What is the standard matrix for T: R^2-->R2?

[Gramsci]

Homework Statement

T: R^2-->R2 first rotates points through -3pi/4 radian clockwise and then reflects points through the horizontal x1-axis. Find the standard matrix of T.

Homework Equations

-

The Attempt at a Solution

Vector 1 is:
(-1sqrt2
- 1 sqrt2)
before the reflection, and after the reflection it is:
(-1 sqrt 2
1 sqrt 2)
This I knew since we can use:
(cos insert radian
sin radian)
for v1.

But for v2, I have some problems.
If I use:
(-sin insert radian
cos insert radian)
to solve this, my answer doesn't seem to be correct.
My attempt:
v2
(1sqrt 2
1 sqrt 2)
before the reflection, since sin is negative here two negatives equals a plus. Cos is plus as we can see.
After the reflection:
v2
(-1sqrt 2
1 sqrt 2)

this however, isn't the answer. Can anyone please help me? I'm having enormous troubles.
/Gramsci

 

[gabbagabbahey]

Let's call the matrix corresponding to a reflection throught the x-axis, \overleftrightarrow{A}. What does \overleftrightarrow{A} do to the components of a vector \vec{v} =(v_x,v_y)^T?

 

[Gramsci]

It transforms it into:
(1 0
0 -1) , right? I can't seem to derive any further info from this though.

 

[gabbagabbahey]

It transforms it into:
(1 0
0 -1) , right? I can't seem to derive any further info from this though.

No, it transforms v_x \rightarrow v_x and v_y \rightarrow -v_y

\Rightarrow \left( \begin{array}{c} v_x \\ -v_y \end{array} \right) = \overleftrightarrow{A} \left( \begin{array}{c} v_x \\ v_y \end{array} \right) = \left( \begin{array}{cc} A_{11} &A_{12} \\ A_{21} &A_{22} \end{array} \right) \left( \begin{array}{c} v_x \\ v_y \end{array} \right)

You should be able to determine A_{11}, A_{12}, A_{21}, and A_{22} from this.

 

[Gramsci]

I think I've got it.
Since:
(-sin -3pi/4
cos -3pi/4)

can be compared to a linear reflection (can it?) It's supposed to be:
(1sqrt 2
- 1 sqrt 2)

I derived the latter entry from the fact that a linear reflection does the things we stated in the previous post. After that, we transform it back and therefore we can take away the minus from the cos and our final answer is:
v2=
(1 sqrt 2
1 sqrt 2)

and v1
-1 sqrt 2
1 sqrt 2

It doesn't feel right though. Thank you so much for helping me and excuse me if I am a little slow, I'm not used to these kinds of problems.

Anything I should think about further on this question? Did I do anything wrong?
/Gramsci

 

[gabbagabbahey]

I think I've got it.
Since:
(-sin -3pi/4
cos -3pi/4)

can be compared to a linear reflection (can it?) It's supposed to be:
(1sqrt 2
- 1 sqrt 2)

I derived the latter entry from the fact that a linear reflection does the things we stated in the previous post. After that, we transform it back and therefore we can take away the minus from the cos and our final answer is:
v2=
(1 sqrt 2
1 sqrt 2)

and v1
-1 sqrt 2
1 sqrt 2

It doesn't feel right though. Thank you so much for helping me and excuse me if I am a little slow, I'm not used to these kinds of problems.

Anything I should think about further on this question? Did I do anything wrong?
/Gramsci

what are v1 and v2 ? are they vectors or matrices?

 

[Gramsci]

Vectors

 

[gabbagabbahey]

Vectors

A linear transformation in R2 is a 2X2 matrix, not a vector. A rotation is a linear transformation, and hence also a 2X2 matrix; and the same is true for a reflection.

 

[Gramsci]

Well, then the complete matrix should be:
(-1 sqrt 2, 1 sqrt 2
1 sqrt 2, 1 sqrt 2)

Right?
Did I do something wrong in how I got these answers?

 

[gabbagabbahey]

Well, then the complete matrix should be:
(-1 sqrt 2, 1 sqrt 2
1 sqrt 2, 1 sqrt 2)

Right?
Did I do something wrong in how I got these answers?

This is close, but not correct. Let's start with the matrix for just the reflection, \overleftrightarrow{A} , what do you get for that?

 

[Gramsci]

This is close, but not correct. Let's start with the matrix for just the reflection, \overleftrightarrow{A} , what do you get for that?

Hm, it seems like I forgot a sign. What I meant was:
(-1/sqrt 2, 1/ sqrt 2
1 / sqrt2, 1/ sqrt 2)

This is however, correct, no?

However, the reflection matrix should be(?):

(A11 -A12
A21 -A22)?

 

[gabbagabbahey]

Hm, it seems like I forgot a sign. What I meant was:
(-1/sqrt 2, 1/ sqrt 2
1 / sqrt2, 1/ sqrt 2)

This is however, correct, no?

However, the reflection matrix should be(?):

(A11 -A12
A21 -A22)?

Still incorrect.

What are the values of A11, A12, A21, A22?

 

[Gramsci]

Are you sure it's incorrect? Because according to my solutions manual, that is the solution.

 

[Gramsci]

However, are the values supposed to be:
(1, -1
1, -1)?
Excuse me if I'm slow.

 

[gabbagabbahey]

Are you sure it's incorrect? Because according to my solutions manual, that is the solution.

I get:

\overleftrightarrow{T} \rightarrow \left( \begin{array}{cc} \frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right)

 

[gabbagabbahey]

However, are the values supposed to be:
(1, -1
1, -1)?
Excuse me if I'm slow.

For the matrix representing just the reflection, I get:

<br /> \overleftrightarrow{A} \rightarrow \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right) <br />

And for just the rotation I get:

<br /> \overleftrightarrow{R} \rightarrow \left( \begin{array}{cc} \frac{-1}{\sqrt{2}} &amp; \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &amp; \frac{-1}{\sqrt{2}} \end{array} \right) <br />

 

[Gramsci]

Hm, according to LAYs "Linear algebra and its applications" it's supposed to be what I previously wrote. Strange. Do you have any form of gmail or msn by the way, it would be easier to discuss through this. Care to show how you came to your answer?

 

[Gramsci]

It transforms it into:
(1 0
0 -1) , right? I can't seem to derive any further info from this though.

I thought I wrote that the reflection matrix was:

(1, 0
0, -1)

up there?

 

[gabbagabbahey]

I thought I wrote that the reflection matrix was:

(1, 0
0, -1)

up there?

Yes, I thought you were saying that that was what the vector became though.

I just realized that the rotation is by an angle -3pi/4, I thought it was +3pi/4, so your T is correct.

 

[Gramsci]

Otherwise, did I do everything right in how I got to the answer?

 

[gabbagabbahey]

Otherwise, did I do everything right in how I got to the answer?

I'm not sure, I don't quite understand your notation...perhaps it is a language barrier?

 

[Gramsci]

Perhaps. Tell me if you want me to clarify something.