Matrix of reflection

The Pointer

Sep 4, 2016 05:17

Can anyone help me with a linear algebra problem please? I've spent literally tens of hours on it and am desperate.

It is a simple problem but I have issues understanding the trigonometry aspect.

Martin Sleziak

@ThePointer As the room description says: "Just ask; don't ask to ask."

The Pointer

Ok

Using the standard basis of ℝ2 to determine the matrix of the following linear transformation

Martin Sleziak

Simply state your problem and maybe somebody will help maybe not. (And certainly link to the question if you posted it on main, too.)

The Pointer

1

Q: Using the standard basis of R2, determine the matrix of each of the following linear transformations

I've asked questions about these problems before but thus far, no one has been able to help me understand. I've been attempting these problems for tens of hours over a couple of weeks now. I'm desperate for some help. The problem are as follows: The issue is not the linear transformation aspe...

Martin Sleziak

The reason for this suggestion that without seeing the problem it is difficult to say whether I will be able to help you.

The Pointer

Sep 4, 2016 05:25

This is the problem. I am having issues with solving (2)

I have solved (1).

Martin Sleziak

So to find a matrix of the reflection w.r.t. the line which has the angle $\theta/2$ to the x-axis.

How did you want to deal with the problem?

The Pointer

I was using trig ratios to try and calculate the values for (x,y)

Based on the image that was provided

Martin Sleziak

I am not sure what you mean by values for (x,y).

The Pointer

sorry, (x,y) values for f(e_1) and f(e_2)

Martin Sleziak

Ok, so you have $e_1=(1,0)$.

The Pointer

Sep 4, 2016 05:28

I calculated f(e_1) to be (cos(theta), sin(theta))

Martin Sleziak

The you immediately see that $f(e_1)$ will have unit length and angle $\theta$.

Exactly as you wrote.

The Pointer

However, I am not getting the correct answer for f(e_2)

Martin Sleziak

Now you need to find $f(e_2)$.

$e_2$ is at angle $\pi/2$. The line is at angle $\theta$. So the angle between them is $\pi/2-\theta$.

So to me it looks like we need the new angle $\frac\pi2-2\left(\frac\pi2-\theta\right)=2\theta-\frac\pi2$.

Is $2\theta-\frac\pi2$ the correct result @ThePointer?

The Pointer

I see a lot of $ symbols in your text, making it hard to understand. Am i doing something wrong?

$\frac\

Martin Sleziak

To get math rendered, you can use bookmarklet from here: math.ucla.edu/~robjohn/math/mathjax.html (The link is both in the room description and in the rules of this chatroom.)

Anyway my results was 2*theta-pi/2. Is that what you are supposed to get?

Sorry, I used $\theta$ instead of $\theta/2$ as the angle of the given line.

So the result changes to $\theta-\pi/2$ if we use $\theta/2$.

The Pointer

Sep 4, 2016 05:38

Ok i see it now

working

The solution is

imgur.com/a/VN0Sr

Not that :(

Martin Sleziak

Well, that's exactly what we got, just written differently.

The Pointer

Ahh, I see.

Martin Sleziak

We got $f(e_2)=(\cos\left(\theta-\frac\pi2\right),\sin\theta-\frac\pi2)$, right?

Le't have a look whether we can simplify it.

$\cos\left(\theta-\frac\pi2\right)=\cos\left(\frac\pi2-\theta\right)=\sin\theta$

$\sin\left(\theta-\frac\pi2\right)=-\sin\left(\frac\pi2-\theta\right)= -\cos\theta$

The Pointer

Ahh, I see.

So since we are reflecting both points on either side of the line

it creates an angle of pi/2?

between f(e_1) and f(e_2)?

Martin Sleziak

No, that's not true. Why should it be $\pi/2$?

Just draw a few pictures and you will see that the angle between the two images can be various, depending on $\theta$.

The Pointer

Sep 4, 2016 05:45

You said e_2 is at angle pi/2

Martin Sleziak

Yes, that's true.

And angle between $e_1$ and $e_2$ is $\pi/2$.

The Pointer

Oh, I see.

Martin Sleziak

But the angle between $f(e_1)$ and $f(e_2)$ can be different.

The Pointer

I thought you were referring to f(e_1) and f(e_2) my apologies

Hmm

Is there any way to calculate these using trig ratios?

Martin Sleziak

I don't know what are trig ratios.

The Pointer

Sep 4, 2016 05:51

That is how I calculated the previous ones but for some reason am not getting the correct answer for f(e_2)

sin(theta) = opposite/hypotenuse, ect

Martin Sleziak

Isn't that what we just did? We found angle for $f(e_2)$ and then expressed it using cos and sin.

The Pointer

Unfortunately, I do not understand :(

Martin Sleziak

BTW in this answer you have a different solution, which uses projection matrix.

@ThePointer So let us try once again.

