Find all orthogonal 3x3 matrices of the form

[pyroknife]

Homework Statement

Find all orthogonal 3x3 matrices of the form

\begin{array}{cc}
a & b & 0 \\
c & d & 1\\
e & f & 0 \\\end{array}

Homework Equations

There are many properties of an orthogonal matrix. The one I chose to use is:
An n x n matrix is an orthogonal matrix IFF $$A^{T}A = I$$. That is, the transpose of A multiplied by A is equal to the n x n identity matrix.

The Attempt at a Solution

$$A^T*A$$ =
\begin{array}{cc}
a^2+c^2+e^2 & ab+cd+ef & c \\
ab+cd+ef & b^2+d^2+f^2 & d\\
c & d & 1 \\\end{array}
=
\begin{array}{cc}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1 \\\end{array}

From this we can see right away that c=d=0.

From the rest, we have left with 3 equations $$a^2+e^2=1; ab+ef=0; b^2+f^2=1$$

Is it possible to obtain a equation solution given the above 3 equations? I don't think so.
I am at a lost about how to solve this problem. Anyone have any suggestions?

[/B]

 

[vela]

That's a good start. If you consider the columns of A to be vectors, $a^2+e^2=1$ says that the first column is normalized. Similarly, $b^2+f^2=1$ says the second column is normalized. Can you come up with an interpretation for $ab+ef=0$?

 

[LCKurtz]

Actually, that third equation (equivalently normalizing the second column) should be $b^2+d^2+f^2 = 1$. Lots of extra variables here so there should be a solution.

 

[vela]

The OP omitted $d$ because the off-diagonal entries require $d=0$.

 

[Ray Vickson]

Homework Statement

Find all orthogonal 3x3 matrices of the form

\begin{array}{cc}
a & b & 0 \\
c & d & 1\\
e & f & 0 \\\end{array}

Homework Equations

There are many properties of an orthogonal matrix. The one I chose to use is:
An n x n matrix is an orthogonal matrix IFF $$A^{T}A = I$$. That is, the transpose of A multiplied by A is equal to the n x n identity matrix.

The Attempt at a Solution

$$A^T*A$$ =
\begin{array}{cc}
a^2+c^2+e^2 & ab+cd+ef & c \\
ab+cd+ef & b^2+d^2+f^2 & d\\
c & d & 1 \\\end{array}
=
\begin{array}{cc}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1 \\\end{array}

From this we can see right away that c=d=0.

From the rest, we have left with 3 equations $$a^2+e^2=1; ab+ef=0; b^2+f^2=1$$

Is it possible to obtain a equation solution given the above 3 equations? I don't think so.
I am at a lost about how to solve this problem. Anyone have any suggestions?
[/B]

Having $a^2+e^2=1$ suggests setting $a = \sin(\theta), e = \cos(\theta)$ --- or maybe $a = \cos(\theta), e = \sin(\theta)$ --- for some $\theta$, and having $b^2 + f^2 = 1$ suggests setting $b = \sin(\phi), f = \cos(\phi)$ for some $\phi$ (or swap the $\sin$ and $\cos$). Now, what would the remaining condition give you?

 

[LCKurtz]

The OP omitted $d$ because the off-diagonal entries require $d=0$.

Ahh.., missed that. Makes it easier too.

 

[pyroknife]

That's a good start. If you consider the columns of A to be vectors, $a^2+e^2=1$ says that the first column is normalized. Similarly, $b^2+f^2=1$ says the second column is normalized. Can you come up with an interpretation for $ab+ef=0$?[/QUOTE

That's a good start. If you consider the columns of A to be vectors, $a^2+e^2=1$ says that the first column is normalized. Similarly, $b^2+f^2=1$ says the second column is normalized. Can you come up with an interpretation for $ab+ef=0$?

