Find all orthogonal 3x3 matrices of the form

[pyroknife]

Yes. You're getting close. You have $(a,e) = (\cos \theta, \sin\theta)$ for the point at angle $\theta$. So for the point $(b,f)$ at angle $\theta +\pi / 2$ you have $(b,f) = (\cos(\theta+\frac \pi 2), --)$ (you need the $y$ coordinate too).

Then do the same thing for $(b,f)$ at $\theta - \pi /2$.

Then you will want to simplify your $(b,f)$ expressions and you will be able to fill in your matrix. What you should find interesting is that when you complete your conversation with Ray you should have the exact same answers.

Yes, so for (a,e) = ($cos(\theta)$, $sin(\theta)$)
(b,f) can be either ($cos(\theta+\pi/2)$, $sin(\theta+\pi/2)$)
or
($cos(\theta-\pi/2)$, $sin(\theta-\pi/2)$)

So for the chosen (a,e), we can have 2 sets of (b,f).

Back to one of my earlier questions.
Couldn't (a,e) ALSO be any of the following 7:
1) (a,e) = ($cos(\theta)$, $-sin(\theta)$)
2) (a,e) = ($-cos(\theta)$, $sin(\theta)$)
3) (a,e) = ($-cos(\theta)$, $-sin(\theta)$)
4) (a,e) = ($sin(\theta)$,$cos(\theta)$)
5) (a,e) = ($-sin(\theta)$,$-cos(\theta)$)
6) (a,e) = ($-sin(\theta)$,$cos(\theta)$)
7) (a,e) = ($sin(\theta)$,$-cos(\theta)$)

 

[LCKurtz]

Note: I fixed your latex below. You want to use double # signs instead of $ signs for inline tex, and you can do the whole expressions at once.

Yes, so for $(a,e) = (\cos(\theta), \sin(\theta))$, $(b,f)$ can be either $(\cos(\theta+\pi/2), \sin(\theta+\pi/2))$ or $(\cos(\theta-\pi/2), \sin(\theta-\pi/2))$.

So for the chosen $(a,e)$, we can have 2 sets of $(b,f)$.

Exactly right. But simplify those expressions to get them in terms of sines or cosines without the $\pi /2$ expressions.

Back to one of my earlier questions.
Couldn't (a,e) ALSO be any of the following 7:(snipped)

You don't need to worry about other ways to do it. Any point on the unit circle can be written $(\cos\theta,\sin\theta)$, standard polar coordinates. So that notation itself covers all possibilities.

 

[pyroknife]

Note: I fixed your latex below. You want to use double # signs instead of $ signs for inline tex, and you can do the whole expressions at once.
Exactly right. But simplify those expressions to get them in terms of sines or cosines without the $\pi /2$ expressions.
You don't need to worry about other ways to do it. Any point on the unit circle can be written $(\cos\theta,\sin\theta)$, standard polar coordinates. So that notation itself covers all possibilities.

Thanks!
Okay, I will simplify it to omit the $\pi /2$.

I understand that any point on the unit circle can be written as $(\cos\theta,\sin\theta)$, but I don't understand why this would cover the other possibilities. Wouldn't the other possibilities yield a distinctively different matrix?

 

[LCKurtz]

Thanks!
Okay, I will simplify it to omit the $\pi /2$.

I understand that any point on the unit circle can be written as $(\cos\theta,\sin\theta)$, but I don't understand why this would cover the other possibilities. Wouldn't the other possibilities yield a distinctively different matrix?

No. They will give the same, or equivalent, answers. I encourage you to finish your conversation with Ray and Dick since you are almost done with their approach. And Ray used $(\sin\phi,\cos\phi)$ in his approach which is different. You will see that you get the same answer.

Also I would like to see your final two matrices that you get for your answers. Here's how you do a nice matrix in latex. You can copy it and put your values in. (Right click on it to see the tex). And surround it with double $ signs.
$$\begin{bmatrix}
a & b & c \\
u & v & w \\
r & s & t
\end{bmatrix}$$

 

[pyroknife]

For the

No. They will give the same, or equivalent, answers. I encourage you to finish your conversation with Ray and Dick since you are almost done with their approach. And Ray used $(\sin\phi,\cos\phi)$ in his approach which is different. You will see that you get the same answer.

