Set is closed as for multiplication of matrices

mathmari

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MHB Site Helper
Hey!!

We have the matrices \begin{equation*}s:=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \ d:=\frac{1}{2}\begin{pmatrix}-1 & -\sqrt{3} \\ \sqrt{3} & -1\end{pmatrix}\end{equation*} and the points \begin{equation*}p:=\begin{pmatrix}2 \\ 0 \end{pmatrix}, \ q:=\begin{pmatrix}-1 \\ \sqrt{3} \end{pmatrix}, \ r:=\begin{pmatrix}-1 \\ -\sqrt{3} \end{pmatrix}\end{equation*}

I draw the points $p, q, r$ and calculate also the points $sp, sq, sr, dp, dq, dr$ and I noticed that $s$ is a reflection as for the $x$-axis and $d$ is a rotation of $\frac{2\pi}{3}$.

1. Consider $G:=\{d, d^2, d^3, sd, sd^2, sd^3\}\subseteq \mathbb{R}^{2\times 2}$ and show that $G$ is closed as for multiplication of matrices, i.e. $gh\in G$ for all $g,h\in G$.
2. Show that the elements of $G$ are invertible and for each $g\in G$ there is $g^{-1}\in G$.
3. What is the geometric interpretation of $G$ ?
4. Let $z=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ and $H:=G\cup \{zg\mid g\in G\}$. Show that $H\subseteq \mathbb{R}^2$ is closed as for multiplication of matrices.

Let's start with question 1. Do we have to consider all possible combinations of the elements of $G$ and show that their product is again in $G$ ?
For example, do we have to do the following?
\begin{align*}d\cdot d^2&=d^3\in G \\ d\cdot d^3&=d^4=4-\text{times rotation about }120^{\circ}=\text{ratotaion about }480^{\circ}=\text{rotation about }360^{\circ}+120^{\circ}=\text{rotation about }120^{\circ}\\ & =d\in G \\ d\cdot s\cdot d&=\text{rotation about }120^{\circ}\text{ then reflection about x-axis and then rotation about }120^{\circ}\\ & =\text{reflection abour ax-axis}=s\in G\end{align*}
Or is there is shorter way?

At question 2 do we have to show that the matrices are invertible? Or do we have to consider what the matrices represent?

Klaas van Aarsen

MHB Seeker
Staff member
Hi mathmari !!

The short way is to recognize that a reflection and a rotation by $\frac{2\pi}3$ generate the symmetries of a triangle $D_3$, which is know to be a group.

Alternatively, first consider that $d^3$ is the identify matrix, and that $s^2$ is also the identity matrix.
All elements are either of the form $s d^i$ or of the form $d^j$. If we for instance multiply them in this order, we get $s d^i d^j = s d^{(i+j) \bmod 3}$, which is again an element.
We have the same argument for other combinations, and the only thing we need to check, is what happens when we evaluate $d s$. You should find that is equal to $s d^2$, after which the same argument applies.

At question 2 do we have to show that the matrices are invertible? Or do we have to consider what the matrices represent?
Both $s$ and $d$ are invertible aren't they?
We can verify by noticing that $s^2=\text{id}$ and $d^3=\text{id}$.
Consequently every product of them is also invertible.

mathmari

Well-known member
MHB Site Helper
Alternatively, first consider that $d^3$ is the identify matrix, and that $s^2$ is also the identity matrix.
All elements are either of the form $s d^i$ or of the form $d^j$. If we for instance multiply them in this order, we get $s d^i d^j = s d^{(i+j) \bmod 3}$, which is again an element.
We have the same argument for other combinations, and the only thing we need to check, is what happens when we evaluate $d s$. You should find that is equal to $s d^2$, after which the same argument applies.
We have that $m=ds \Rightarrow ms=ds^2 \Rightarrow ms=d \Rightarrow d^2ms=d^3 \Rightarrow d^2ms=1_{\mathbb{R}^{2\times 2}}$. How do we get from there that $m=sd^2$ ?

Both $s$ and $d$ are invertible aren't they?
We can verify by noticing that $s^2=\text{id}$ and $d^3=\text{id}$.
Consequently every product of them is also invertible.
And to show that the inverse of each element is also in $G$ do we have to check again all possible forms of the elements?

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Klaas van Aarsen

MHB Seeker
Staff member
We have that $m=ds \Rightarrow ms=ds^2 \Rightarrow ms=d \Rightarrow d^2ms=d^3 \Rightarrow d^2ms=1_{\mathbb{R}^{2\times 2}}$. How do we get from there that $m=sd^2$ ?
We can do the matrix multiplication.

And to show that the inverse of each element is also in $G$ do we have to check again all possible forms of the elements?
That is not necessary.
We have the matrix property that the product of invertible matrices is also invertible.
So it suffices to verify that both $s$ and $d$ are invertible. Then we can conclude that all elements are invertible.

mathmari

Well-known member
MHB Site Helper
We have the matrix property that the product of invertible matrices is also invertible.
So it suffices to verify that both $s$ and $d$ are invertible. Then we can conclude that all elements are invertible.
We have that $s^2=\text{id}$ so the inverse of $s$ is $s$ itself, but $s$ is not in $G$, but we have that $s=sd^3\in G$, si this is the inverse in $G$.

