$A\in M_{m\times n}(k)$ is one one iff row rank is $n$.
Here $k$ is a field.
My Attempt 1):
$A$ is $1-1$
$\iff$ker $A$ = $\{0_{m\times1}\}$
$\iff$All cols are linearly independant
$\iff$Col Rank is $n$
My Attempt 2):
row rank is $n$
$\iff \{R_1,R_2,\cdots,R_m\}$ has $n$ indep row vectors
In both these attempts I'm not able to proceed because matrix A is defined as tranformation from column vectors $k^n \to k^m$. If it were designed as transformation from row vector $k^m\to k^n$ Then I'd be done. However, then I'd have problem with $A$ is onto iff col rank is $m$
Please note that this is a homework and we are not allowed to utilize row rank = col rank.
Edit:
Attempt 3 I guess: (Continuation of attempt 2 actually)
$R_i\in (k^n)^*$
Thus $Ax = [R_1x\,\,\,R_2x\,\,\,\cdots\,\,\,R_mx]^T$
Now, since $R_i$ have $n$ lin indep vectors $(n\le m)$, can it be shown that if $R_ix = 0\,\forall i\implies x=0$?
Edit 2: I did it. I'll Add it as an answer
Row rank is $n$
$\iff [R_1\,\,R_2\,\,\dots\,\,R_m]^T$ has $n$ indep rows
$\iff \left([R_1\,\,R_2\,\,\dots\,\,R_m]^Tx=0\implies x=0\right)$
$\iff R_ix=0$ for $i=1,2\dots ,n$ Assuming first $n$ rows are indep
$\iff e_i^*x=0$ for $i=1,2,\dots,n$
$\iff x =0_{n\times1}$
$\iff ker A =$ singleton set $\{x\}$
$\iff A$ is $1-1$
Note: $2^{nd}$ iff statement holds because if it had only $n-1$ indep rows, it is possible that $x \ne 0$ because of the way dual basis are defined. Also $R_i\in (k^n)^*$