Set is closed as for multiplication of matrices

• MHB
• mathmari
In summary, the points $p, q, r$ are in a position such that $s=dp=dq=dr=1_{\mathbb{R}^{2\times 2}}$.
mathmari
Gold Member
MHB
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?
Or is there is shorter way? :unsure:

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

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.

Klaas van Aarsen said:
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 $m=sd^2$ ? :unsure:
Klaas van Aarsen said:
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? :unsure:

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mathmari said:
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.

Klaas van Aarsen said:
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? :unsure:

mathmari said:
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. (Nod)

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? :unsure:

mathmari said:
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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Klaas van Aarsen said:
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?

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

:unsure:

mathmari said:
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 $j$?

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

Klaas van Aarsen said:
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? :unsure:

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

Klaas van Aarsen said:
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? :unsure:

mathmari said:
Ah yes! So it will be a power of $d$, or not?
Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

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? :unsure:

Yep. (Nod)

Great! Thank you! (Handshake)

1. What does it mean for a set to be closed as for multiplication of matrices?

When a set is closed as for multiplication of matrices, it means that when two matrices from the set are multiplied together, the resulting matrix is also in the set. In other words, the set is closed under the operation of matrix multiplication.

2. How is closure under matrix multiplication important in linear algebra?

Closure under matrix multiplication is important in linear algebra because it allows for the creation of new matrices by combining existing ones. This is essential for solving systems of linear equations and for performing other operations such as matrix inversion and diagonalization.

3. Can a set be closed for addition but not for multiplication of matrices?

Yes, a set can be closed for addition but not for multiplication of matrices. This means that when two matrices from the set are added together, the resulting matrix is also in the set, but when they are multiplied together, the resulting matrix may not be in the set.

4. How can I determine if a set is closed as for multiplication of matrices?

To determine if a set is closed as for multiplication of matrices, you can multiply every possible combination of matrices from the set and check if the resulting matrices are also in the set. If they are, then the set is closed for multiplication of matrices.

5. What are the benefits of working with a closed set for multiplication of matrices?

Working with a closed set for multiplication of matrices allows for simpler and more efficient calculations. It also ensures that the results of matrix operations stay within the set, which is important for maintaining the properties and characteristics of the original set.

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