The inverse of $d$ is $d^2$ since $d^3=\text{id}$.

Since all the other elements are products of these elements it follows that their inverses will also be in $G$.

Is everything correct?

Klaas van Aarsen

MHB Seeker
Staff member
We have that $s^2=\text{id}$ so the inverse of $s$ is $s$ itself, but $s$ is not in $G$, but we have that $s=sd^3\in G$, si this is the inverse in $G$.

The inverse of $d$ is $d^2$ since $d^3=\text{id}$.

Since all the other elements are products of these elements it follows that their inverses will also be in $G$.

Is everything correct?
Yep.

mathmari

Well-known member
MHB Site Helper
As for question 3 : What is the geometric interpretation of $G$ ? That it contains the matrices for reflections and rotations?

As for question 4 : Let $g=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ then $zg=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}=\begin{pmatrix}-a & -b \\ -c & -d\end{pmatrix}$ and $gz=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}=\begin{pmatrix}-a & -b \\ -c & -d\end{pmatrix}$ and so we get $zg=gz$.
So every possible element of $G$ is $d^j$ or $sd^i$. Since $zg=gz$ we have to check only one order of multiplication, right?

Is the geometric interpretation of $H$ the reflection and rotation?

Klaas van Aarsen

MHB Seeker
Staff member
As for question 3 : What is the geometric interpretation of $G$ ? That it contains the matrices for reflections and rotations?
Yes, and more specifically the reflections and rotations that make up the symmetries of the triangle.
It's the dihedral group $D_3$.

As for question 4 : Let $g=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ then $zg=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}=\begin{pmatrix}-a & -b \\ -c & -d\end{pmatrix}$ and $gz=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}=\begin{pmatrix}-a & -b \\ -c & -d\end{pmatrix}$ and so we get $zg=gz$.
So every possible element of $G$ is $d^j$ or $sd^i$. Since $zg=gz$ we have to check only one order of multiplication, right?

Is the geometric interpretation of $H$ the reflection and rotation?
Matrix $z$ represents a rotation by $\pi$. The composition $zd^2$ is therefore a rotation by $\frac\pi 3$.
So we get twice as many rotations.
The result is the group of symmetries of a hexagon - the dihedral group $D_6$.

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mathmari

Well-known member
MHB Site Helper
Alternatively, first consider that $d^3$ is the identify matrix, and that $s^2$ is also the identity matrix.
All elements are either of the form $s d^i$ or of the form $d^j$. If we for instance multiply them in this order, we get $s d^i d^j = s d^{(i+j) \bmod 3}$, which is again an element.
We have the same argument for other combinations, and the only thing we need to check, is what happens when we evaluate $d s$. You should find that is equal to $s d^2$, after which the same argument applies.
Are all possible products the following?

- $sd^isd^i=1_{\mathbb{R}^{2\times 2}}=d^3\in G$
- $sd^id^j=s d^{(i+j) \bmod 3}\in G$
- $d^jsd^i=sd^i\in G$
- $d^jd^j=d^{(i+j) \bmod 3}\in G$

Klaas van Aarsen

MHB Seeker
Staff member
Are all possible products the following?

- $sd^isd^i=1_{\mathbb{R}^{2\times 2}}=d^3\in G$
The products $sd^isd^j$ with $i\ne j$ seem to be missing.

- $sd^id^j=s d^{(i+j) \bmod 3}\in G$
- $d^jsd^i=sd^i\in G$
What happened to $j$?

- $d^jd^j=d^{(i+j) \bmod 3}\in G$

mathmari

Well-known member
MHB Site Helper
The products $sd^isd^j$ with $i\ne j$ seem to be missing.

What happened to $j$?
It should be :

- $sd^isd^j=1_{\mathbb{R}^{2\times 2}}=d^3\in G$
- $sd^id^j=s d^{(i+j) \bmod 3}\in G$
- $d^jsd^i=sd^k\in G$ for some $k\in \{1,2,3\}$
- $d^id^j=d^{(i+j) \bmod 3}\in G$

Or not?

Klaas van Aarsen

MHB Seeker
Staff member
- $sd^isd^j=1_{\mathbb{R}^{2\times 2}}=d^3\in G$
Suppose $i=3$ and $j=1$, then $sd^isd^j=d\ne d^3$ isn't it?

mathmari

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MHB Site Helper
Suppose $i=3$ and $j=1$, then $sd^isd^j=d\ne d^3$ isn't it?
Ah yes! So it will be a power of $d$, or not?

Staff member

mathmari

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MHB Site Helper
Ok! So we get:

- $sd^isd^j=d^k\in G$ for some $k\in \{1,2,3\}$
- $sd^id^j=s d^{(i+j) \bmod 3}\in G$
- $d^jsd^i=sd^k\in G$ for some $k\in \{1,2,3\}$
- $d^id^j=d^{(i+j) \bmod 3}\in G$

right?

Staff member
Yep.

mathmari

Well-known member
MHB Site Helper
Great!! Thank you!!