2
$\begingroup$

Consider $$ M:=\Big\{ \left[ {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right] :n \in \mathbb Z \Big\}$$

Is $M$ a group under the operation of matrix multiplication?

I know that we have to show that $M$ is closed under matrix multiplication, which is associative in $M$, and there is an identity element in M and each element in $M$ has an inverse. I think that because the determinant will be positive, it is closed under matrix multiplication.

Moreover, as matrix multiplication is associative in general, it holds true here. and: $\ e= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right] $ is a clear identity. Further: any element $\ q= \left[ {\begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} } \right] \in M, \ e= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right] $ is an inverse(I think??) And therefore it forms a group. QED

Any Thoughts?

$\endgroup$
  • 2
    $\begingroup$ The set of matrices of the form $\pmatrix{1&n\\0&1}$ for $n\in\Bbb Z$ is a group. You need to prove that the inverse of each such matrix is also in the set. $\endgroup$ – Angina Seng Feb 17 '18 at 3:40
  • 3
    $\begingroup$ Just pick generic matrices $\pmatrix{1 & n \\ 0 & 1}$ and $\pmatrix{1 & m \\ 0 & 1}$, and compute their product -- I think you'll have a much better understanding of what you're up against. $\endgroup$ – pjs36 Feb 17 '18 at 3:42
  • 1
    $\begingroup$ $e$ is not an inverse-- it's the identity element. $\endgroup$ – coffeemath Feb 17 '18 at 3:44
4
$\begingroup$

Closure under matrix multiplication:

$$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & m \\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & m+n \\ 0 & 1\end{bmatrix}$$

Since $m+n\in \mathbb{Z}$, it is closed under matrix mulitplication. That is if $A \in M$ and $B \in M$, then $AB \in M$.

  • It is irrelevant to the determinant.

Identity and associativity inherits from the normal matrix operations.

  • matrix $q$ again is irrelevant, afterall, in general , $q \notin M$.

For matrix inverse, try to evaluate $$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & -n \\ 0 & 1\end{bmatrix}$$

$\endgroup$
1
$\begingroup$

This looks isomorphic to the group $(\mathbb Z,+)$ to me.

$\endgroup$
0
$\begingroup$

Another way to prove that $M$ is a group is to notice that $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$(you can check this by induction)

So $M=<\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}>$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.