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Stretching and Shearing away from Coordinate Axes

Date: 08/27/2012 at 16:40:11
From: Costa
Subject: Linear Transformations - Stretching and Shearing

Hi!  

After an introduction to 2x2 matrices, the chapter I am studying goes on
to transformations. I am having great difficulty with these four 
questions -- in particular, the last three.

Find the matrices corresponding to the following linear transformations:

   (a) Shear with y = x fixed and (1, 0) --> (0, -1)
   (b) Stretch (times 3) perpendicular to y = -x and with y = -x fixed
   (c) Shear with y = x fixed and (0, 2) --> (4, 6)
   (d) Shear y = 2x fixed and (0, 4) --> (2, 0)

The main principle I am working from is that every linear transformation
of the plane has an associated 2x2 matrix. This can be determined by
examining the images of the vectors (1 0) and (0 1). 

The chapter has only explained the more basic rules. That's all well and
good when I am told to find matrices corresponding to linear
transformations such as "180 degrees about the origin" and "enlargement
scale factor 3, center (0, 0)." But I find it very difficult to understand
stretching and shearing when the fixed line is not the x- or y-axis. In
fact, I am a little confused as to what shearing is to begin with.

More specifically, looking at question (b), I don't understand how to
quantify the distances that the images (1 0) and (0 1) are meant to
stretch, as the stretch is occurring at an angle to the x axis. The same
goes with the other questions, I suppose. I tried to use coordinate
geometry to calculate the distances, and wound up going on quite a tangent.

I feel completely lost. Please offer some advice! Thank you!



Date: 08/27/2012 at 17:23:51
From: Costa
Subject: Linear Transformations - Stretching and Shearing

I managed to get the correct answer to the first question, but only by
drawing the question to scale on graph paper and using this formula I
found:

                       distance moved by image due to shear
   scale factor = ----------------------------------------------
                   distance moved by object from invariant line

I imagine there is a more appropriate method to use. Without the formula,
I can see (0 1) moves parallel to y = x. But I am unable to see why it
moves in one direction instead of the other.



Date: 08/27/2012 at 19:19:55
From: Costa
Subject: Linear Transformations - Stretching and Shearing

I have also managed to make the last two questions easier. For example, in
question (c), I multiplied the matrices as follows:

    |a b| |0| = |4|
    |c d| |2|   |6|

This obtained values for b and d, and thereby revealed that (0 1) is 
mapped to (2 3). 

I did the same thing in the last question, thus making these questions
similar to the first. 

Still, I am not sure where to go from here.



Date: 08/27/2012 at 21:28:47
From: Doctor Schwa
Subject: Re: Linear Transformations - Stretching and Shearing

Hi Costa,

From your later messages, it looks like you're doing some great work
toward figuring these things out!

I'm going to suggest two methods: one based on the work you've already
done, and one based on a different idea. In each case, I'll use problem
(c) as my example.

Method 1: As you noticed, if (0, 2) goes to (4, 6), then you can figure
out some things about the matrix.

You also have the fact that y = x is fixed, so (1, 1) goes to (1, 1).

Using those two facts, you can find exactly what the matrix must be. A
similar approach should work for your other problems!

Computationally, I think this method is the easiest; but in terms of
understanding what's going on, I like this method:

Method 2: First, rotate your axes so that the fixed line is the x-axis.
For (c), that means a 45 degree clockwise rotation. Let's call this 
matrix R.

Now you know that the x-axis is fixed, and after rotating your points, you
have (sqrt(2), sqrt(2)) going to (5sqrt(2), sqrt(2)) or -- equivalently,
after a rescaling -- (1, 1) goes to (5, 1).

This shows that when you're one unit above the x-axis, you shear 4 units
to the right (and in general if you're y units above the x-axis, you shear
4y units to the right).

So, let's call the matrix for this S = (1 4 0 1).

Now we need to rotate back, so we multiply by the matrix for a 45 degree
counterclockwise rotation. Let's call this R'.

Probably, based on your notation, you want to multiply them in the order
R'SR to get the result (because you put the point on the right when you're
multiplying, yes?).

This general process of doing something (R), doing something else (S), and
then undoing the first thing (R'), is called conjugation, and it's
incredibly useful in all kinds of places. You can think of conjugation as
the kind of order of events required to walk into a house -- you open the
door first, then go in, and finally close the door.

