I have like two ways of doing this, but I like to understand the formal one

[flyingpig]

Homework Statement

http://img219.imageshack.us/img219/3434/transp.th.png

Uploaded with ImageShack.us

The Attempt at a Solution

Personally, I always have trouble doing these type of problems algebraically and using the def of Linear Transformation.

I always resort to drawing pictures and see where they go then writ out my standard matrix

So I did that and I got

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP159819f5fef5cd3be33100005ce6h6eh806b03h8?MSPStoreType=image/gif&s=63&w=55&h=36

Now if I were to do this using algebra, I have to find the standard matrix for each transformation

So starting with my unit vectors

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP212219f5fci0014e55a80000102e73aga167a9ih?MSPStoreType=image/gif&s=62&w=44&h=36

For a reflection the standard matrix for this transformation is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP165419f5fef5cd3be331000047gig59d7ga41029?MSPStoreType=image/gif&s=63&w=44&h=36

For a rotation 90 degrees the standard matrix for this transformation is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP240719f5fcfg088hgc650000205f87i63i2h04fg?MSPStoreType=image/gif&s=57&w=55&h=36

So e₁ -> e₂ -> e₂

e₂ -> e₁ -> -e₁

So the standard matrix I get is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP240719f5fcfg088hgc650000205f87i63i2h04fg?MSPStoreType=image/gif&s=57&w=55&h=36

 

[LCKurtz]

Homework Statement

http://img219.imageshack.us/img219/3434/transp.th.png

Uploaded with ImageShack.us

The Attempt at a Solution

Personally, I always have trouble doing these type of problems algebraically and using the def of Linear Transformation.

I always resort to drawing pictures and see where they go then writ out my standard matrix

So I did that and I got

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP159819f5fef5cd3be33100005ce6h6eh806b03h8?MSPStoreType=image/gif&s=63&w=55&h=36

Now if I were to do this using algebra, I have to find the standard matrix for each transformation

So starting with my unit vectors

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP212219f5fci0014e55a80000102e73aga167a9ih?MSPStoreType=image/gif&s=62&w=44&h=36

For a reflection the standard matrix for this transformation is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP165419f5fef5cd3be331000047gig59d7ga41029?MSPStoreType=image/gif&s=63&w=44&h=36

For a rotation 90 degrees the standard matrix for this transformation is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP168519f5fffc532c41310000549gaci25232dg63?MSPStoreType=image/gif&s=64&w=55&h=36

So e₁ -> e₂ -> e₂

e₂ -> e₁ -> -e₁

So the standard matrix I get is [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP240719f5fcfg088hgc650000205f87i63i2h04fg?MSPStoreType=image/gif&s=57&w=55&h=36[/QUOTE]

Your 90 degree rotation matrix is wrong.

 

[flyingpig]

Fixed, copied off the wrong matrix

 

[LCKurtz]

It is getting confusing which matrix is which. You have the standard rotation by 90 degrees as

\left[ \begin{array}{cc}<br /> <br /> 0&amp; 1\\<br /> -1 &amp; 0<br /> \end{array}<br /> \right]

That isn't right and I don't see what you corrected. You should get the same answer as you did at the top of your post.

 

[flyingpig]

No the matrix I fixed is this [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP240719f5fcfg088hgc650000205f87i63i2h04fg?MSPStoreType=image/gif&s=57&w=55&h=36

My standard matrix for the final answer is still right

 

[Mark44]

No the matrix I fixed is this [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP240719f5fcfg088hgc650000205f87i63i2h04fg?MSPStoreType=image/gif&s=57&w=55&h=36

My standard matrix for the final answer is still right

Bad link in the image above.
Instead of posting images, why don't you put the matrices inline? There's really not much to it. Here's the one that LCKurtz posted, using the simpler bmatrix environment.

