So, I'm attempting to write a proof for the following theorem:
Let $f:V \to W$ be an isomorphism. If $(v_1,v_2,\ldots,v_r)$ is an r-tuple of linearly independent vectors in $V$, then the $r$-tuple of vectors $(f(v_1),f(v_2),\ldots,f(v_r))$ in $W$ is also linearly independent.
Proof Attempt:
Let $I = \{1,2,\ldots,r\}$. Consider the following linear combination of vectors:
$\sum_{k = 0}^{r} \alpha_k f(v_k) = 0$
By the linearity of $f$, the following chain of implications hold:
$\sum_{k=0}^{r} \alpha_k f(v_k) = 0$
$\implies \sum_{k=0}^{r} f(\alpha_k v_k) = 0$
$\implies f[\sum_{k=0}^{r} \alpha_k v_k] = 0$
$\implies \sum_{k=0}^{r} \alpha_k v_k \in \ker(f)$
By a previously proved result, we know that since $f$ is bijective, $\ker(f) = \{0\}$. Hence:
$\sum_{k=0}^{r} \alpha_k v_k = 0$
$\implies \forall k \in I: \alpha_k = 0$
That proves that the initial list of vectors in $W$ is a linearly independent list.
Can someone check if my argument above is correct? It feels correct but I don't think I used the bijectivity of $f$. Like, I feel like there's a stronger statement that can be made, where f can be injective only and the above statement still holds. Also, I'd appreciate any criticism whatsoever on the way I've written my proof.