2x2 Shear Matrices General form Watch

johnniel
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Hi,

TLDR - I'm trying to create my own shear matices, is there a general form for a shear parallel to the line y=mx?

I'm attempting to create some questions for finding invariant lines of 2x2 shear matrices. For example matrices like {(4, 3), (-3, 2)}. Short of copying past paper questions I'm struggling to come up with my own matrices which are shears parellel to the line y=mx.

I've searched the internet for a general form of a shear 2x2 matrices but they only seem to be parallel to x or y axis.

I've tried taking these matrices and rotating them but the numbers don't work out very nice.

If there was a general form it would be easy to create 'nice' matrices that lead to nice invariant lines.

Any help would be much appreicated.

Thanks
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Pangol
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(Original post by johnniel)
Hi,

TLDR - I'm trying to create my own shear matices, is there a general form for a shear parallel to the line y=mx?
I can't see a nice form for this.

If I am correct - and I make no strong claims that I am! - the matrix for a shear parallel to y = (tan a) x is

1 - k cos a sin a, k cos^2 a
-k sin^2 a, 1 + k cos a sin a

(where k is the "scale factor" of the shear), which I've not been able to find a much tidier version of. It can be written in terms of 2a instead of a, which also removes the squares from the trig functions, but it still doesn't look very nice.

So not much help, I'm afraid. Sorry!
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alow
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(Original post by Pangol)
I can't see a nice form for this.

If I am correct - and I make no strong claims that I am! - the matrix for a shear parallel to y = (tan a) x is

\begin{bmatrix}1 - k \cos a \sin a & k \cos^2 a \\

-k \sin^2 a & 1 + k \cos a \sin a \end{bmatrix}

(where k is the "scale factor" of the shear), which I've not been able to find a much tidier version of. It can be written in terms of 2a instead of a, which also removes the squares from the trig functions, but it still doesn't look very nice.

So not much help, I'm afraid. Sorry!
FTFY
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Pangol
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(Original post by alow)
FTFY
That does look nicer, thank you! (But life's too short to learn LaTex when the meaning is clear, and I'm on holiday!)
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johnniel
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(Original post by Pangol)
That does look nicer, thank you! (But life's too short to learn LaTex when the meaning is clear, and I'm on holiday!)
This form seems to work for the few examples I've got so your maths looks good.

Thank you!
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johnniel
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(Original post by Pangol)
I can't see a nice form for this.

If I am correct - and I make no strong claims that I am! - the matrix for a shear parallel to y = (tan a) x is

1 - k cos a sin a, k cos^2 a
-k sin^2 a, 1 + k cos a sin a

(where k is the "scale factor" of the shear), which I've not been able to find a much tidier version of. It can be written in terms of 2a instead of a, which also removes the squares from the trig functions, but it still doesn't look very nice.

So not much help, I'm afraid. Sorry!
Originally I was checking shears parallel to y=x and y=-x where cos^2(x) and sin^2(x) are identical. However, I've run into a slight problem using other shears. Your general form seems to work if I swap the sin and cos around like following. Do you have your working out on how you derived this matrix in the first place?

It is highly possible I have made a mistake here, but just want to make sure.

Name:  Shear matrix.PNG
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Size:  5.1 KB

Thanks again, I fully appreicate the help.
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Pangol
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(Original post by johnniel)
Do you have your working out on how you derived this matrix in the first place?
Instead of directly shearing along y = (tan a) x, I rotated clockwise by a, sheared parallel to the x-axis, and then rotated anticlockwise by a. The matrices for each of the pieces are easy to get, and then they have to be multiplied together in the correct (i.e. backwards) order.

I think that my version is correct. When we put a = 0, we should get a shear parallel to the x-axis. Try this with both versions and see what you get.
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johnniel
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(Original post by Pangol)
Instead of directly shearing along y = (tan a) x, I rotated clockwise by a, sheared parallel to the x-axis, and then rotated anticlockwise by a. The matrices for each of the pieces are easy to get, and then they have to be multiplied together in the correct (i.e. backwards) order.

I think that my version is correct. When we put a = 0, we should get a shear parallel to the x-axis. Try this with both versions and see what you get.
Yes you are indeed correct, I got the same result deriving it and have since found the mistake in my original working (always the most simple error too, ha)

Thanks
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Pangol
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(Original post by johnniel)
Yes you are indeed correct, I got the same result deriving it and have since found the mistake in my original working (always the most simple error too, ha)
Easily done. The version i got before my first post was wrong too, but i spotted my error in exactly the same way (put a= 0 and didn't get what I was expecting!).
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