Symmetry Operations of a Cube: Geometric Descriptions and Matrix Representations

[mathmari]

Hey! :giggle:

Let $\displaystyle{W:=\left \{\begin{pmatrix}x\\ y\\ z\end{pmatrix}\in \mathbb{R}^3\mid x,y,z\in \{-1,1\}\right \}}$.

Draw the set $W$ in a coordinate system. Let $v=\neq w$ and $v,w\in W$. If they differ only at one coordinate connect these points by a line.

With this description we get a cube with vertices $(\pm 1 , \pm 1 , \pm 1)$, right? :unsure: Let \begin{equation*}d:=\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\ , \ s:=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \ , \ D:=\left \{\begin{pmatrix}e_1 & 0 & 0 \\ 0 & e_2 & 0 \\ 0 & 0 & e_3 \end{pmatrix} \mid e_i\in \{-1,1\}\right \}\end{equation*}
Show that for all $a\in \{d,s\}\cup D$ the map $w\mapsto aw$ is a symmetry of the cube and descibe the symmetry gemetrically.

Do we have tomultiply all vertices of the cube with each of these matrices? Or what are we supposed to do? :unsure: The following moves are the symmetries of the cube $W$:
1. Rotation at the axis $\mathbb{R}\begin{pmatrix}0 \\ 0\\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$.
2. Rotation at the axis $\mathbb{R}\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$.
3. Rotation at the axis $\mathbb{R}\begin{pmatrix}1 \\ 1\\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$.
4. Rotation at the axis $\mathbb{R}\begin{pmatrix}1 \\ -1\\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$.
5. Reflections to the plane, that is spanned by the vectors $\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$.

How can we find for these symmetries a matrix $a$ so that the map $w\mapsto aw$ describes each symmetry? :unsure:

 

[I like Serena]

Let \begin{equation*}d:=\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\ , \ s:=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \ , \ D:=\left \{\begin{pmatrix}e_1 & 0 & 0 \\ 0 & e_2 & 0 \\ 0 & 0 & e_3 \end{pmatrix} \mid e_i\in \{-1,1\}\right \}\end{equation*}
Show that for all $a\in \{d,s\}\cup D$ the map $w\mapsto aw$ is a symmetry of the cube and descibe the symmetry gemetrically.
Do we have tomultiply all vertices of the cube with each of these matrices? Or what are we supposed to do?

Hi mathmari!

A symmetry of an object in Euclidean space is an isometry (transformation that preserves distances) that maps the object to itself.
So it suffices if we verify if all matrices are isometries and if they map all vertices of the cube to vertices of the cube.

For instance the matrix $d$ is orthogonal (all columns are pairwise orthogonal and of unit length), which implies that it is an isometry.
Let $(x,y,z)$ be a vertex of the cube, meaning that $x,y,z\in \{-1,1\}$.
Then we have for instance:
\[ d\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}y \\ z \\ x \end{pmatrix} \]
The result has again all coordinates in $\{-1,1\}$, so $d$ maps all vertices of the cube onto the vertices of the cube.
Therefore $d$ is a symmetry of the cube.

The geometric descriptions are the classifations of the matrices as rotations around a specific axis with a specific angle, or as reflections in a specific plane.

The following moves are the symmetries of the cube $W$:

How can we find for these symmetries a matrix $a$ so that the map $w\mapsto aw$ describes each symmetry?

If we can find 3 independent vectors and their images, we can construct a matrix from that.
Can you find for instance the image of $(0,0,1)$ when rotated with transformation 1?
How about the image of $(1,0,0)$?

If we have the 3 independent vectors $u,v,w$ and there images $u',v',w'$, we get:
\[ au=u', av=v', aw=w' \implies a\begin{pmatrix}u&v&w\end{pmatrix}=\begin{pmatrix}u'&v'&w'\end{pmatrix}\implies a=\begin{pmatrix}u'&v'&w'\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}^{-1} \]

 

[mathmari]

For instance the matrix $d$ is orthogonal (all columns are pairwise orthogonal and of unit length), which implies that it is an isometry.
Let $(x,y,z)$ be a vertex of the cube, meaning that $x,y,z\in \{-1,1\}$.
Then we have for instance:
\[ d\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}y \\ z \\ x \end{pmatrix} \]
The result has again all coordinates in $\{-1,1\}$, so $d$ maps all vertices of the cube onto the vertices of the cube.
Therefore $d$ is a symmetry of the cube.

We consider now the matrix $s$, it is orthogonal (all columns are pairwise orthogonal and of unit length), which implies that it is an isometry.
Let $(x,y,z)$ be a vertex of the cube, meaning that $x,y,z\in \{-1,1\}$.
Then we have:
\[ s\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}x \\ z \\ y \end{pmatrix} \]
The result has again all coordinates in $\{-1,1\}$, so $s$ maps all vertices of the cube onto the vertices of the cube.
Therefore $s$ is a symmetry of the cube.

