# 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.

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This holds even when $k$ is a commutative ring with unity: artofproblemsolving.com/Forum/viewtopic.php?f=349&t=124137 – darij grinberg Nov 8 '11 at 1:13

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?

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Does your hint apply to both directions of the proof or just one side? In particular I do not see how the rank nullity theorem imply the condition about the minors. – user7980 Nov 6 '11 at 23:37
@user7980: What's the relation between the minors and the rank of the matrix? – Jyrki Lahtonen Nov 7 '11 at 5:19
I am not aware of a theorem relating minors to matrix rank could you please elaborate on what fact you are talking about and if this is just a consequence of rank nullify – user7980 Nov 7 '11 at 7:39
It is a rather basic result in linear algebra that the rank of the matrix equals the size of its largest non-vanishing minor. Never heard of it? – Jyrki Lahtonen Nov 7 '11 at 7:57
An argument here: by rank-nullity $\ker f=0\Leftrightarrow \dim f(V)= \dim V$, so $rank\;A=\dim V$. If you (or your book) sets things up in the way that the images of basis vectors of $V$ are listed as columns of $A$, then the columns must be linearly independent. Therefore there are $\dim V$ linearly independent rows. Therefore $\dim W\ge\dim V$ and those rows give a non-vanishing minor. – Jyrki Lahtonen Nov 7 '11 at 8:06