The Student Room Group

Matrices and transformations

Question is as follows:

Let A be the matrix (cosθsinθsinθcosθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}

(a) Interpret A geometrically

Basically, I'm stuck on the first part; the answer in the back of the book is a reflection in the line y=xtan12θy = x tan\frac{1}{2}\theta but I have no clue as to how you arrive at this - would be great if I could have some insight :biggrin:
Original post by Femto
Question is as follows:

Let A be the matrix (cosθsinθsinθcosθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}

(a) Interpret A geometrically

Basically, I'm stuck on the first part; the answer in the back of the book is a reflection in the line y=xtan12θy = x tan\frac{1}{2}\theta but I have no clue as to how you arrive at this - would be great if I could have some insight :biggrin:


Look at what happens to (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}
Reply 2
Original post by Get me off the £\?%!^@ computer
Look at what happens to (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}


Sorry I'm a bit confused. My teacher has rushed through the transformations and I didn't understand it all to be honest.
(edited 13 years ago)
(cosθsinθsinθcosθ)(10)=(cosθsinθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}

(cosθsinθsinθcosθ)(01)=(sinθcosθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} \sin \theta \\ -\cos \theta \end{pmatrix}

Sketch the vectors for some angle theta and you will see that the given answer is correct.

Sorry this is not a good explanation. I'm feeling like **** today. Signing off now. Someone else will explain better.
(edited 13 years ago)
Reply 4
Original post by Get me off the £\?%!^@ computer
(cosθsinθsinθcosθ)(10)=(cosθsinθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}

(cosθsinθsinθcosθ)(01)=(sinθcosθ)\begin{pmatrix} cos \theta & sin \theta \\sin \theta & -cos \theta \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} \sin \theta \\ -\cos \theta \end{pmatrix}

Sketch the vectors for some angle theta and you will see that the given answer is correct.

Sorry this is not a good explanation. I'm feeling like **** today. Signing off now. Someone else will explain better.


Ok, sorry for bothering you :biggrin: Hope you get better.

I kind of understand it :s-smilie: Not entirely though.
Reply 5
I just don't understand how you sketch those vectors to be honest and how the line equation comes into it.. :/
Reply 6
Its determinant is -1 so it must be a reflection with no scaling. You can work out the line of reflection by plotting the points (1,0) and the image of (1,0) under the transformation. Draw a line through the origin and (1,0) (i.e. the x-axis) and a line through the origin and its image; what is the angle between these two lines? The line of reflection must bisect this angle; work out the gradient (and hence equation) of this line. Et voilà.
Original post by nuodai
Its determinant is -1 so it must be a reflection with no scaling.


Really?
Reply 8
Original post by Get me off the £\?%!^@ computer
Really?


*Cough* and it's orthogonal.
Original post by nuodai
*Cough* and it's orthogonal.


which I knew you meant :wink:
Reply 10
Original post by nuodai
Its determinant is -1 so it must be a reflection with no scaling. You can work out the line of reflection by plotting the points (1,0) and the image of (1,0) under the transformation. Draw a line through the origin and (1,0) (i.e. the x-axis) and a line through the origin and its image; what is the angle between these two lines? The line of reflection must bisect this angle; work out the gradient (and hence equation) of this line. Et voilà.


I don't understand how you plot cos theta on an axis :frown:

If (1,0) becomes (cos, sin) how am I meant to plot that?

Sorry for short hand writing by the way, writing on iPod.
(edited 13 years ago)
Reply 11
Original post by Femto
I don't understand how you plot cos theta on an axis :frown:

If (1,0) becomes (cos, sin) how am I meant to plot that?

Sorry for short hand writing by the way, writing on iPod.


Draw a circle of radius 1 centred at the origin. What are the coordinates of a point at an angle θ\theta round the circle (starting at (1,0) and going anticlockwise)? Hint: draw a right-angled triangle to work out the coordinates.
Reply 12
Original post by nuodai
Draw a circle of radius 1 centred at the origin. What are the coordinates of a point at an angle θ\theta round the circle (starting at (1,0) and going anticlockwise)? Hint: draw a right-angled triangle to work out the coordinates.


Right so that point would be (cos theta, sin theta)? :smile:
Reply 13
Original post by Femto
Right so that point would be (cos theta, sin theta)? :smile:


Yup. Also look at where (0,1) goes (it maps to a point on the circle). What does this tell you about the linear transformation represented by the matrix?
Reply 14
Original post by nuodai
Yup. Also look at where (0,1) goes (it maps to a point on the circle). What does this tell you about the linear transformation represented by the matrix?


Hmm well (0,1) goes to (sin theta, -cos theta).

I'm a bit confused as to what this tells me about the transformation. In really sorry about this, but it is pretty difficult for me to understand what the hell is going on :wink:
(edited 13 years ago)

Quick Reply