Show that the matrix $A$ with integer entries is injective on the reals to $\mathbb{R}^m$ iff it is injective on the integer lattice.

Show that the $m \times n$ matrix $A$ with integer entries is an injective linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ iff it is injective as a linear map from $\mathbb{Z}^n$ to $\mathbb{Z}^m$.

One direction is quite obvious, it is injective on the reals, the kernel is empty and intersecting that with $\mathbb{Z}^n$ is still empty so the map restricted to the lattice has trivial kernel and is therefore injective.

In the other direction I can't seem to make progress. I have been trying to think about it in two different ways. The first, consider the columns of the matrix. For it to be injective as a map between lattices, these columns need to be linearly independent, i.e. they satisfy no non trivial relation ($\sum r_iv_i =0$ implies $r_i = 0$). However, the key is that they don't satisfy an non trivial relation over the integers, but this may not be the case over the reals; there may be some non-zero real numbers $r'_i$ that cause $\sum r_iv_i =0$ and thus the map is not injective as a map between real spaces. This is where we need to use the fact that entries in the column vectors are also integers. Clearly injectivity over integers implies it for the rationals because if we could find special rationals $r'_i$ that cause $\sum r_iv_i =0$ we could just clear denominators and derive a contradiction.

The other way I've been trying to think about it is by proving the converse. Suppose the map wasn't injective we have to show that it can't be both composed of integer entries and injective as map between lattices. For it to be injective as a lattice map though would mean that the none of standard basis elements of $\mathbb{R}$ are in the kernel, and that no integer (or rational and just clear denominators) combination of them are in the kernel either, thus for something to be in the kernel it has to be linear combination of the basis vectors with at least one coefficient irrational. I think this proves it, but I feel unsatisfied, so I am not sure if something is wrong.

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If $A$ is injective from $Z^n$ to $Z^m$, then $A$ is injective from $Q^n$ to $Q^m$, and hence $A$ has a left inverse over $Q$....

Alternately, you can use the fact that the reduced row echelon form of $A$ has all the entries rational, thus if the kernel is non-trivial, it has some vector in $Q^n$. Then multiply by lcm of denominators...

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I already had the idea that $A$ was injective over the rationals. What does the fact there is an left inverse over the rationals do for the problem? –  Steven-Owen Apr 16 '12 at 2:18
@jake Well since $A$ has a left inverse over the rationals, it has a left inverse over $R$. And then if $Ax=0$ then $BAx=0$ where $B$ is the left inverse ;) –  N. S. Apr 16 '12 at 2:20
It's a matrix over a field: left inverse = right inverse. But in any case, any function which has a left inverse is injective: if f(x)=f(y) the x=g(f(x))=g(f(y))=y, where g is the left inverse of f. –  user1306 Apr 16 '12 at 2:21
@user1306 Nope, that is true only for square matrices. A $n \times m$ matrix can have a left/right inverse without having the other. Actually it has a left inverse iff the rank is $n$ and a right inverse iff the rank is $m$. This is basically the left invertible iff injective and right invertible iff surjective property for functions. –  N. S. Apr 16 '12 at 2:21
@jake If $B$ is the left inverse of $A$ over the rationals, it has rational thus real entries. If $BA=I$ over $Q$ is also true over $R$. ;) –  N. S. Apr 16 '12 at 2:57
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Tell me if this is a little too high-powered

Take an integral matrix $A$. You are then defining its action on $\mathbb{R}^n$ by $A(r_1,\cdots,r_n)=r_1 A(1,0,\cdots,)+\cdots+r_nA(0,\cdots,1)$--let's call this map on $\mathbb{R}^n$ $\widetilde{A}$. We then have the following commutative diagram

$$\begin{array}{ccc}\mathbb{R}\otimes_\mathbb{Z}\mathbb{Z}^n & \overset{1\otimes A}{\longrightarrow} & \mathbb{R}\otimes_\mathbb{Z}\mathbb{Z}^m\\ \big\uparrow & & \big\downarrow\\ \mathbb{R}^n & \underset{\widetilde{A}}{\longrightarrow} & \mathbb{R}^m\end{array}$$

where the vertical arrows are the usual isomorphisms. Thus, it suffices to show that $1\otimes A$ is injective, but $\mathbb{R}$ is torsion free, and so flat, and thus this follows from the injectivity of $A$.

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Ya, this is definitely too high powered. I see that it suffices to show that $1\otimes A$ is injective, but everything you said after that it was beyond my knowledge. Could you explain in more elementary terms? –  Steven-Owen Apr 16 '12 at 1:32
Which parts? So, you understand tensors? There is a general theorem which says (as a small portion) that tensoring $1\otimes A$ will be preserve the injectivity of $A$ if the thing you are tensoring with is the additive group of a (characteristic zero) field. Does that make sense to you? –  Alex Youcis Apr 16 '12 at 1:34
Yes, I am fine with tensors. I know that the reals when though of as a Z-module is flat so that it preserves the injectivity when tensoring, But this machinery is beyond the scope of what I am supposed to use (and beyond what I understand so far) so I was hoping to find something more elementary. –  Steven-Owen Apr 16 '12 at 2:41
@AlexYoucis Wow +1 This for me is the ultimate use of thermonuclear weapons!! –  fpqc Apr 16 '12 at 4:36
@BenjaminLim Not sure if I should take that as a compliment haha –  Alex Youcis Apr 16 '12 at 22:30
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