MHB Set of 2-dimensional orthogonal matrices equal to an union of sets

[mathmari]

Hey! :giggle:

The set of $2$-dimensional orthogonal matrices is given by $$O(2, \mathbb{R})=\{a\in \mathbb{R}^{2\times 2}\mid a^ta=u_2\}$$ Show the following:

(a) $O(2, \mathbb{R})=D\cup S$ and $D\cap S=\emptyset$. It holds that $D=\{d_{\alpha}\mid \alpha\in \mathbb{R}\}$ and $S=\{s_{\alpha}\mid \alpha\in \mathbb{R}\}$, where $d_{\alpha}=\begin{pmatrix}\cos (\alpha) & -\sin (\alpha) \\ \sin (\alpha ) & \cos (\alpha )\end{pmatrix}$ and $s_{\alpha}=\begin{pmatrix} \cos (\alpha )& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)\end{pmatrix}$.

(b) For all $\alpha\in \mathbb{R}$ is $B_{\alpha}$ an orthonormal basis of $\mathbb{R}^2$. It holds that $B_{\alpha}=(e_{\alpha}, f_{\alpha})$, where $e_{\alpha}\begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right )\end{pmatrix}$ and $f_{\alpha}\begin{pmatrix}-\sin \left (\frac{\alpha}{2}\right ) \\ \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}$

(c) Calculate $M_{B_{\alpha}}(\sigma_{\alpha})$,where $\sigma_{\alpha}(x)=s_{\alpha}x$.

I have done the following :

(a) To show that $D\cap S=\emptyset$, we assume that this is not true, i.e. that there is a matrix that belongs to $D$ and to $S$. Then for some $\alpha\in \mathbb{R}$ it must hold that $-\sin (\alpha)= \sin (\alpha) \Rightarrow \sin (\alpha)=0$ and that $\cos (\alpha)= -\cos (\alpha) \Rightarrow \cos (\alpha)=0$. There is no such $\alpha$ and therefore the intersection is empty.

How can we show that $O(2, \mathbb{R})=D\cup S$ ?

(b) We have to show that $e_{\alpha}$ and $f_{\alpha}$ are linearly independent, so that we can say that $B_{\alpha}$ is a basis of $\mathbb{R}^2$, right? To show also that itis an orthonormal basis, we have to show that the vectors $e_{\alpha}$ and $f_{\alpha}$ are orthogonal, i.e. their dot product is equal to $0$ and that it is normal, i.e. that both vectors have length $1$, right?

(c) Do we have to write the columns of $s_{\alpha}$ as a linear combination of $e_{\alpha}$ and $f_{\alpha}$ ? :unsure:

 

[I like Serena]

How can we show that $O(2, \mathbb{R})=D\cup S$ ?

Hey mathmari!

Suppose we consider a matrix $a$ with column vectors $\vec a$ and $\vec b$.
What can we deduce from the condition $a^t a = u_2$?

(b) show that the vectors $e_{\alpha}$ and $f_{\alpha}$ are orthogonal, i.e. their dot product is equal to $0$ and that it is normal, i.e. that both vectors have length $1$, right?

Yep. (Nod)

(c) Do we have to write the columns of $s_{\alpha}$ as a linear combination of $e_{\alpha}$ and $f_{\alpha}$ ?

What we need, is to find the images of $e_\alpha$ and $f_\alpha$ in terms of $e_\alpha$ and $f_\alpha$.
It may help to write $s_{\alpha}$ as a linear combination of $e_{\alpha}$ and $f_{\alpha}$. :unsure:

 

[mathmari]

Suppose we consider a matrix $a$ with column vectors $\vec a$ and $\vec b$.
What can we deduce from the conditions $a^t a = u_2$?

