Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
“Write these ... as ..” assumes that the $v_i$ span $V$ (and hence form a basis). –  Carsten Schultz Mar 23 at 15:03

1 Answer 1

"$f \in \mathcal{L}(V, W)$" 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.