Please let me know if you have any further questions (or if I've made any
computation errors here!).

Enjoy,

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/28/2012 at 08:33:01
From: Costa
Subject: Linear Transformations - Stretching and Shearing

Wow, thanks for your help.

With regards to Method 1, I understand that completely. However, with
Method 2, there's the odd thing I'm stuck on. 

I wasn't sure what matrices to use to rotate the axis. Rotating (0, 2) 45
degrees clockwise was fine, as I just used basic trig. However, for 
(4, 6), I had to look ahead in the chapter in my book to discover this 
matrix:

   |cosx -sinx|
   |sinx  cosx|

As I was rotating clockwise, I used the angle (360 - 45 =) 315. I then
understood the rest of what you were saying ... up until the conjugation
part. I see it like this: I rotated the axis 45 degrees clockwise, found
the relevant matrix (what you called matrix S), then rotated the axis back
to its original position (45 degrees anti-clockwise from where we are now)
using the matrix you called R'. However, if I do this, I then need to
multiply my result again by R (which rotates the axis 45 degrees
clockwise), and I don't understand this. Please can you explain this to
me?

I have a further question from the next set of problems. I'm essentially
meant to do the inverse of what I had been doing so far: given a
transformation matrix, come up with a geometrical description. 

I can do the basic ones, but when I am given a matrix like this, I am
unsure how I am meant to identify the transformation:

   |-2 3|
   |-3 4|

I draw the graph and look at how the points (1, 0) and (0, 1) have
transformed to (-2, -3) and (3, 4), but I can only estimate some kind of
shearing has taken place.

Thanks!



Date: 08/28/2012 at 16:27:08
From: Doctor Schwa
Subject: Re: Linear Transformations - Stretching and Shearing

Costa,

I agree, it gets hard to describe transformations geometrically beyond
"the unit square turns into this parallelogram" and things like that.

You can always "simplify" a transformation into the product of a rotation
-- or maybe a reflection, a scaling, and a shearing. But I'm not sure if
that helps. Here's a good source for this:

   http://en.wikipedia.org/wiki/Affine_transformation 

(For matrices, we can't have translation, but any affine transformation
that leaves the origin fixed is fair game.)

First of all you can look at the determinant, which gives the area of the
parallelogram (which here is still 1) to see what the scaling should be;
in this case, it has none. Also the determinant is positive 1, not
negative 1, so there's no reflection.

So this is a rotation or a shear. That should be geometrically simple
enough! Looking at the picture, it sure doesn't look like a rotation, so
let's try to find out what line through the origin we're shearing along.
This line stays fixed, so if we look at the point (x, mx), we should still
have that same ratio of m. (And if there's no value of m that makes the
line stay fixed, then we know we have a rotation after all.)

(x, mx) gets mapped to ((-2 + 3m)x, (-3 + 4m)x), so

        (-3 + 4m)
   m = ------------
        (-2 + 3m)

This is solved by m = 1. So it's the y = x line that's staying fixed.

Now you can probably see that (1, 0) is 1/sqrt(2) units below the line,
and it goes to (-2, -3), moving 3sqrt(2) units. So ... assuming I haven't
made any calculation errors here ... the shear factor along that line is 

     3sqrt(2)
   -------------   = 6
    (1/sqrt(2))

And, you can see that this transformation is purely a shear by seeing that
(0, 1), which is 1/sqrt(2) units above the line, moves 3sqrt(2) units
along the line in the opposite direction.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/30/2012 at 10:27:23
From: Costa
Subject: Linear Transformations - Stretching and Shearing

Thanks again for all your help.

Regarding the first question, I'm not sure why I didn't get this at first.
Rotate ... shear ... rotate back. Of course! That makes sense.

Regarding the second question: Thanks, I understand this fine now, except
for one detail. How did you know the line we were shearing along was
through the origin?



Date: 08/30/2012 at 15:21:31
From: Doctor Schwa
Subject: Re: Linear Transformations - Stretching and Shearing

Hi Costa,

If it's a matrix transformation M, then M(0, 0) = (0, 0). That means that
one of the fixed points must be the origin. So if it's a shear with a
fixed line and all other points moving, the line has to go through the
origin.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Coordinate Plane Geometry
High School Linear Algebra

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