$V$ and $W$ are vector spaces over $F$.
$(v_1,\ldots,v_n)$ is a basis of $V$
$f: V \to W$ is a linear map
I have shown that $(f(v_1),\ldots,f(v_n))$ is linearly independent if $f$ is injective (from def. of linear independence and def. of $\operatorname{ker}(f)$)
I am stuck on how to prove the other direction ie suppose $(f(v_1),\ldots,f(v_n))$ is linearly independent, show $f$ is injective.
Best Answer
Take $x \in V$ and suppose that $f(x)=0$. As $(v_1, \dots,v_n)$ spans $V$, it exists $\lambda_1, \dots \lambda_n$ such that $$x=\lambda_1 v_1+ \dots +\lambda_n v_n$$
Which implies $$0=f(x)=\lambda_1 f(v_1)+ \dots +\lambda_n f(v_n)$$
and as $(f(v_1), \dots,f(v_n))$ is supposed to be linearly independent, we get $\lambda_1=\dots=\lambda_n=0$ and finally $x=0$, proving that $\ker f$ is reduced to the zero vector and that $f$ is injective.