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Where can I learn more about shear matrices?

The Wikipedia article is not enough, and sadly it does not have any references.

I understand they are linear transformations. Do they form a group? How do they look like for n-dimensional vectors? How many independent shears can I do in n-dimensions? Are they related to the orthogonal group SO(D), perhaps contained?

Google takes me to this other website. It is almost enough, but again no references.

I am trying to do some physics with this, so I do not think you will tolerate my explanation. ;-)

edited Nov 4 2009 at 22:47

asked Nov 4 2009 at 15:42

M. E. Irizarry-Gelpí
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What do you mean by SO(2,D)? I would usually interpret it as the group of linear transformations preserving a quadratic form of signature (2,D), but then the answer is clearly no, e.g., set D = 0 and notice that angles are not preserved in the picture on that Wikipedia page. – Reid Barton Nov 4 2009 at 16:19

Yes, I meant that group. Yes, angles are not preserved so no relation to conformal transformations. Thanks! Still, any idea where I can get a more formal treatment of these type of linear transformations? – M. E. Irizarry-Gelpí Nov 4 2009 at 20:36

"Shear transformations" isn't really a subject, so you won't find a book with that title. Could you please clarify what it is that you want to know about shears, or what you're trying to do with them? As stated, I don't think this question has an answer. – Anton Geraschenko♦ Nov 4 2009 at 21:56

I edited the question. – M. E. Irizarry-Gelpí Nov 4 2009 at 22:54

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No, they don't form a group. However you define a shear, any upper (or lower) triangular matrix with 1s on the diagonal is a composition of shears. But any determinant 1 matrix can be written as a product of such a lower and upper triangular matrix, and it's certainly not true that any determinant 1 matrix is a shear (for example, no non-identity element of SO(n) is a shear, since non-trivial shears always change some angles). You should really just say what you're trying to do; many math people do lots of physics, and this question is pretty impossible to answer without any motivation. – Anton Geraschenko♦ Nov 5 2009 at 2:22

Anton Geraschenko · Accepted Answer · 2009-11-05 04:43:32Z UTC

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It might be helpful to note that, in two dimensions, a shear transformation is exactly one whose Jordan canonical form is $\begin{pmatrix}1&1\0&1\end{pmatrix}$ or $\begin{pmatrix}1&0\0&1\end{pmatrix}$ if you include the identity as a shear transformation. These are exactly the transformations whose eigenvalues are all 1, so-called unipotent matrices. The unipotent 2x2 matrices do not form a group: $\begin{pmatrix}1&1\0&1\end{pmatrix}\times\begin{pmatrix}1&0\1&1\end{pmatrix}$ is not unipotent.

edited Nov 5 2009 at 4:43

Anton Geraschenko♦♦

answered Nov 5 2009 at 2:23

Reid Barton
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	@Reid: I hope you don't mind I took the liberty of prettying up your matrices using our new equation editor. – Anton Geraschenko♦ Nov 5 2009 at 4:44
	@Anton: The matrices are now invisible... – Igor Rivin Oct 24 at 2:48

Deane Yang · Answer 2 · 2009-11-05 03:03:11Z UTC

As Reid Barton's example shows, the set of all shear transformations is not a group.

However, the set of all, say, upper triangular matrices with all 1's along the diagonal, which I believe is a group comprised of only shear transformations, does.

Shear transformations

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