Click the matrix to see the LaTeX code.

\begin{bmatrix}0&amp; 1\\-1 &amp; 0\end{bmatrix}

 

[flyingpig]

No I copied it off wolfram because i got lazy off writing my own tex code lol

 

[flyingpig]

Here I will do it again since PF won't let me edit it anymore and the image link is broken now

For a 90 degree counter clockwise rotation, we get

\begin{bmatrix}<br /> 0 &amp; -1\\ <br /> 1 &amp; 0<br /> \end{bmatrix}

For a reflection about y = x, we get

\begin{bmatrix}<br /> 0 &amp; 1\\ <br /> 1 &amp; 0<br /> \end{bmatrix}

Normally this is what a textbook would do

e_1 \to e_2 \to e_2

e_2 \to -e_1 \to e_1

Which should yield the matrix \begin{bmatrix}0 &amp; -1 \\ 1 &amp; 0\end{bmatrix}

Now the way I did it is that I drew a picture with my unit vectors and I just follow where my vectors go according to the instructions

I will even draw a picture for you

http://img703.imageshack.us/img703/6665/unledhg.th.png

Uploaded with ImageShack.us

So at the end, my vectors ended up as \begin{bmatrix}<br /> -1 &amp; 0 \\ <br /> 0 &amp; 1<br /> \end{bmatrix}

Now my question is, who is right?

 

[flyingpig]

Mark or Kurts, here try seeing this

 

[flyingpig]

Oh good it worked lol

 

[Mark44]

I'm not sure what your question is. If you're looking for the matrix that represents a rotation by 90° CCW followed by a reflection across the line y = x, that would be:
\begin{bmatrix}0 &amp; 1\\ 1 &amp; 0\end{bmatrix}\begin{bmatrix}0 &amp; -1\\ 1 &amp; 0\end{bmatrix} = \begin{bmatrix}1 &amp; 0\\ 0 &amp; -1\end{bmatrix}

If that's not what you're asking about, please restate your question.

 

[flyingpig]

Yes that is what I want, but why did you multiply the matrix...?

 

[Mark44]

Because the composition of linear transformations is equivalent to multiplying the matrices that represent the transformations.

 

[LCKurtz]

Because the composition of linear transformations is equivalent to multiplying the matrices that represent the transformations.

The original problem was to be a reflection followed by a rotation so they would be multiplied the other way:

<br /> \begin{bmatrix}0 &amp; -1\\ 1 &amp; 0\end{bmatrix}\begin{bmatrix}0 &amp; 1\\ 1 &amp; 0\end{bmatrix} = \begin{bmatrix}-1 &amp; 0\\ 0 &amp; 1\end{bmatrix}<br />

which agrees with his intuitive answer [that used to be] in his first post.

 

[flyingpig]

Does that mean I am wrong? Because my textbook does the following and I tried multiplying matrices like you did, but it isn't right.

T: \mathbb{R}^2 \mapsto \mathbb{R}^2 first reflects points through the horizontal x₁ axis and then reflects points through the line x₂ = x₁

My book has the matrix \begin{bmatrix}<br /> 0 &amp; -1\\ <br /> 1 &amp; 0<br /> \end{bmatrix} as the answerNow I also got that as my answer using the matrix transformation or drawing the pictures.

If I were to multiply the two transmations like you did

\begin{bmatrix}1 &amp; 0\\ 0 &amp; -1\end{bmatrix} \begin{bmatrix} 0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix} = \begin{bmatrix} 0 &amp; 1 \\ -1 &amp; 0 \end{bmatrix}

Which is wrong.

Now in my textbook, they do transformations like this

Now going back my original question, I did it both ways, I did it the way my textbook did and the way I usually do it (with pictures). So why do they give me different answers?

 

[LCKurtz]

Does that mean I am wrong? Because my textbook does the following and I tried multiplying matrices like you did, but it isn't right.



My book has the matrix \begin{bmatrix}<br /> 0 &amp; -1\\ <br /> 1 &amp; 0<br /> \end{bmatrix} as the answer


Now I also got that as my answer using the matrix transformation or drawing the pictures.