We consider now a matrix of $D$, say $D_i$, it is orthogonal (all columns are pairwise orthogonal and of unit length), which implies that it is an isometry.
Let $(x,y,z)$ be a vertex of the cube, meaning that $x,y,z\in \{-1,1\}$.
Then we have:
\[ D_i\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}\pm x \\ \pm y \\ \pm z \end{pmatrix} \]
The result has again all coordinates in $\{-1,1\}$, so $D_i$ maps all vertices of the cube onto the vertices of the cube.
Therefore $D_i$ is a symmetry of the cube.To see the geometric interpretation we have to see that graphically, or not? :unsure:

If we can find 3 independent vectors and their images, we can construct a matrix from that.
Can you find for instance the image of $(0,0,1)$ when rotated with transformation 1?
How about the image of $(1,0,0)$?

If we have the 3 independent vectors $u,v,w$ and there images $u',v',w'$, we get:
\[ au=u', av=v', aw=w' \implies a\begin{pmatrix}u&v&w\end{pmatrix}=\begin{pmatrix}u'&v'&w'\end{pmatrix}\implies a=\begin{pmatrix}u'&v'&w'\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}^{-1} \]

Let's consider the transformation 1, is the image of $(0,0,1)$ the $(0,0,0)$ ? :unsure:

 

[I like Serena]

To see the geometric interpretation we have to see that graphically, or not?

Graphing can help to understand what happens.

We need to classify the matrices as rotations and reflections.
A rotation has a vector on its axis that is transformed to itself, and a vector perpendicular to that axis is transformed to another perpendicular vector with a fixed angle between them, which is the angle of rotation.
A reflection has a vector that is normal to the plane of reflection that is transformed to its opposite. And a vector perpendicular to that normal vector is transformed to itself.

We can find which is which by for instance finding the eigenvalues and eigenvectors.

Or alternatively we can graph it and deduce what the geometric interpretation is.

Let's consider the transformation 1, is the image of $(0,0,1)$ the $(0,0,0)$ ?

I don't think so. (Shake)

The vector $(0,0,1)$ is the axis of rotation, which means that it is transformed to itself.

 

[mathmari]

Graphing can help to understand what happens.

We need to classify the matrices as rotations and reflections.
A rotation has a vector on its axis that is transformed to itself, and a vector perpendicular to that axis is transformed to another perpendicular vector with a fixed angle between them, which is the angle of rotation.
A reflection has a vector that is normal to the plane of reflection that is transformed to its opposite. And a vector perpendicular to that normal vector is transformed to itself.

We can find which is which by for instance finding the eigenvalues and eigenvectors.

I gt stuck right now. How are the eigenvalues and eigenvectors related?

For example for thematrix $d$ the eigenvalues are $\lambda_1=1$, $\lambda_2=\frac{1}{2}(-1+i\sqrt{3})$, $\lambda_3=\frac{1}{2}(-1-i\sqrt{3})$.The respective eigenvectors are $v_1=(1,1,1)$, $v_2=\left (\frac{1}{2}(-1+i\sqrt{3}), \frac{1}{2}(-1-i\sqrt{3}), 1\right )$, $v_3=\left (\frac{1}{2}(-1-i\sqrt{3}), \frac{1}{2}(-1+i\sqrt{3}), 1\right )$.

:unsure:

 

[I like Serena]

I gt stuck right now. How are the eigenvalues and eigenvectors related?

For example for thematrix $d$ the eigenvalues are $\lambda_1=1$, $\lambda_2=\frac{1}{2}(-1+i\sqrt{3})$, $\lambda_3=\frac{1}{2}(-1-i\sqrt{3})$.The respective eigenvectors are $v_1=(1,1,1)$, $v_2=\left (\frac{1}{2}(-1+i\sqrt{3}), \frac{1}{2}(-1-i\sqrt{3}), 1\right )$, $v_3=\left (\frac{1}{2}(-1-i\sqrt{3}), \frac{1}{2}(-1+i\sqrt{3}), 1\right )$.

A rotation matrix must have exactly one eigenvalue 1.
The corresponding eigenvector is a vector along the axis of rotation, which is transformed to itself (hence eigenvalue 1).
The other 2 eigenvalues are usually imaginary and each others conjugate. They are $\cos\phi\pm i\sin\phi$ where $\phi$ is the angle of rotation.

So we can see that the axis of rotation is along the vector $(1,1,1)$.
And the rotation has an angle $\phi$ with $\cos\phi=-\frac 12$ and $\sin\phi=\pm \frac 12\sqrt 3$.
It follows that $\phi=\pm\frac{2\pi}{3}$.

