Suppose that $T(v):=Av$ is one-to-one from $F^n$ to $F^n$ (A is a matrix) That is not surjective.
Does that mean that $A$ is regular?
On the one hand:
$T$ is One-To-One $\implies\ker(T)=\{0\} \implies$ only $v=0$ solves $Av=0 \implies A$ is regular
On the other hand:
$T$ is not surjective $\implies$ there exists $b$ such that for each $v$, $T(v)=Av\neq b$. Therefore, $A$ is not regular.
Which of these proofs is wrong and why?
Thanks!
The assumption that there exists a one-to-one linear transformation that is not surjective from $F^n$ to $F^n$ is false. You can see this by using the rank-nullity theorem and the fact that the kernel is trivial. Since they don't exist both statements are vacuously true.