]
Hmmm I am still struggling with this. But if we use what Ray Vickson stated, then it would come out to be any of the following 4:
sin(θ)*sin(ϕ)+cos(θ)*cos(ϕ)=0
sin(θ)*cos(ϕ)+cos(θ)*sin(ϕ)=0
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Actually, it reduces to the following 2:
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

 

[pyroknife]

Hmmm I am still struggling with this. But if we use what Ray Vickson stated, then it would come out to be any of the following 4:
sin(θ)*sin(ϕ)+cos(θ)*cos(ϕ)=0
sin(θ)*cos(ϕ)+cos(θ)*sin(ϕ)=0
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Actually, it reduces to the following 2:
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

 

[pyroknife]

Also, a,b,c,d can be positive or negative sin or cos...

I am still unsure how to write the expressions for variables a,b,e,f

 

[Ray Vickson]

Hmmm I am still struggling with this. But if we use what Ray Vickson stated, then it would come out to be any of the following 4:
sin(θ)*sin(ϕ)+cos(θ)*cos(ϕ)=0
sin(θ)*cos(ϕ)+cos(θ)*sin(ϕ)=0
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Actually, it reduces to the following 2:
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Remember: addition formulas for sin and cos!

Hmmm I am still struggling with this. But if we use what Ray Vickson stated, then it would come out to be any of the following 4:
sin(θ)*sin(ϕ)+cos(θ)*cos(ϕ)=0
sin(θ)*cos(ϕ)+cos(θ)*sin(ϕ)=0
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Actually, it reduces to the following 2:
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Can't you simplify $\cos(\theta) \sin(\phi) + \sin(\theta) \cos(\phi)$?

 

[pyroknife]

Remember: addition formulas for sin and cos!Can't you simplify $\cos(\theta) \sin(\phi) + \sin(\theta) \cos(\phi)$?

Hmmm, I don't see a trig identity that would put this into a simpler form?

 

[pyroknife]

Using a trig identity, I was able to obtain that
$$cos(\theta)sin(\phi)+sin(\theta)cos(\phi) = sin(\theta+\phi)-sin(\theta-\phi)$$

But going back to the question, since a,b,c,d, can be +/-cos or +/-sin, wouldn't there be many matrices that is a function of the 2 angles?

 

[LCKurtz]

@pyroknife: You have $(a,e)$ and $(b,f)$ both on the unit circle. You have been asked what $ab +ef=0$ does for you, and as far as I can see, you haven't addressed that question. It is the missing link in finishing this question. Think about what that equation tells you about the position vectors to your two points.

 

[pyroknife]

Oh hmmm. From post #12, we see that $$sin(\theta+\phi)=sin(\theta-\phi).$$
It seems this condition may only be satisfied if $$\phi=0$$
Is this the missing link?

 

[Ray Vickson]

Using a trig identity, I was able to obtain that
$$cos(\theta)sin(\phi)+sin(\theta)cos(\phi) = sin(\theta+\phi)-sin(\theta-\phi)$$

But going back to the question, since a,b,c,d, can be +/-cos or +/-sin, wouldn't there be many matrices that is a function of the 2 angles?

No. See, eg., http://www.sosmath.com/trig/Trig5/trig5/trig5.html .

 

[LCKurtz]

Oh hmmm. From post #12, we see that $$sin(\theta+\phi)=sin(\theta-\phi).$$
It seems this condition may only be satisfied if $$\phi=0$$
Is this the missing link?

No. Here's what I asked from you:

@pyroknife:Think about what that equation tells you about the position vectors to your two points.

That is the question you haven't addressed and need to answer.

 

[pyroknife]

I'm having trouble with this. If a,e indicate the x coordinate on a unit circle, and b,f indicate the y-coordinate on a unit circle, then the equation is saying
the sum of the product of the 2 x-coordinates and the 2 y-coordinates will yield 0. But I don't think this is what you're looking for?

 

[pyroknife]

No. Here's what I asked from you:That is the question you haven't addressed and need to answer.

I'm having trouble with this. If a,e indicate the x coordinate on a unit circle, and b,f indicate the y-coordinate on a unit circle, then the equation is saying
the sum of the product of the 2 x-coordinates and the 2 y-coordinates will yield 0. But I don't think this is what you're looking for?