More generally a rotation matrix is similar to a matrix of the form:
$$\begin{pmatrix}1\\&\cos\phi&-\sin\phi\\&\sin\phi&\cos\phi\end{pmatrix}$$
which is a rotation around the x-axis by an angle of $\phi$.
It's a bit easier to see what the eigenvalues must be from this matrix.
Also note that the trace (which is invariant under a similarity transform) is $1+2\cos\phi$, so we can deduce the angle directly from the trace.

 

[mathmari]

A rotation matrix must have exactly one eigenvalue 1.
The corresponding eigenvector is a vector along the axis of rotation, which is transformed to itself (hence eigenvalue 1).
The other 2 eigenvalues are imaginary and each others conjugate. They are $\cos\phi\pm i\sin\phi$ where $\phi$ is the angle of rotation.

So we can see that the axis of rotation is along the vector $(1,1,1)$.
And the rotation has an angle $\phi$ with $\cos\phi=-\frac 12$ and $\sin\phi=\pm \frac 12\sqrt 3$.
It follows that $\phi=\pm\frac{2\pi}{3}$.

For the matrix $s$ we have the eigenvalues $\lambda_1=-1$, $\lambda_2=1$, $\lambda_3=1$ and the corresponding eigenvectors are $v_1=(0,-1,1)$, $v_2=(0,1,1)$, $v_3=(1,0,0)$.

Since we have have two eigenvalues "1" this means that here we don't have a rotation, that means that it must be a reflection, right? :unsure:
As for the set $D$ do we have to do the same procedure for each combination of $\pm 1$ ? :unsure:

 

[I like Serena]

For the matrix $s$ we have the eigenvalues $\lambda_1=-1$, $\lambda_2=1$, $\lambda_3=1$ and the corresponding eigenvectors are $v_1=(0,-1,1)$, $v_2=(0,1,1)$, $v_3=(1,0,0)$.

Since we have have two eigenvalues "1" this means that here we don't have a rotation, that means that it must be a reflection, right?

Yes. (Nod)
Those are 2 vectors that span the reflection plane. Since they are in the plane, their reflection in the plane is again the same vector in the plane (hence eigenvalue 1).
And 1 vector that is normal to the plane. Its reflection is on the other side of the plane at the same length (hence eigenvalue -1).

As for the set $D$ do we have to do the same procedure for each combination of $\pm 1$ ?

I think that is the easiest way to go yes. (Nod)

 

[mathmari]

Yes. (Nod)
Those are 2 vectors that span the reflection plane. Since they are in the plane, their reflection in the plane is again the same vector in the plane (hence eigenvalue 1).
And 1 vector that is normal to the plane. Its reflection is on the other side of the plane at the same length (hence eigenvalue -1).

So do we have a reflection in the plane $v_2\times t_3=(0,1,1)\times (1,0,0) =(0,1,-1)$ ? :unsure:

I think that is the easiest way to go yes. (Nod)

If $e_1=e_2=e_3=1$ then the eigenvalues are $\lambda_1=\lambda_2=\lambda_3=1$ and the eigenvectors are $v_1=(0,0,1), v_2=(0,1,0), v_3=(1,0,0)$.

If $e_1=e_2=1$ and $e_3=-1$ then the eigenvalues are $\lambda_1=-1, \lambda_2=\lambda_3=1$ and the eigenvectors are $v_1=(0,0,1), v_2=(0,1,0), v_3=(1,0,0)$.

If $e_1=e_3=1$ and $e_2=-1$ then the eigenvalues are $\lambda_1=-1, \lambda_2=\lambda_3=1$ and the eigenvectors are $v_1=(0,1,0), v_2=(0,0,1), v_3=(1,0,0)$.

If $e_2=e_3=1$ and $e_1=-1$ then the eigenvalues are $\lambda_1=-1, \lambda_2=\lambda_3=1$ and the eigenvectors are $v_1=(1,0,0), v_2=(0,0,1), v_3=(0,1,0)$.

If $e_2=e_3=-1$ and $e_1=1$ then the eigenvalues are $\lambda_3=1, \lambda_1=\lambda_2=-1$ and the eigenvectors are $v_1=(0,0,1), v_2=(0,1,0), v_3=(1,0,0)$.

If $e_1=e_3=-1$ and $e_2=1$ then the eigenvalues are $\lambda_3=1, \lambda_1=\lambda_2=-1$ and the eigenvectors are $v_1=(0,0,1), v_2=(1,0,0), v_3=(0,1,0)$.

If $e_1=e_2=-1$ and $e_3=1$ then the eigenvalues are $\lambda_3=1, \lambda_1=\lambda_2=-1$ and the eigenvectors are $v_1=(0,1,0), v_2=(1,0,0), v_3=(0,0,1)$.