The Pointer

Yes. However, I am not supposed to solve using the projection matrix.

Martin Sleziak

If it does no help, maybe you can spend some time looking at other similar problems on the main, I listed a few here:

The Pointer

Sep 4, 2016 05:55

Ok will do

Martin Sleziak

So we have $e_2=(0,1)$ and want to calculate $f(e_2)$, right?

The Pointer

Yes. It is in the 4th quadrant

Martin Sleziak

First, the angle between x-axis and $e_2$ is $\pi/2$, do you agree with this?

The Pointer

Yes

Martin Sleziak

And we have the axis of reflection, which is the line with the angle $\theta/2$.

The Pointer

Sep 4, 2016 05:57

Yes

Martin Sleziak

So the angle between $e_2$ and the axis of reflection is $\pi/2-\theta/2$.

So far, ok?

The Pointer

Yes

Martin Sleziak

In fact, let us denote $\alpha=\pi/2-\theta/2$ for simplicity.

The Pointer

Ok

Martin Sleziak

So $\alpha$ is the angle between the vector we want to transform and the reflection axis.

Now we should ask how the angle changes if we look at $f(e_2)$ instead of $e_2$.

We have to subtract the angle $2\alpha$.

The Pointer

Sep 4, 2016 05:59

well f(e_2) is in quadrant 4

Yes

Martin Sleziak

Notice that if we rotate $e_2$ by the angle $-\alpha$, we get a vector which is in the direction of the reflection axis.

If we do it once again, it is the result of reflectoin.

So we need to calculate $\pi/2-2\alpha$.

Is this ok?

The Pointer

Yes

Martin Sleziak

Now it is some simple algebraic manipulation $$\frac\pi2-2\alpha = \frac\pi2-2\left(\frac\pi2-\frac\theta2\right) = \theta-\frac\pi2.$$

We know that the angle is $\theta-\pi/2$ and the vector has unit length, which means
$$f(e_2)=\left(\cos\left(\theta-\frac\pi2\right),\sin\left(\theta-\frac\pi2\right)\right).$$

Now it only remains to simplify $\cos\left(\theta-\frac\pi2\right)$ and $\sin\left(\theta-\frac\pi2\right)$.

The Pointer

I got most of it

The part where you say rotate e_2 by angle -a

and then we do it once again, won't that be along the x-axis?

Martin Sleziak

Not necessarily.

You probably drew the picture where the $\theta=\pi/4$.

Try to draw several picture, with various reflecion axes.

What you wrote is true only if the reflection axis is exactly in the middle between $e_1$ and $e_2$.

The Pointer

Sep 4, 2016 06:15

Yes

That is true

so pi/2 - 2a gets us to f(e_2)?

Martin Sleziak

Yes.

The Pointer

Hmm. This method is difficult for me to understand conceptually and generalise.

How would one do this using purely the sin, cos ratios?

And drawing the image?

Martin Sleziak

I think that we did exactly what you said.

The Pointer

Hmm this must be the only way

Yes I think so

Martin Sleziak

If you draw the image you could see that the the reflection works like this.

@ThePointer That is certainly not the only way.

I don't think you can say about any mathematical problem that there is only one way to solve it.

BTW I have posted an answer summarizing what we discussed here in chat.

The Pointer

Sep 4, 2016 06:23

Yes it is very elaborate thank you

I am reading it now

Martin Sleziak

And I have also tried to retype your question. (It is better not to use images if possible.)

The Pointer

I was going well all semester

until I got to these two questions

I've spent literally tens of hours trying to understand these. There must be a deficiency in my previous knowledge

Feels very bad :(

Martin Sleziak

Do you mind if I point out some things which I think the question you posted is lacking @ThePointer?

The Pointer

Yes please do

Martin Sleziak

1) You knew what the solution is supposed to be. (You mention this in chat.) But you did not say so in the post. (That's why I have edited your question.)

2) You did not mention where the problem comes from.

It is always good to give all the available details which might be useful for answerers.

The Pointer

Sep 4, 2016 06:26

My apologies. I will do this in the future.

Martin Sleziak

It might be useful to read How to ask a good question?. The things I mentioned are in the part about context

@ThePointer Why in the future. You can still improve this post by adding at least the source of the problem.

You can also remove the image if you are satisfied with the transcription.

The Pointer

Ok.

Martin Sleziak

@ThePointer I know looked at the list of your questions and I see that you have asked the same question as part 2 of this one here.

You should not post the same question twice.

The Pointer

Yes. I became desperate :(

I will not do it again

Martin Sleziak

The only difference seems to be that the other question does not require calculation of the determinant.

Probably the best thing to do is if I delete my answer and post it to the other question - it is more suitable there.

The Pointer

Sep 4, 2016 06:33

Ok that is fine

Yes I am reading your answer and it is making complete sense now

$Mathematics$ Mathematics

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