If I were to multiply the two transmations like you did

\begin{bmatrix}1 &amp; 0\\ 0 &amp; -1\end{bmatrix} \begin{bmatrix} 0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix} = \begin{bmatrix} 0 &amp; 1 \\ -1 &amp; 0 \end{bmatrix}

Which is wrong.

Yes. That is wrong because you are multiplying them the wrong way. When you take matrices A and B multiply Bx first then multiply by A you get ABx. In matrix multiplication, the first one to operate is on the right. Multiply them in the other order.

 

[flyingpig]

I used wolframalpha http://www4a.wolframalpha.com/input/?i={{1%2C0}%2C{0%2C-1}}*{{0%2C1}%2C{1%2C0}}

Also, can you tell me what the book is doing?

Also, going back to my very original question, the one where my intuition was right but my textbook method was wrong, why?

 

[LCKurtz]

Your textbook method is correct and you get the same answers with matrices if you multiply them in the correct order. If you are still having problems and you want me to continue this discussion with you, post a specific example you have worked both ways in your next post. Don't refer to older posts or links.

 

[flyingpig]

No the original problem is the problem where I worked both ways and I arrived with different answers. Check post #8 of this thread

 

[Mark44]

Here I will do it again since PF won't let me edit it anymore and the image link is broken now

For a 90 degree counter clockwise rotation, we get

\begin{bmatrix}<br /> 0 &amp; -1\\ <br /> 1 &amp; 0<br /> \end{bmatrix}

For a reflection about y = x, we get

\begin{bmatrix}<br /> 0 &amp; 1\\ <br /> 1 &amp; 0<br /> \end{bmatrix}

Normally this is what a textbook would do

e_1 \to e_2 \to e_2

e_2 \to -e_1 \to e_1

There are errors in both of these. I'm assuming that the transformations are 1)reflect across the line y = x, and then 2) rotate by 90 deg. CCW, in that order. This is what you had in your first post.

In the first line above, e₁ --> e₂, but then e₂ --> -e₁.
In the second line above, e₂ --> e₁, and then e₁ --> e₂. As a result of both transformations, the vectors {e₁, e₂} become {-e₁, e₂}

The matrix that represents the composition of both transformations is
\begin{bmatrix} -1&amp;0\\0&amp;1\end{bmatrix}

Notice that this is the same as when you carry out this multiplication
\begin{bmatrix} 0&amp;-1\\1&amp;0\end{bmatrix}\begin{bmatrix} 0&amp;1\\1&amp;0\end{bmatrix} <br />

The second matrix in the product above is the reflection across the line y = x. The first matrix in the product is the rotation.

Which should yield the matrix \begin{bmatrix}0 &amp; -1 \\ 1 &amp; 0\end{bmatrix}

Now the way I did it is that I drew a picture with my unit vectors and I just follow where my vectors go according to the instructions

I will even draw a picture for you

http://img703.imageshack.us/img703/6665/unledhg.th.png

Uploaded with ImageShack.us

So at the end, my vectors ended up as \begin{bmatrix}<br /> -1 &amp; 0 \\ <br /> 0 &amp; 1<br /> \end{bmatrix}

Now my question is, who is right?

 

[LCKurtz]

Your textbook method is correct and you get the same answers with matrices if you multiply them in the correct order. If you are still having problems and you want me to continue this discussion with you, post a specific example you have worked both ways in your next post. Don't refer to older posts or links.

No the original problem is the problem where I worked both ways and I arrived with different answers. Check post #8 of this thread

I knew I would regret re-entering this thread. Post #8 does not state what the problem is. It matters which operation you are doing first and you are getting your operations mixed up. For the last time, state in one post the statement of the problem and your two solutions which don't give the same answer. If you won't do that, consider this as goodbye.

 

[flyingpig]

No never mind, this is whta it should have been

S: = first operation

T: = second operation

S(x) = Ax

T(S(x)) = T(Ax) = B(A)x, so take the reverse product.