If $e_1=e_2=e_3=-1$ then the eigenvalues are $\lambda_1=\lambda_2=\lambda_3=-1$ and the eigenvectors are $v_1=(0,0,1), v_2=(0,1,0), v_3=(1,0,0)$. So each of these cases we have a reflection with reflection plane the cross product of the vectors that correspond to the eigenvalues $1$, right?
But what happens whenwe have alleigenvalues equal to $1$, or all equal to $-1$ ? Or if two are $-1$ and just $1$, do we take in this case the cross product of teh vectors that correspond to the eigenvalues $-1$ ?

:unsure:

 

[I like Serena]

So do we have a reflection in the plane $v_2\times t_3=(0,1,1)\times (1,0,0) =(0,1,-1)$ ?

Your notation is a bit confusing.
The reflection plane can be written as $\{\lambda(0,1,1)+\mu(1,0,0)\mid \lambda,\mu\in\mathbb R\}$, or alternatively as $\{(x,y,z)\mid \langle(x,y,z),(0,1,-1)\rangle=0\}$.
And we have that $(0,1,-1)$ is a vector that must be on the same line as $(0,1,1)\times (1,0,0)$, but it's a bit coincidental that in this case the vector is exactly that cross product.

So each of these cases we have a reflection with reflection plane the cross product of the vectors that correspond to the eigenvalues $1$, right?
But what happens whenwe have alleigenvalues equal to $1$, or all equal to $-1$ ? Or if two are $-1$ and just $1$, do we take in this case the cross product of teh vectors that correspond to the eigenvalues $-1$ ?

Actually, not all orthogonal 3x3 matrices are either rotations or reflections. We'll get back to that.
Either way, if we have eigenvalues 1,1,-1, then we have a reflection.
If we have the eigenvalue 1 and the other two are conjugates of each other, then we have a rotation.
Note that -1 and -1 are also conjugates of each other. Can we find a rotation angle in that case?

Let's take a look at what we have if all eigenvalues are 1.
It's $\begin{pmatrix}1\\&1\\&&1\end{pmatrix}$.
Does that matrix look familiar?
Is it a rotation or a reflection?
I guess 1 and 1 are also conjugates of each other. If we consider the possibility that it's a rotation matrix, what would be the angle?
Or is it something else?

 

[mathmari]

Let's take a look at what we have if all eigenvalues are 1.
It's $\begin{pmatrix}1\\&1\\&&1\end{pmatrix}$.
Does that matrix look familiar?
Is it a rotation or a reflection?
I guess 1 and 1 are also conjugates of each other. If we consider the possibility that it's a rotation matrix, what would be the angle?
Or is it something else?

In that case we have the identity, which can be also expressed by rotation and by refelction if we want, right?

In the case of $\lambda_1=\lambda_2=\lambda_3=-1$ then we have that the cube is reflected with plane the origin, or not?

What do we have in the case two eigenvalues are $-1$ and one is $1$ ? :unsure:

 

[I like Serena]

In that case we have the identity, which can be also expressed by rotation and by reflection if we want, right?

It cannot be a reflection since there is no vector that is mapped to its opposite.
We might call it a rotation around an arbitrary axis with rotation angle zero.
I prefer to distinguish it from other rotations and simply call it the identity though. :geek:

In the case of $\lambda_1=\lambda_2=\lambda_3=-1$ then we have that the cube is reflected with plane the origin, or not?

The origin is not a plane...
It's called an inversion, or an inversion in a point, or a point reflection.
It's not a "normal" (plane) reflection, since there is no plane that is invariant.
It's one of those orthogonal matrices that is neither a rotation nor a (plane) reflection.

What do we have in the case two eigenvalues are $-1$ and one is $1$ ?

Could it be a rotation around an axis?
If it is, what would be the angle?

 

[mathmari]

Could it be a rotation around an axis?
If it is, what would be the angle?

Is it a rotation with angle $\pi$ ? :unsure:

 

[I like Serena]

Is it a rotation with angle $\pi$ ?

Yep. (Nod)

 

[mathmari]

The following moves are the symmetries of the cube $W$:
1. Rotation at the axis $\mathbb{R}\begin{pmatrix}0 \\ 0\\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$.

$
\begin{pmatrix}0 &\cos\frac{\pi}{2}&-\sin\frac{\pi}{2}\\ 0 &\sin\frac{\pi}{2}&\cos\frac{\pi}{2} \\ 1 & 0 & 0\end{pmatrix}
$

Is that matrix, corect?

2. Rotation at the axis $\mathbb{R}\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$.

$
\begin{pmatrix}\cos\frac{\pi}{2} & 0&-\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ \sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}
$

Is that matrix, corect?

