T(y)$ for some $x$ and $y$ we must have that $x = y$. (The property you referr to is being 'onto').

Now, in your case, we have (as $T$ is linear and hence $T(0) = 0$, that $$T \left(\sum_{i=1}^n a_is_i\right) = 0 = T(0) \Rightarrow \sum_{i=1}^n a_i s_i = 0 .$$

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Suppose $T$ is not one-to-one (I prefer "injective") so that there exist $v_1\neq v_2$ with $T(v_1)=T(v_2)$. Then by linearity $T(v_1-v_2)=0$, so that $\{v_1-v_2\}$ is a linearly independent subset whose image $\{0_W\}$ is linearly dependent. This proves the "if" part.

Now suppose $T$ is injective, $v_1,\ldots,v_k$ a linearly independent, and $c_iT(v_1)+\cdots+c_kT(v_k)=0$ a relation between their images. Then by linearity $T(c_iv_1+\cdots+c_kv_k)=c_iT(v_1)+\cdots+c_kT(v_k)=0_W=T(0_V)$ (the last equality is what you missed), so by injectivity $c_iv_1+\cdots+c_kv_k=0_v$ and by linear independence of the $v_i$ all coefficients $c_i$ have to be $0$; this shows that $T(v_1),\ldots,T(v_k)$ are linearly independent. This proves the "f" part.

I did write the proof so that the independent set was finite, but your proposed answer shows how that can be avoided.

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