# Images of basis vectors under injective linear map form a linearly independent set

I missed this question on a quiz:

Prove that if $\{v_1, \ldots v_n \}$ is a basis for $V$ and $f\,:\,V\rightarrow W$ is an injective linear map, then $\{f(v_1), \ldots f(v_n)\}$ is linearly independent in W.

Apparently, my original proof was backwards: I started with $0\in V$ and went from there. In his comments, my professor gave me the first line: Let $0 = a_1 f(v_1) + \ldots + a_n f(v_n)$, with $a_i \in \mathbb{F}$. Here is how I proceeded:

Since $f$ is linear, we can write $0=f(a_1 v_1 + \ldots + a_n v_n)$. Therefore, $a_1 v_1 + \ldots + a_n v_n \in \text{null}(f)$. Since $f$ is injective, $a_1 v_1 + \ldots +a_n v_n = 0$. And since $\{v_1, \ldots v_n \}$ is linearly independent, $a_i = 0$ for all $a_i$.

This implies that for any $v \in V$ such that $v = b_1 v_1 + \ldots b_n v_n$, if there exists $b_i \neq 0$, then $v \notin \text{null}(f)$. Therefore, whenever $0 = b_1 f(v_1) + \ldots + b_n f(v_n)$, $b_i = 0$, for all $b_i$. In other words, $\{f(v)1), \ldots f(v_n)\}$ is linearly independent.

I think this works, but for some reason I feel like I'm waving my hands a little in the penultimate step; maybe it's just that I haven't built up enough intuition yet. Is there anything that's missing? Is there anything that could be more concise?

You should've started with $0 \in W$ in which case, your proof would be complete and valid when you said: "And since $\{v_1, \ldots v_n \}$ is linearly independent, $a_i = 0$ for all $a_i$".
"This implies that for any $v \in V$ such that $v = b_1 v_1 + \ldots b_n v_n$, if there exists $b_i \neq 0$, then $v \notin \text{null}(f)$. Therefore, whenever $0 = b_1 f(v_1) + \ldots + b_n f(v_n)$, $b_i = 0$, for all $b_i$. In other words, $\{f(v)1), \ldots f(v_n)\}$ is linearly independent."
is extraneous to the proof and unclear. In particular, the sentence "Therefore, whenever $0 = b_1 f(v_1) + \ldots + b_n f(v_n)$, $b_i = 0$, for all $b_i$", is not a complete sentence.
• Right, I could have be clearer about $0 \in W$. I left it out, though, since $f \in \mathcal{L}(V,W)$; so if $0$ is a linear combination of the image vectors, that means we're talking about $W$. Beyond that, why is the second paragraph extraneous? I don't address the independence of the image vectors until that point. (That one sentence is complete, by the way; I'm an English teacher :).) – dmk Mar 20 '14 at 17:03
• @dmk Sorry. I misread that. Yeah, it looks correct given you start with $0 \in W$ as you said. However, note that the statement "This implies that for any $v \in V$ …. " is a restatement of the fact that $f$ is injective and $\{v_1 , \dots, v_n\}$ is linearly independent. – Ink Mar 20 '14 at 20:15
• @dmk I'm saying that it's enough to show that $a_i = 0$ for all $i$. That completes the argument. – Ink Mar 20 '14 at 20:18
• Sorry for the delay. I've reread the second graf and I see what you mean: I'm following a thread rather than tying disparate things together. It wasn't clear to me at the time that my original vector was the only way to write $0$ in terms of the image vectors. Thanks! – dmk Mar 21 '14 at 16:45