3. Rotation at the axis $\mathbb{R}\begin{pmatrix}1 \\ 1\\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$.

How can we get that martix?

4. Rotation at the axis $\mathbb{R}\begin{pmatrix}1 \\ -1\\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$.

How can we get that martix?

5. Reflections to the plane, that is spanned by the vectors $\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$.

In this case we have to have more than one eigenvalue $1$, right?:unsure:

 

[I like Serena]

$
\begin{pmatrix}0 &\cos\frac{\pi}{2}&-\sin\frac{\pi}{2}\\ 0 &\sin\frac{\pi}{2}&\cos\frac{\pi}{2} \\ 1 & 0 & 0\end{pmatrix}
$

Is that matrix, correct?

Does it map the axis along (0,0,1) to itself?

$
\begin{pmatrix}\cos\frac{\pi}{2} & 0&-\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ \sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}
$

Is that matrix, correct?

The usual convention is that sign of the angle of a rotation follows the right hand rule.
In this case the axis is in the direction of (0,1,0), which means that (0,0,1) should be mapped to (1,0,0).
Currently that is not the case.

3. Rotation at the axis $\mathbb{R}\begin{pmatrix}1 \\ 1\\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$.

How can we get that matrix?

There are multiple ways to do this.

1. Find 3 independent vectors $u,v,w$ and their images $u',v',w'$.
   Then the matrix is $\begin{pmatrix}u'&v'&w'\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}^{-1}$.
   We can use Gaussian elimination with column reductions on $\left(\begin{array}{ccc|ccc}u&v&w&u'&v'&w'\end{array}\right)$ to easily find it.
   The axis $u=(1,1,-1)$ must be mapped to $u$.
   We need 2 more vectors.
   We can either pick a vertex $v$ with its image. A drawing would help to figure that out.
   Or we can pick a vector $v$ perpendicular to $u$ and find a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=\cos\frac{2\pi}{3}=-\frac 12$.
   We can construct the third vector $w$ by taking the cross product $w=u\times v$.
2. Start with the rotation matrix $R$ with axis $(1,0,0)$ and angle $\frac{2\pi}3$ and use an orthogonal matrix $B$ to achieve a basis transformation.
   The desired matrix is then $BRB^T$.
   The matrix $B$ must map $(1,0,0)$ to $u=\frac{1}{\sqrt 3}(1,1,-1)$. That means that its first column must be $u$. The other 2 columns are arbitrary with only the restriction that all 3 columns must be pairwise orthogonal and they must be of unit length. We may need to take one column negative to achieve the correct direction of rotation though.
3. We can construct the matrix directly from axis and angle as explained in this wiki section.
4. Use trial and error since we already have matrix $d$ that has axis $(1,1,1)$ and rotation angle $\frac{2\pi}3$.
   So the desired matrix must be similar with the restriction that the axis $u=(1,1,-1)$ must be mapped to $u$.

(Sweating)

5. Reflections to the plane, that is spanned by the vectors $\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$.

In this case we have to have more than one eigenvalue $1$, right?

Both vectors must be eigenvectors of eigenvalue 1.
And their cross product must be an eigenvector of eigenvalue -1.

 

[mathmari]

Does it map the axis along (0,0,1) to itself?

Ah no... So we have to consider the matrix $
\begin{pmatrix}\cos\frac{\pi}{2}&\sin\frac{\pi}{2} & 0 \\ -\sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}
$ or not? :unsure:

The usual convention is that sign of the angle of a rotation follows the right hand rule.
In this case the axis is in the direction of (0,1,0), which means that (0,0,1) should be mapped to (1,0,0).
Currently that is not the case.

I got stuck right now. Are the signs wrong at the matrix?
Now we get the vector $(-1,0,0)$ instead of $(1,0,0)$.

:unsure:

There are multiple ways to do this.

1. Find 3 independent vectors $u,v,w$ and their images $u',v',w'$.
   Then the matrix is $\begin{pmatrix}u'&v'&w'\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}^{-1}$.
   We can use Gaussian elimination with column reductions on $\left(\begin{array}{ccc|ccc}u&v&w&u'&v'&w'\end{array}\right)$ to easily find it.
   The axis $u=(1,1,-1)$ must be mapped to $u$.
   We need 2 more vectors.
   We can either pick a vertex $v$ with its image. A drawing would help to figure that out.
   Or we can pick a vector $v$ perpendicular to $u$ and find a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=\cos\frac{2\pi}{3}=-\frac 12$.
   We can construct the third vector $w$ by taking the cross product $w=u\times v$.
2. Start with the rotation matrix $R$ with axis $(1,0,0)$ and angle $\frac{2\pi}3$ and use an orthogonal matrix $B$ to achieve a basis transformation.
   The desired matrix is then $BRB^T$.
   The matrix $B$ must map $(1,0,0)$ to $u=(1,1,-1)$. That means that its first column must be $u$. The other 2 columns are arbitrary with only the restriction that all 3 columns must be pairwise orthogonal and they must be of unit length. We may need to take one column negative to achieve the correct direction of rotation though.
3. We can construct the matrix directly from axis and angle as explained in this wiki section.
4. Use trial and error since we already have matrix $d$ that has axis $(1,1,1)$ and rotation angle $\frac{2\pi}3$.
   So the desired matrix must be similar with the restriction that the axis $u=(1,1,-1)$ must be mapped to $u$.