We have that $\vec{a}\cdot \vec{a}=\vec{b}\cdot \vec{b}=1$ and $\vec{a}\cdot \vec{b}=0$, right? :unsure:

What we need, is to find the images of $e_\alpha$ and $f_\alpha$ in terms of $e_\alpha$ and $f_\alpha$.
It may help to write $s_{\alpha}$ as a linear combination of $e_{\alpha}$ and $f_{\alpha}$. :unsure:

Will the coefficients of the linear combination be matrices? :unsure:

 

[I like Serena]

We have that $\vec{a}\cdot \vec{a}=\vec{b}\cdot \vec{b}=1$ and $\vec{a}\cdot \vec{b}=0$, right?

Yep.

Will the coefficients of the linear combination be matrices?

Perhaps we should consider the matrix given by $B_\alpha$.
It transforms the unit vectors to $e_\alpha$ and $f_\alpha$.
So we should be able to construct $M_{B_\alpha}(\sigma_\alpha)$ using $B_\alpha$ and its inverse.

 

[mathmari]

Yep.

This properties are satisfied by the columns of the mayrices of $D$ and $S$, right? So does the desired result just follow then? :unsure:

Perhaps we should consider the matrix given by $B_\alpha$.
It transforms the unit vectors to $e_\alpha$ and $f_\alpha$.
So we should be able to construct $M_{B_\alpha}(\sigma_\alpha)$ using $B_\alpha$ and its inverse.

We have that $B_{\alpha}= \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & - \sin \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}$.

Do you mean that $M_{B_\alpha}(\sigma_\alpha)=B_\alpha^{-1}B_\alpha$ ? :unsure:

 

[I like Serena]

This properties are satisfied by the columns of the matrices of $D$ and $S$, right? So does the desired result just follow then?

Not necessarily.
That is what we still have to prove.
We have to prove that any $\vec a$ can be written as $(\cos\alpha,\sin\alpha)$. Can it?
And additionally that any $\vec b$ is either $(-\sin\alpha,\cos\alpha)$ or $(\sin\alpha,-\cos\alpha)$.
Can we prove that?

Do you mean that $M_{B_\alpha}(\sigma_\alpha)=B_\alpha^{-1}B_\alpha$ ?

Close... but that right side is just identity isn't it. Something is missing... (Sweating)

 

[mathmari]

Not necessarily.
That is what we still have to prove.
We have to prove that any $\vec a$ can be written as $(\cos\alpha,\sin\alpha)$. Can it?
And additionally that any $\vec b$ is either $(-\sin\alpha,\cos\alpha)$ or $(\sin\alpha,-\cos\alpha)$.
Can we prove that?

Do we prove that by proving that these two vectors are orthognal to each otherand have length $1$ ? :unsure:

Close... but that right side is just identity isn't it. Something is missing... (Sweating)

Ah yes. Should it be $M_{B_\alpha}(\sigma_\alpha)=B_\alpha^{-1}(s_\alpha)B_\alpha$ ? :unsure:

 

[I like Serena]

Do we prove that by proving that these two vectors are orthognal to each otherand have length $1$ ?

We should observe that any vector of length 1 must be on the unit circle, which implies that it can be written as $(\cos\alpha,\sin\alpha)$ for some $\alpha$.
And there are only 2 unit vectors that are orthogonal in 2 dimensions (dot product 0).
That is either $(\sin\alpha,-\cos\alpha)$ or $(-\sin\alpha,\cos\alpha)$.

Ah yes. Should it be $M_{B_\alpha}(\sigma_\alpha)=B_\alpha^{-1}(s_\alpha)B_\alpha$ ?

Yep. (Nod)

And since $B_\alpha$ is an orthogonal matrix, its inverse is the same as its transpose.

 

[mathmari]

We should observe that any vector of length 1 must be on the unit circle, which implies that it can be written as $(\cos\alpha,\sin\alpha)$ for some $\alpha$.
And there are only 2 unit vectors that are orthogonal in 2 dimensions (dot product 0).
That is either $(\sin\alpha,-\cos\alpha)$ or $(-\sin\alpha,\cos\alpha)$.

