2
$\begingroup$

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. ;-)

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – Reid Barton
    Nov 4, 2009 at 16:19
  • $\begingroup$ 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? $\endgroup$
    – M. E. Irizarry-Gelpí
    Nov 4, 2009 at 20:36
  • $\begingroup$ "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. $\endgroup$
    – Anton Geraschenko
    Nov 4, 2009 at 21:56
  • $\begingroup$ I edited the question. $\endgroup$
    – M. E. Irizarry-Gelpí
    Nov 4, 2009 at 22:54
  • 2
    $\begingroup$ 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. $\endgroup$
    – Anton Geraschenko
    Nov 5, 2009 at 2:22

2 Answers 2

Reset to default
1
$\begingroup$

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 $2 \times 2$ 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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @Reid: I hope you don't mind I took the liberty of prettying up your matrices using our new equation editor. $\endgroup$
    – Anton Geraschenko
    Nov 5, 2009 at 4:44
  • $\begingroup$ @Anton: The matrices are now invisible... $\endgroup$
    – Igor Rivin
    Oct 24, 2012 at 2:48
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.