(Sweating)

Let's consider the first way.

A vector $v$ perpendicular to $u$ is for example $(-1,0,1)$, or not?
The image is a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=\cos\frac{2\pi}{3}=-\frac 12$. So let $v'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $v\cdot v'=-\frac 12$ we get $-x+z=-\frac{1}{2}$.
So we get for example $v'=\left (1, -\frac{1}{2}, \frac{1}{2}\right )$, right?

The third vector $w$ is $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$.
The image is a vector $w'$ that is also perpendicular to $u$ and has $w\cdot w'=\cos\frac{2\pi}{3}=-\frac 12$. So let $w'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $w\cdot w'=-\frac 12$ we get $x+z=-\frac{1}{2}$.
So we get for example $w'=\left (0, -\frac{1}{2}, -\frac{1}{2}\right )$, right? :unsure:

Both vectors must be eigenvectors of eigenvalue 1.
And their cross product must be an eigenvector of eigenvalue -1.

The cross product is $\begin{pmatrix}-1 \\ -1 \\ 0\end{pmatrix}$.

Now the desired matrixis $A=PDP^{-1}$ where $D$ is the digonal matrix with the eigenvalues of the diagonal and $P$ is the matrix with the eigenvectors as the columns, right? :unsure:

 

[I like Serena]

Ah no... So we have to consider the matrix $
\begin{pmatrix}\cos\frac{\pi}{2}&\sin\frac{\pi}{2} & 0 \\ -\sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}
$ or not?

The vector along the axis in invariant now. Good.
That leaves the orientation.
We have the axis vector (0,0,1). According to the right hand rule we should map (1,0,0) to (0,1,0). Is that the case? :unsure:

I got stuck right now. Are the signs wrong at the matrix?
Now we get the vector $(-1,0,0)$ instead of $(1,0,0)$.

It means that we need to rotate in the opposite direction.
We can achieve that by replacing $\frac \pi 2$ by $-\frac\pi 2$.
Or alternatively be taking the inverse of the matrix, which is simply the transpose because it is an orthogonal matrix.

A vector $v$ perpendicular to $u$ is for example $(-1,0,1)$, or not?
The image is a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=\cos\frac{2\pi}{3}=-\frac 12$. So let $v'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $v\cdot v'=-\frac 12$ we get $-x+z=-\frac{1}{2}$.
So we get for example $v'=\left (1, -\frac{1}{2}, \frac{1}{2}\right )$, right?

I forgot to mention an additional condition: both $v$ and $v'$ must have the same length, since otherwise it wouldn't be an isometry (preserves lengths).
And the dot product should actually be $v\cdot v' = \|v\|\|v'\|\cos \frac{2\pi}{3}=-\frac 12v^2$. (Blush)

The third vector $w$ is $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$.
The image is a vector $w'$ that is also perpendicular to $u$ and has $w\cdot w'=\cos\frac{2\pi}{3}=-\frac 12$. So let $w'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $w\cdot w'=-\frac 12$ we get $x+z=-\frac{1}{2}$.
So we get for example $w'=\left (0, -\frac{1}{2}, -\frac{1}{2}\right )$, right?

We do have an extra condition that I hadn't mentioned yet: we need to rotate $w$ in the same direction as $v$.
As it is, we have two choices for $w'$ and we need to pick the right one.
An easy way to do that, is to pick $w'$ in the direction of $u\times v'$. That way we preserve the orthogonality of the vectors when we map $u,v,w$ to $u',v',w'$, which is required for an isometry.

Oh, and $w'$ needs to have the same length as $w$, which is also required for an isometry. So it should be $w'=\|w\|\frac{u\times v'}{\|u\times v'\|}$.

The cross product is $\begin{pmatrix}-1 \\ -1 \\ 0\end{pmatrix}$.

Now the desired matrixis $A=PDP^{-1}$ where $D$ is the digonal matrix with the eigenvalues of the diagonal and $P$ is the matrix with the eigenvectors as the columns, right

Yep.
And if we normalize the eigenvectors to unit length, then $P$ is an orthogonal matrix, so that $P^{-1}=P^T$.

