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In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?

I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated

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  • $\begingroup$ One of your fractions should have a denominator with an a in it; perhaps someone with higher karma than me can edit that. $\endgroup$ Commented Aug 30, 2010 at 19:23
  • $\begingroup$ @Asaf: Sl_n are those of det 1 typically. It's a simple fact that an integer matrix $A$ is in $Gl_n(\mathbb{Z})$ iff $det A = \pm 1$, though. $\endgroup$ Commented Aug 30, 2010 at 19:36
  • $\begingroup$ @Jason: Take an integer $d\ne 1$ and take $dI_n$ for the scalar matrix of which all the diagonal values are $d$. The determinant is non-zero, therefore it is in $GL_n$, it is an integer matrix, and it is not of determinant $1$. $\endgroup$
    – Asaf Karagila
    Commented Aug 30, 2010 at 23:30
  • $\begingroup$ @Asaf. It is in $GL_n(\mathbb{R})$, but it is not in $GL_n(\mathbb{Z})$. Said more explicitly, the matrix you are describing is invertible if real (or even rational) values are allowed, but if only integer values are allowed, it's not invertible. $\endgroup$ Commented Aug 31, 2010 at 0:02
  • $\begingroup$ Thanks for the comments, I fixed all the errors, pointed out. I'll try to get better at proof-reading $\endgroup$
    – WWright
    Commented Aug 31, 2010 at 0:57

4 Answers 4

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As others have pointed out, the key here is that the determinant of $M^{-1}$ has to be equal to $\frac{1}{det(M)}$ on the one hand, by the properties of the determinant, and also that since $M$ and $M^{-1}$ both have integer coefficients then their determinants must be integers. So you need an integer $e=det(M)$ such that both $e$ and $\frac{1}{e}$ are integers, and the only possibility is $e=\pm 1$.

If you don't consider arguments using the determinant to be "elementary", perhaps the following will do: note that $ad-bc$ must divide each of $a$, $b$, $c$, and $d$. Let $D=ad-bc$; we can then write $a=Da'$, $b=Db'$, $c=Dc'$, $d=Dd'$; then $D = ad-bc=D^2(a'd'-b'c')$, so $D=D^2(a'd'-b'c')$. Therefore, since $D\neq 0$, we must have $D(a'd'-b'c')=1$, so $D|1$, hence $D=ad-bc=\pm 1$, which is what you wanted to show.

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    $\begingroup$ Your argument could be shortened by referring to the coefficients of the inverse matrix provided in the question itself: computing their determinant shows the determinant of the inverse matrix equals 1/(ad-bc) and we're done, because determinants of integral matrices are obviously integral. $\endgroup$
    – whuber
    Commented Aug 30, 2010 at 20:42
  • $\begingroup$ I assume you are referring to the first part, the one that uses determinants? Yes; using determinants is the nice, elegant, easy, generalizable way of doing it. The second part, on the other hand, does not talk about determinants at all, just about how requiring the coefficients of the inverse (which can be found by solving a system of linear equations over $\mathbb{Q}$ without using determinants at all) to be integers forces the quantity ad-bc to be either 1 or -1. $\endgroup$ Commented Aug 31, 2010 at 0:45
  • $\begingroup$ Sure your second part talks about determinants, Arturo: it's entirely about the expression D = ad - bc, which clearly is integral whenever a, b, c, d all are. A simple calculation establishes that the corresponding expression for the inverse (as explicitly given in the question) equals D/D^2 = 1/D, whence D must be a unit, QED. $\endgroup$
    – whuber
    Commented Aug 31, 2010 at 14:27
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    $\begingroup$ Fair enough; I was trying to address the part of the question where WWright is asking why can we not have all the given fractions integers with the denominator not equal to $1$ and $-1$. Even if you did not know they came from matrices or inverse matrices, or knew nothing whatsoever about matrices, you can still deduce that the denominator must be either $1$ or $-1$. $\endgroup$ Commented Aug 31, 2010 at 14:54
  • $\begingroup$ Brilliant! I love the 2nd argument, treating just the fractions - fascinating how the introduction of new symbols that add no new information make the argument so clear. Where did you learn about this kind of approach? Also, what is meant by the notation “D|1” ? $\endgroup$
    – Rax Adaam
    Commented Dec 15, 2020 at 20:34
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The map $\det:M_2(\mathbb Z)\to\mathbb Z$ is multiplicative, and the determinant of the identity matrix is $1$. It follows from this that the determinant of any inversible element of $M_2(\mathbb Z)$ must be an inversible element of $\mathbb Z$. There are only two such elements in $\mathbb Z$, v.g. $1$ and $-1$.

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HINT $\;\;$ Multiplicative maps preserve units: $\rm\; MN = 1 \;\;\Rightarrow\;\; d(M)\:d(N) = 1$

NOTE $\rm\;\; d(1) = 1\;$ via apply $\rm d$ to $1\cdot 1 = 1\:$ then cancel $\rm d(1)\ne 0,$ valid since $\mathbb Z$ has cancellation.

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The determinant of the inverse is the inverse of the determinant, whence the determinant must be an integral unit.

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