Can we just say that any vector of length 1 must be on the unit circle, which implies that it can be written as $(\cos\alpha,\sin\alpha)$ for some $\alpha$, or do we have to prove that? :unsure:

And since $B_\alpha$ is an orthogonal matrix, its inverse is the same as its transpose.

So we have that
\begin{align*}M_{B_\alpha}(\sigma_\alpha)&=B_\alpha^T s_\alpha B_\alpha \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos (\alpha )& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)\end{pmatrix} \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & - \sin \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos (\alpha )\cos \left (\frac{\alpha}{2}\right )+\sin (\alpha)\sin \left (\frac{\alpha}{2}\right )& -\cos (\alpha )\sin \left (\frac{\alpha}{2}\right )+\sin (\alpha)\cos \left (\frac{\alpha}{2}\right ) \\ \sin (\alpha )\cos \left (\frac{\alpha}{2}\right )-\cos (\alpha)\sin \left (\frac{\alpha}{2}\right ) & -\sin (\alpha )\sin \left (\frac{\alpha}{2}\right )-\cos (\alpha)\cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos \left (\frac{\alpha}{2}\right )& \sin \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}
\\ & = \begin{pmatrix} \cos^2 \left (\frac{\alpha}{2}\right )+\sin^2 \left (\frac{\alpha}{2}\right )& 2\cos\left (\frac{\alpha}{2}\right )\sin \left (\frac{\alpha}{2}\right ) \\ 0 & -\sin^2 \left (\frac{\alpha}{2}\right )+\cos^2 \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix} 1& 2\cos\left (\frac{\alpha}{2}\right )\sin \left (\frac{\alpha}{2}\right ) \\ 0 & -\sin^2 \left (\frac{\alpha}{2}\right )+\cos^2 \left (\frac{\alpha}{2}\right )\end{pmatrix} \end{align*}

:unsure:

 

[I like Serena]

Can we just say that any vector of length 1 must be on the unit circle, which implies that it can be written as $(\cos\alpha,\sin\alpha)$ for some $\alpha$, or do we have to prove that?

Yes, we can just state that. (Nod)

So we have that
\begin{align*}M_{B_\alpha}(\sigma_\alpha)&=B_\alpha^T s_\alpha B_\alpha \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos (\alpha )& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)\end{pmatrix} \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & - \sin \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos (\alpha )\cos \left (\frac{\alpha}{2}\right )+\sin (\alpha)\sin \left (\frac{\alpha}{2}\right )& -\cos (\alpha )\sin \left (\frac{\alpha}{2}\right )+\sin (\alpha)\cos \left (\frac{\alpha}{2}\right ) \\ \sin (\alpha )\cos \left (\frac{\alpha}{2}\right )-\cos (\alpha)\sin \left (\frac{\alpha}{2}\right ) & -\sin (\alpha )\sin \left (\frac{\alpha}{2}\right )-\cos (\alpha)\cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) & \sin \left (\frac{\alpha}{2}\right ) \\ -\sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix} \begin{pmatrix} \cos \left (\frac{\alpha}{2}\right )& \sin \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right ) & \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}
\\ & = \begin{pmatrix} \cos^2 \left (\frac{\alpha}{2}\right )+\sin^2 \left (\frac{\alpha}{2}\right )& 2\cos\left (\frac{\alpha}{2}\right )\sin \left (\frac{\alpha}{2}\right ) \\ 0 & -\sin^2 \left (\frac{\alpha}{2}\right )+\cos^2 \left (\frac{\alpha}{2}\right )\end{pmatrix} \\ & = \begin{pmatrix} 1& 2\cos\left (\frac{\alpha}{2}\right )\sin \left (\frac{\alpha}{2}\right ) \\ 0 & -\sin^2 \left (\frac{\alpha}{2}\right )+\cos^2 \left (\frac{\alpha}{2}\right )\end{pmatrix} \end{align*}

Looks about right, although I didn't check the calculations.
We can still simplify the result a bit more can't we? We can use the double angle formulas.