 

[pyroknife]

No. See, eg., http://www.sosmath.com/trig/Trig5/trig5/trig5.html .

Yes, that is the link I used to obtain the trig identities I used. Did you provide the link to answer the question in post #12 or was my trig identity used incorrectly?

 

[LCKurtz]

$(a,e)$ and $(b,f)$ are the coordinates of the two points. Anyway, do you recognize $ab + ef$ as a vector operation?

 

[Ray Vickson]

Yes, that is the link I used to obtain the trig identities I used. Did you provide the link to answer the question in post #12 or was my trig identity used incorrectly?

Yes, you mis-read the identity and used it incorrectly.

 

[pyroknife]

$(a,e)$ and $(b,f)$ are the coordinates of the two points. Anyway, do you recognize $ab + ef$ as a vector operation?

Oh...It is the dot product.

I actually had a question specifically on the dot product. If u and v are 2 vectors of the same size. I was wondering if the following is an 'if and only if' or 'if' condition:
u and v are perpendicular to each other IFF (or if?) their dot product is zero?
I always thought it is IFF and not just 'if'

If it is IFF, then that would mean then column #1 and column #2 would have to be perpendicular to one another.

 

[LCKurtz]

Oh...It is the dot product

And it is zero. What does that tell you about the two position vectors? And if you, following Ray's suggestion, write $(a,e) = (\cos\theta,\sin\theta)$, what possibilities does that give you for $(b,f)$? I have to run now but hopefully that will get you going.

 

[pyroknife]

Yes, you mis-read the identity and used it incorrectly.

Ahh, yes I did. It should be
$$cos(\theta)sin(\phi)+sin(\theta)cos(\phi)=sin(\theta+\phi)$$)

 

[pyroknife]

I still can't figure out how to proceed with the problem.
We have a normalization condition for the 2 unknown columns of A, and the dot product of any 2 vectors of the column of A should be zero.

We can let b=$$cos(\phi)$$ and f=$$sin(\phi)$$, and find the values of the 2 angles where the dot product is zero.

But the problem I see is, b can also be $$sin(\phi), -cos(\phi), -sin(\phi)$$. It just seems there are many different matrices where the elements is a function of cos() and sin()...
Or am I misunderstanding something?

 

[Ray Vickson]

Ahh, yes I did. It should be
$$cos(\theta)sin(\phi)+sin(\theta)cos(\phi)=sin(\theta+\phi)$$)

Exactly. And setting that to 0 gives you information.

 

[LCKurtz]

With both Ray and myself communicating with you, you need to quote whatever post to which you are replying. I'm guessing this is a reply to my post #23. See my comments below.

I still can't figure out how to proceed with the problem.
We have a normalization condition for the 2 unknown columns of A, and the dot product of any 2 vectors of the column of A should be zero.

We can let b=$$cos(\phi)$$ and f=$$sin(\phi)$$, and find the values of the 2 angles where the dot product is zero.

But the problem I see is, b can also be $$sin(\phi), -cos(\phi), -sin(\phi)$$. It just seems there are many different matrices where the elements is a function of cos() and sin()...
Or am I misunderstanding something?

You didn't answer my questions:
1. What does the dot product of nonzero vectors being zero tell you about the two vectors? I'm talking about any two vectors, including the two in this problem. Please quote this message and answer that.

2. If $(a,e) = (\cos\theta,\sin\theta)$, given the answer to 1, what are the possible angles for $(b,f)$?

 

[pyroknife]

Exactly. And setting that to 0 gives you information.

Yes, setting $$sin(\theta+\phi)=0$$, would mean that $$\theta+\phi = k*\pi$$, where k = 0, +/-1, +/-2, +/-3, ...
But this is for the sum of $$\theta$$ and $$\phi$$. Each element of matrix A will only be a function of either $$\theta$$ or $$\phi$$. Does this mean the angles are dependent on each other?