Also I would like to see your final two matrices that you get for your answers. Here's how you do a nice matrix in latex. You can copy it and put your values in. (Right click on it to see the tex). And surround it with double $ signs.
$$\begin{bmatrix}
a & b & c \\
u & v & w \\
r & s & t
\end{bmatrix}$$

I obtain the following two matrices:
$$\begin{bmatrix}
\cos(\theta) & -\sin(\theta) & 0 \\
0 & 0 & 1 \\
\sin(\theta) & \cos(\theta) & 0
\end{bmatrix}$$

and

$$\begin{bmatrix}
\cos(\theta) & \sin(\theta) & 0 \\
0 & 0 & 1 \\
\sin(\theta) & -\cos(\theta) & 0
\end{bmatrix}$$

 

[pyroknife]

For the

I obtain the following two matrices:
$$\begin{bmatrix}
cos(\theta) & -sin(\theta) & 0 \\
0 & 0 & 1 \\
sin(\theta) & cos(\theta) & 0
\end{bmatrix}$$

and

$$\begin{bmatrix}
cos(\theta) & sin(\theta) & 0 \\
0 & 0 & 1 \\
sin(\theta) & -cos(\theta) & 0
\end{bmatrix}$$

which satisfies the normalization and orthogonality conditions.

 

[Dick]

which satisfies the normalization and orthogonality conditions.

Now that sounds right. You get two families of matrices depending on a rotation angle. One has determinant 1 and the other has determinant -1. One is a pure rotation and the other is a rotation combined with a reflection.

 

[LCKurtz]

Looking good. Note that if you want a trig function in latex, you can use \sin and \cos which will give a more mathematical typeface for them. Try editing that last post and put a \ in front of each trig function.

And, really, follow up with Ray's method. I think Dick's suggestion in post #31 will help you finish it.

 

[pyroknife]

For the

I obtain the following two matrices:
$$\begin{bmatrix}
\cos(\theta) & -\sin(\theta) & 0 \\
0 & 0 & 1 \\
\sin(\theta) & \cos(\theta) & 0
\end{bmatrix}$$

and

$$\begin{bmatrix}
\cos(\theta) & \sin(\theta) & 0 \\
0 & 0 & 1 \\
\sin(\theta) & -\cos(\theta) & 0
\end{bmatrix}$$

 

[pyroknife]

Looking good. Note that if you want a trig function in latex, you can use \sin and \cos which will give a more mathematical typeface for them. Try editing that last post and put a \ in front of each trig function.

And, really, follow up with Ray's method. I think Dick's suggestion in post #31 will help you finish it.

I just edited the last post with the 2 matrices. Looks good.

Yeah, I am trying to look at Dick's suggestion and Ray's method as I'm posting this.Also, I guess I don't understand this quite well just yet.
Say $(a,e)$ was $(\sin\theta, cos\theta)$ instead.

Then the matrices become

$$\begin{bmatrix}
\sin(\theta) & \cos(\theta) & 0 \\
0 & 0 & 1 \\
\cos(\theta) & -\sin(\theta) & 0
\end{bmatrix}$$

and

$$\begin{bmatrix}
\sin(\theta) & -\cos(\theta) & 0 \\
0 & 0 & 1 \\
\cos(\theta) & \sin(\theta) & 0
\end{bmatrix}$$

Aren't these matrices considered to be different from the two matrices resulting from $(a,e)$ = $(cos\theta, \sin\theta)$?

 

[LCKurtz]

I just edited the last post with the 2 matrices. Looks good.

Yeah, I am trying to look at Dick's suggestion and Ray's method as I'm posting this.Also, I guess I don't understand this quite well just yet.
Say $(a,e)$ was $(\sin\theta, cos\theta)$ instead.

Then the matrices become

$$\begin{bmatrix}
\sin(\theta) & \cos(\theta) & 0 \\
0 & 0 & 1 \\
\cos(\theta) & -\sin(\theta) & 0
\end{bmatrix}$$

and

$$\begin{bmatrix}
\sin(\theta) & -\cos(\theta) & 0 \\
0 & 0 & 1 \\
\cos(\theta) & \sin(\theta) & 0
\end{bmatrix}$$

Aren't these matrices considered to be different from the two matrices resulting from $(a,e)$ = $(cos\theta, \sin\theta)$?

They look different but they are just a different parameterization. Suppose you let $\theta = \pi/2 - \alpha$ in your polar expression $(\cos\theta,\sin\theta)$. This gives you $(\cos(\pi /2 - \alpha),\sin(\pi /2 - \alpha)) = (\sin\alpha,\cos\alpha)$. You are just using a different angle and orientation but it's covered with standard polar.

 

[pyroknife]

They look different but they are just a different parameterization. Suppose you let $\theta = \pi/2 - \alpha$ in your polar expression $(\cos\theta,\sin\theta)$. This gives you $(\cos(\pi /2 - \alpha),\sin(\pi /2 - \alpha)) = (\sin\alpha,\cos\alpha)$. You are just using a different angle and orientation but it's covered with standard polar.

Ahh that makes sense.
I think I am making sense out of Ray&Dick's version of obtaining the solution. I'll communicate with them tomorrow about it to see if I understand it correctly (too tired today).

Thank you for all the help.