# Why doesn't a linearly independent set of image vectors imply an injection?

While researching a question I had, I came across this post. Without reading the answers, I started working on it myself and eventually came up with a proof for this statement:

Let $f \in \mathcal{L}(V, W)$ and $\{v_1, \ldots v_n\}$ be a linearly independent set in $V$. Then $\{f(v_1), \ldots f(v_n)\}$ is linearly independent if and only if $f$ is injective.

Once I finished, I started reading the responses and saw that, actually, the statement is not true. Specifically, B. S. gives a counterexample of a linear map that's not injective yet maps a linearly independent set to another linearly independent set. I assume that means there's something wrong with my proof. As for what, I have no idea. So here's my proof:

Assume $f$ is not injective. Then there exists $v, \, v' \in V$ such that $v \neq v'$ but $f(v) = f(v')$. Write these two vectors as $v = a_1 v_1 + \ldots + a_n v_n$ and $v' = b_1 v_1 + \ldots + b_n v_n$, where at least one $a_i \neq b_i$. Since $f(v) = f(v')$,

\begin{aligned} \ \\ f(a_1 v_1 + \ldots + a_n v_n) &= f(b_1 v_1 + \ldots + b_n v_n) \ \\ a_1 f(v_1) + \ldots + a_n f(v_n) &= b_1 f(v_1) + \ldots + b_n f(v_n) \ \\ (a_1 - b_1)f(v_1) + \ldots + (a_n - b_n)f(v_n) &= 0 \ \end{aligned}

where the coefficient for at least one $f(v_i)$ is not $0$. Therefore, $\{f(v_1), \ldots f(v_n)\}$ is linearly dependent. By the contrapositive, then, if $\{f(v_1), \ldots f(v_n)\}$ is linearly independent, $f$ must be injective.

So where did I go wrong?

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“Write these ... as ..” assumes that the $v_i$ span $V$ (and hence form a basis). – Carsten S Mar 23 '14 at 15:03

"$f \in \mathcal{L}(V, W)$" implies that even vectors not in the span of $\{v_1,\ldots,v_n\}$ have images under $f$. Suppose one of thos, $v$, has the same image that $v_1$ has. Then $f$ is not injective and you can't write $v$ as a linear combination of $v_1,\ldots,v_n$. If you can do that step in your proof, then I think it would be otherwise correct.