# Can all affine transformations be just expressed as a combination of the common transformations we are taught?

(At the time I was writing these questions, I forgot about Projection, and was focusing on isomorphic transformations, so I suspect I may have made some mistake with my presumption in 1. — please correct me if I have. And also, if non-isomorphic transformations wreck my questions, then I still want to know the answers to them for isomorphic transformations only.)

Here, I ask a set of questions as I have asked myself as I considered the question asked in the title. As I didn’t know the answer to them for sure, (although I suspect they are true), I hope someone could enlighten me… Also, throughout the questions, I imply that I consider Affine Transformations to be Linear Transformations + Translations. Is this a correct interpretation (as I haven’t really studied Affine Transformations)?

We learn about several types of Linear Transformations:

• Stretching
• Rotation
• Shearing
• (Reflection)
• (Projection - which reduces the dimension of the imageon



1. If we were to consider all of the possible linear transformations in $R^2$, would they all be some combination of the top 3 (since reflection is just negative stretching, and projection is some combination of stretching with a factor of 0 and some other transformations, if I’m not mistaken)?

2. Would the answer be the same in $R^n$? (I imagine that for $R^1$ there is only stretching, whilst for $R^0$ there is only the identity transformation)

And then when we learn about Affine Transformations, we add Translation to this list.

1. If we were to then consider all of the possible affine transformations in $R^2$, would they all be some combination of the top 3 and Translation?

2. Would the answer be the same in $R^n$? (And then for $R^1$ there would only be stretching and translating)

• Reflection is not just negative scaling. At least, not by the definition of scaling that I'm used to. – Thomas Andrews Oct 26 '15 at 18:01
• But surely a scale by scale factor -1 is the same as a reflection? – Shuri2060 Oct 26 '15 at 18:03
• Is the map $(x,y)\to(x,-y)$ negative scaling? Only if you define scaling as directional. – Thomas Andrews Oct 26 '15 at 18:03
• But surely you can scale in the $x$ direction, or the $y$ direction? EDIT: Oh - rereading my textbook, what I really meant by scaling is stretching. – Shuri2060 Oct 26 '15 at 18:04
• I recommend not asking 8 questions in a single post. There has to be a way to make it less monolithic. – rschwieb Oct 26 '15 at 18:07

Start with a toolbox that contains just the stretches parallel to the $x$-axis and the rotations.

You can stretch parallel to the $y$-axis by rotating one quarter turn, stretching, and then rotating back.

Moreover, as explained on another stackexchange site, you can perform any shear transformation by a sequence of rotations and stretches in the $x$- and $y$-directions.

Hence stretching along one axis, plus rotation, gives you all linear transformations.

The order in which you perform the primitive transformations in order to achieve any particular linear transformation will not be commutative in general, however, so this does not reduce linear transformations to two dimensions.

There are various other ways to express all linear transformations by sequences of simple transformations selected from a very short list, allowing each transformation in the list to take one parameter. For example, you could take reflections (using the direction as a parameter) and stretching in the $x$-direction, since the reflections will give you all rotations.

The stretches in the $x$-direction and the stretches in the $y$-direction are not sufficient to generate all other transformations, but if you allow two parameters for each stretch (a direction and a stretch factor) then you can generate all reflections across lines through the origin if negative stretch factors are allowed. With these reflections you can generate all rotations, with which (along with stretches) you can generate all shears. This implies that if you start with just stretches of arbitrary (possibly negative) stretch factors in arbitrary directions, you can generate all linear transformations by composition.

• I was told here in an answer that the shear could not be formed by a sequence of transformations...?? And can everything really be formed from stretches!? math.stackexchange.com/questions/1498899/… – Shuri2060 Oct 27 '15 at 20:30
• Nobody said you could not make a shear by composing other transformations. You cannot make it from transformations that preserve angles. Stretches do not preserve angles. Also, for "all transformations from stretching alone" you need to use all possible stretches, not just stretchs in the x and y directions. – David K Oct 27 '15 at 22:08
• Wow that's amazing - thank you! I really never expected that the shear could be decomposed to rotations and stretches. (Or stretches in any direction) – Shuri2060 Oct 28 '15 at 9:33