 

[mathmari]

I forgot to mention an additional condition: both $v$ and $v'$ must have the same length, since otherwise it wouldn't be an isometry (preserves lengths).
And the dot product should actually be $v\cdot v' = \|v\|\|v'\|\cos \frac{2\pi}{3}=-\frac 12v^2$. (Blush)
We do have an extra condition that I hadn't mentioned yet: we need to rotate $w$ in the same direction as $v$.
As it is, we have two choices for $w'$ and we need to pick the right one.
An easy way to do that, is to pick $w'$ in the direction of $u\times v'$. That way we preserve the orthogonality of the vectors when we map $u,v,w$ to $u',v',w'$, which is required for an isometry.

Oh, and $w'$ needs to have the same length as $w$, which is also required for an isometry. So it should be $w'=\|w\|\frac{u\times v'}{\|u\times v'\|}$.

A vector $v$ perpendicular to $u$ is for example $(-1,0,1)$ with $\|v\|=\sqrt{2}$.
The image is a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=-\frac 12v^2=-\frac{1}{2}\cdot 2=-1$. So let $v'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $v\cdot v'=-1$ we get $-x+z=-1$.
So we get for example $v'=\left (1, -1, 0\right )$, right?

The third vector $w$ is $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$. This is perpendicular to $v$ and not the same direction, right? :unsure:

 

[I like Serena]

A vector $v$ perpendicular to $u$ is for example $(-1,0,1)$ with $\|v\|=\sqrt{2}$.
The image is a vector $v'$ that is also perpendicular to $u$ and has $v\cdot v'=-\frac 12v^2=-\frac{1}{2}\cdot 2=-1$. So let $v'=(x,y,z)$ then since it is perpendicular to $u$ the dot product must be $0$, i.e. $x+y-z=0$.
From $v\cdot v'=-1$ we get $-x+z=-1$.
So we get for example $v'=\left (1, -1, 0\right )$, right?

The third vector $w$ is $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$. This is perpendicular to $v$ and not the same direction, right?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

What about $w$ ? Which vector do we take since we cannot take the cross product of $u$ and $v$ ? :unsure:

 

[I like Serena]

What about $w$ ? Which vector do we take since we cannot take the cross product of $u$ and $v$ ?

Didn't you already pick $w=u\times v$?
That choice is fine.

 

[mathmari]

Didn't you already pick $w=u\times v$?
That choice is fine.

Ahh I thought you meant that was wrong.

You meant my $w'$ was wrong? :unsure:

 

[I like Serena]

Ahh I thought you meant that was wrong.

You meant my $w'$ was wrong?

Which $w'$?

 

[mathmari]

If we take as $w$ the cross product of $u$ and $v$ we get $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$.

Then $$w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{2}\frac{(1,1,-1)\times (1,-1,0)}{\|(1,1,-1)\times (1,-1,0)\|}=\sqrt{2}\frac{(-1,-1,-2)}{\sqrt{6}}=\left (-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}}\right )$$ right?So using these vectors we can now determine the matrix, which is \begin{equation*}\begin{pmatrix}-\frac{1}{2}-\frac{1}{2\sqrt{3}} & 2 & \frac{1}{2}-\frac{1}{2\sqrt{3}} \\ \frac{1}{2}-\frac{1}{2\sqrt{3}} & 0 & -\frac{1}{2}-\frac{1}{2\sqrt{3}} \\ -\frac{1}{\sqrt{3}} & -1 & -\frac{1}{\sqrt{3}}\end{pmatrix}\end{equation*} right? :unsure:

 

[I like Serena]

If we take as $w$ the cross product of $u$ and $v$ we get $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$.
So using these vectors we can now determine the matrix, which is \begin{equation*}\begin{pmatrix}-\frac{1}{2}-\frac{1}{2\sqrt{3}} & 2 & \frac{1}{2}-\frac{1}{2\sqrt{3}} \\ \frac{1}{2}-\frac{1}{2\sqrt{3}} & 0 & -\frac{1}{2}-\frac{1}{2\sqrt{3}} \\ -\frac{1}{\sqrt{3}} & -1 & -\frac{1}{\sqrt{3}}\end{pmatrix}\end{equation*} right? :unsure:

That does not look good.

On closer examination I see that $u$ and $v$ are not orthogonal.
Their dot product is -2 instead of 0.

I checked this with geogebra 3d.
If we enter:

Cube((-1,-1,-1),(1,-1,-1),(1,1,-1))
u=(1,1,-1)
v=(-1,0,1)

and if we rotate the view a bit, then we can see that $u$ and $v$ are not orthogonal. :geek:

 

[mathmari]

I corrected it...

