Condition on minors of a matrix to check when linear transformation is injective I have been reviewing criteria for bijectivity of linear mappings because I am going to have to be able to proof similar results on an upcoming exam and use various theorems to deduce if a linear transformation is one-one, onto, ect.  One theorem I am not so sure how to prove is below. I was wondering if it came up in a standard textbook on linear algebra because it was presented without proof in a set of lecture notes I am using.
Let $V,W$ be vector spaces over some field $F$ and suppose $f$ is a linear transformation between these spaces.

How do we show $f:V\rightarrow W $ is one to one if and only $\dim(V) \leq \dim(W)$ and if $A$ is the matrix corresponding to the linear transformation $f$ (with respect to the standard basis of $V$ and $W$) then there exists no scalar $c \neq 0$ such that the products by $c$ of all minors of order $\dim(V)$ of $A$ are zero.

 A: A linear mapping $f$ is one-to-one, iff $\ker f=0$. The rank-nullity theorem tells you that $$\dim \ker f=0 \Leftrightarrow \dim\; \mathrm{im}\; f= \dim V.$$
Therefore we know that $f$ is one-to-one, iff the matrix $A$ has rank $\dim V$. More specifically (if the usual notation is adopted, and the images of the basis vectors are columns of $A$) we know that this happens, iff the column space of $A$ has the same dimension as the space $V$, iff the row space of $A$ has the same dimension as $V$.
The row space has of $A$ always has a basis consisting of rows of $A$. Remember that each row of $A$ has $\dim V$ components. For there to be $\dim V$ linearly independent rows in $A$:
A) The matrix $A$ must have at least $\dim V$ rows. IOW $\dim W\ge \dim V$.
B) The minor gotten by using that set of rows has to be non-zero.
Clearly conditions $A$ and $B$ imply that the row space of $A$ has dimension $\dim V$,
so we can go the chain backwards also.
I still don't get what's the deal with scalar multiples of minors. Are you sure that this is about vector spaces over fields?