 

[pyroknife]

With both Ray and myself communicating with you, you need to quote whatever post to which you are replying. I'm guessing this is a reply to my post #23. See my comments below.
You didn't answer my questions:
1. What does the dot product of nonzero vectors being zero tell you about the two vectors? I'm talking about any two vectors, including the two in this problem. Please quote this message and answer that.

2. If $(a,e) = (\cos\theta,\sin\theta)$, given the answer to 1, what are the possible angles for $(b,f)$?

1) If the dot product of two vectors is zero, doesn't that mean those two vectors are orthogonal to each one another?

2) sin(θ+ϕ)=0
θ+ϕ=k∗π
, where k = 0, +/-1, +/-2, +/-3, ...

So if $$\phi$$ is the angle associated with (b,f), then it is dependent on $$\theta$$, the angle associated with (a,e).

 

[LCKurtz]

1) If the dot product of two vectors is zero, doesn't that mean those two vectors are orthogonal to each one another?

Yes.

2) sin(θ+ϕ)=0
θ+ϕ=k∗π
, where k = 0, +/-1, +/-2, +/-3, ...

So if $$\phi$$ is the angle associated with (b,f), then it is dependent on $$\theta$$, the angle associated with (a,e).

Now you are mixing up my discussion with Ray's. I haven't been following your and Ray's discussion closely and I'm not sure I agree with $\sin(\theta+\phi)=0$, but maybe your notation is different. Regardless, draw a unit circle with $(a,e)$ at an angle $\theta$. Now there are two points where $(b,f)$ can be to have the position vectors perpendicular. What are these angles in terms of $\theta$? Stick with angles in $[0,2\pi]$, or just 2 non-coterminal angles.

 

[Dick]

Yes, setting $$sin(\theta+\phi)=0$$, would mean that $$\theta+\phi = k*\pi$$, where k = 0, +/-1, +/-2, +/-3, ...
But this is for the sum of $$\theta$$ and $$\phi$$. Each element of matrix A will only be a function of either $$\theta$$ or $$\phi$$. Does this mean the angles are dependent on each other?

If that's true then $cos(\theta)=cos(\phi)$ and $sin(\theta)=-sin(\phi)$, yes? Show that if you don't know it. So you can replace all of the matrix elements with their dependence on a single angle. Right?

 

[pyroknife]

Yes.Now you are mixing up my discussion with Ray's. I haven't been following your and Ray's discussion closely and I'm not sure I agree with $\sin(\theta+\phi)=0$, but maybe your notation is different. Regardless, draw a unit circle with $(a,e)$ at an angle $\theta$. Now there are two points where $(b,f)$ can be to have the position vectors perpendicular. What are these angles in terms of $\theta$? Stick with angles in $[0,2\pi]$, or just 2 non-coterminal angles.

In terms of $$\theta$$, (b,f) would be at an angle of $$\theta + \frac{\pi}{2}$$

 

[LCKurtz]

In terms of $$\theta$$, (b,f) would be at an angle of $$\theta + \frac{\pi}{2}$$

That's one of the possibilities. What would the polar coordinates of that point be?

Now, what about $\pi/2$ the other way? What would its polar coordinates be?

So you have a couple of possibilities, all in terms of $\theta$.

 

[pyroknife]

That's one of the possibilities. What would the polar coordinates of that point be?

Now, what about $\pi/2$ the other way? What would its polar coordinates be?

So you have a couple of possibilities, all in terms of $\theta$.

Yes, sorry, I mean it could be +pi/2 or -pi/2.
So if a was $$cos(\theta)$$ then b would be $$cos(\theta+\pi/2) or cos(\theta-\pi/2)$$.

Is this the right idea?

 

[LCKurtz]

Yes, sorry, I mean it could be +pi/2 or -pi/2.
So if a was $$cos(\theta)$$ then b would be $$cos(\theta+\pi/2) or cos(\theta-\pi/2)$$.

Is this the right idea?

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.