Do we have the following for each case? 1. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$ :

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2}&-\sin\frac{\pi}{2} & 0 \\ \sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}0 &-1 & 0 \\ 1& 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*} 2. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2} & 0&\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ -\sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}=\begin{pmatrix}0 & 0& 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\end{equation*} 3. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u= (1,1, -1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, -1, 0\right )$.
We consider the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (1, -1, 0)}{\|(1,1,-1)\times (1, -1, 0)\|}=\sqrt{6}\frac{(-1,-1,-2)}{\sqrt{6}}=(-1,-1,-2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & -1 & 0\end{pmatrix}\end{equation*} 4. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u=(1,-1,-1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, 1, 0\right )$.
We consider the vector $w=u\times v=(1,-1,-1)\times (-1,0,-1)=(1,2,-1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,-1,-1)\times (1, 1, 0)}{\|(1,-1,-1)\times (1, 1, 0)\|}=\sqrt{6}\frac{(1,-1,2)}{\sqrt{6}}=(1,-1,2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\end{equation*} 5. Reflection to the plance that is spanned by the vectors $\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}$ und $\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$

We have that \begin{equation*}D=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\end{equation*}
and \begin{equation*}P=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\end{equation*}
and so we get\begin{equation*}A=PDP^{-1}=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\end{pmatrix}=\begin{pmatrix}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*} :unsure:

 

[I like Serena]

1. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$ :

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2}&-\sin\frac{\pi}{2} & 0 \\ \sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}0 &-1 & 0 \\ 1& 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}2. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2} & 0&\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ -\sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}=\begin{pmatrix}0 & 0& 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\end{equation*}

Correct. (Nod)

3. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u= (1,1, -1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, -1, 0\right )$.
We consider the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (1, -1, 0)}{\|(1,1,-1)\times (1, -1, 0)\|}=\sqrt{6}\frac{(-1,-1,-2)}{\sqrt{6}}=(-1,-1,-2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & -1 & 0\end{pmatrix}\end{equation*}

How did you get the matrix?

Now we get to the orientation.
The map from $v$ to $v'$ must be according to the right hand rule with respect to the axis vector $u$.
It means that $v\times v'$ must be in the same direction as $u$.
That is equivalent to $(u\times v) \cdot v'\ge 0$.
Since we already had $w=u\times v$, we can check $w\cdot v' = (-1,2,1)\cdot (1, -1, 0) = -3\not\ge 0$.
So the orientation is wrong: we need the transpose of the matrix.

4. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u=(1,-1,-1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, 1, 0\right )$.
We consider the vector $w=u\times v=(1,-1,-1)\times (-1,0,-1)=(1,2,-1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,-1,-1)\times (1, 1, 0)}{\|(1,-1,-1)\times (1, 1, 0)\|}=\sqrt{6}\frac{(1,-1,2)}{\sqrt{6}}=(1,-1,2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\end{equation*}

Checking the orientation of the rotation... $w\cdot v' = (1,2,-1)\cdot (1, 1, 0) = 3 \ge 0$, so this one does have the correct rotation.

5. Reflection to the plance that is spanned by the vectors $\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}$ und $\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$

We have that \begin{equation*}D=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\end{equation*}
and \begin{equation*}P=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\end{equation*}
and so we get\begin{equation*}A=PDP^{-1}=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\end{pmatrix}=\begin{pmatrix}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}

The explanation seems to be missing... but the result is correct! (Nod)

 

[mathmari]

How did you get the matrix?

Now we get to the orientation.
The map from $v$ to $v'$ must be according to the right hand rule with respect to the axis vector $u$.
It means that $v\times v'$ must be in the same direction as $u$.
That is equivalent to $(u\times v) \cdot v'\ge 0$.
Since we already had $w=u\times v$, we can check $w\cdot v' = (-1,2,1)\cdot (1, -1, 0) = -3\not\ge 0$.
So the orientation is wrong: we need the transpose of the matrix.

Why is the orientation wrong? I did the same procedure at the next one, which was correct?
Is the vector of $v$ that I chose wrong? :unsure:

 

[I like Serena]

Why is the orientation wrong? I did the same procedure at the next one, which was correct?
Is the vector of $v$ that I chose wrong?

The vector $v$ is fine.
The problem is that we have 2 possible choices for $v'$ and we need to pick the right one.
That is, the one such that $v\times v'$ is in the same direction as $u$.

 

[mathmari]

The vector $v$ is fine.
The problem is that we have 2 possible choices for $v'$ and we need to pick the right one.
That is, the one such that $v\times v'$ is in the same direction as $u$.

Taking $v'=\left (0, 1, 1\right )$ we have the following:

We have the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and we get $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (0, 1, 1)}{\|(1,1,-1)\times (0, 1, 1)\|}=\sqrt{6}\frac{(2,-1,1)}{\sqrt{6}}=(2,-1,1)}$.

The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}

:unsure:

 

[I like Serena]

The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}

Yep. All correct. (Nod)