Is this an if-and-only-if definition of affine? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:00:15Z http://mathoverflow.net/feeds/question/20207 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine Is this an if-and-only-if definition of affine? Learner 2010-04-03T03:22:44Z 2010-04-06T04:44:44Z <pre><code>x -&gt; A x+ b. </code></pre> <p>Quoted from <a href="http://en.wikipedia.org/wiki/Affine_transformation" rel="nofollow">Affine transformation</a>:</p> <blockquote> <p>In general, an affine transformation is composed of linear transformations (rotation, scaling or shear) and a translation (or "shift").</p> </blockquote> <p>Are these two concepts the same thing:</p> <ol> <li>affine transformation</li> <li>rotation, scaling , shear, translation </li> </ol> <p>If not,is there a complete expansion of affine transformation?</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20213#20213 Answer by S. Carnahan for Is this an if-and-only-if definition of affine? S. Carnahan 2010-04-03T04:28:50Z 2010-04-03T04:28:50Z <p>I think you need to make the second concept in your list more precise. For example, "A transformation given by a sequence of shifts, rotations, dilations, and shears."</p> <p>Assuming you are considering the set of affine transformations generated by the special cases listed, the answer is no, for the reason maxmoo gave in the comments. If you look in the "Representation" section of the Wikipedia article you linked, you find a description of these transformations using matrices, and the transformations listed only yield matrices of rank 1 or $n+1$. You need to include projections, or something equivalent, to get matrices of intermediate rank.</p> <p>Instead of using the list above, you can use <a href="http://en.wikipedia.org/wiki/Singular_value_decomposition" rel="nofollow">singular value decomposition</a> to find that any affine transformation is a composition of translations, stretches (by possibly singular diagonal matrices), and rotations.</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20249#20249 Answer by Leah Wrenn Berman for Is this an if-and-only-if definition of affine? Leah Wrenn Berman 2010-04-03T17:01:24Z 2010-04-03T17:01:24Z <p>There seems to be some disagreement about what precisely the definition of an affine transformation is. For example, <a href="http://books.google.com/books?id=KW4EwONsQJgC&amp;pg=PA169&amp;dq=affine+transformation&amp;lr=&amp;ei=O3K3S7qnIZWKkATS1dygDg&amp;cd=17#v=onepage&amp;q=affine%2520transformation&amp;f=false" rel="nofollow">Martin,</a> in his transformation geometry book, proves that an affine transformation is a map which takes any three noncollinear points to any other three noncollinear points (this is sometimes called <a href="http://books.google.com/books?id=q49lhAzXTFEC&amp;pg=PA73&amp;dq=brannan+affine+theorem&amp;lr=&amp;ei=RnS3S_SjFI7ikwSk_L3wBA&amp;cd=1#v=onepage&amp;q=brannan%2520affine%2520theorem&amp;f=false" rel="nofollow">the fundamental theorem of affine geometry</a>). In particular, this means that every affine transformation may be represented as the composition of an <em>invertible</em> linear transformation and a translation. </p> <p>On the other hand, <a href="http://books.google.com/books?id=wCfWkc_E3GkC&amp;pg=PA122&amp;dq=affine+transformation&amp;ei=G3K3S-iLJIXylQTk0JiTBA&amp;cd=2#v=onepage&amp;q=affine%2520transformation&amp;f=false" rel="nofollow">some references</a> simply define an affine transformation as the composition of a linear transformation and a translation, in which case the transformation need not be invertible.</p> <p>(I'm not familiar with the second reference; it was just handy on a Google Books search on Affine transformations.)</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20264#20264 Answer by Sergei Ivanov for Is this an if-and-only-if definition of affine? Sergei Ivanov 2010-04-03T21:33:00Z 2010-04-03T21:33:00Z <p>If you are looking for a geometric classification of all affine transformations of the Euclidean plane (not to mention higher dimensions and other fields), there is probably no such thing. It is possible to classify them in some sense, but some transformations are too far away from things that have names in elementary geometry. And if you move out of elementary geometry, it is so much easier to just use linear algebra.</p> <p>The worst example is the following type of a linear map: $A=S^{-1}RS$ where $R$ is a rotation (take some nontrivial angle, not $\pi$ or $\pi/2$) and $S$ is a scaling in one direction: $S=diag(k,1)$. The resulting map resembles rotation, but it rotates along ellipses rather than circles.</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20273#20273 Answer by Deane Yang for Is this an if-and-only-if definition of affine? Deane Yang 2010-04-04T02:35:43Z 2010-04-04T02:35:43Z <p>Usually, when we say "affine transformation", we mean an invertible one. Either way, any affine transformation is indeed of the form $x \mapsto Ax + b$, where $A$ is a (invertible) linear transformation and $b$ is a fixed vector.</p> <p>If by "scaling", you mean a scalar multiple of the identity matrix and by a shear, you mean an upper triangular matrix with 1's along the diagonal, then rotations, scalings, shears, and translations do not generate all possible affine transformations, because rotations, scalings, and shears do not generate all possible linear transformations.</p> <p>On the other hand, any linear transformation can always be written as $A = RDS$, where $R$ and $S$ are orthogonal transformations (i.e., rotations) and $D$ is diagonal. The action of a diagonal matrix can be viewed as rescaling by different factors in different amounts in each co-ordinate direction. So compositions of translations, rotations, and co-ordinate scalings generate all affine transformations.</p> <p>Any linear transformation can also be written as $A = RDU$, where $R$ is orthogonal, $D$ is diagonal, and $U$ is upper triangular with all $1$'s along the diagonal. In this sense any invertible affine transformation can be written as a composition of a translation, a rotation, co-ordinate scalings, and a composition of shears.</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20274#20274 Answer by Igor Balla for Is this an if-and-only-if definition of affine? Igor Balla 2010-04-04T02:46:11Z 2010-04-04T02:46:11Z <p>I think the most appropriate definition of an affine transformation is a mapping that preserves the affine structure of a space (see definition of affine space). In that sense, it is the composition of a linear transformation and a translation (so characterizing affine transformations equates to characterizing linear transformations). It would not be right to restrict the linear transformation to be invertible, or else you could not have affine transformations between spaces of different dimensions.</p> http://mathoverflow.net/questions/20207/is-this-an-if-and-only-if-definition-of-affine/20304#20304 Answer by Chris Godsil for Is this an if-and-only-if definition of affine? Chris Godsil 2010-04-04T13:46:32Z 2010-04-04T13:46:32Z <p>An affine space is a projective space with a distinguished hyperplane "at infinity". An affine transformation of the space is a projective transformation that fixes the distinguished hyperplane as a set. If the space is desarguesian (for example, if its dimension is at least three) then our affine space is a vector space over a skew field and an affine transformation is the composition of a linear map and a translation. We can only speak about rotations if we have an inner product space. In this case, using the QR-decomposition, we can write any invertible matrix $A$ as $A=QDS$ where $Q$ is orthogonal, $D$ is diagonal and $S$ is upper triangular with diagonal entries equal to 1. Here $Q$ is a rotation, $D$ a scaling and $S$ is the shear. (A shear is an invertible collineation that fixes each point in the space at infinity, and all lines on some point at infinity---it fixes each parallel class of lines as a set, and fixes each line in some parallel class.)</p>