The reason I asked these 4 questions regarding if all linear/affine transformations could be formed by some combination of:

  • Rotation
  • Stretching
  • Projection
  • Translation

Was that I was speculating if they could hence form some sort of basis for a space. However, in this speculation, I came across some problems — for example the non-commutativity of combinations of linear transformations, amongst other things.

  1. If the answer to 1. is yes, then could the linear transformations in $\Bbb R^2$ be considered to be some sort of non-commutative vector space where the basis is:

    $S_x$ - Stretching in the $x$ direction by a factor of 1

    $S_y$ - Stretching in the $y$ direction by a factor of 1

    $\Bbb R$ - Rotation by 360º

    $T_x$ - Shearing in the $x$ direction by a factor of 1

    $T_y$ - Shearing in the $y$ direction by a factor of 1

    The non-commutativity of this space makes it seem weird… and I haven’t studied any spaces other than $\Bbb R^n$ so I don’t even know if non-commutative spaces can even exist since $a+b ≠ b+a$ generally…

    And then if this was true, then it would imply linear transformations for $\Bbb R^2$ exist in 5 dimensions!? And for affine transformations in $\Bbb R^2$, 7 dimensions (since the $x$ and $y$ Translations would be added)?

  2. Forgetting about the speculation, when we learn about the above types of Linear Transformations, could our teachers (or rather, mathematicians in general), have chosen a different way to categorise the linear transformations so that all linear transformations can be expressed as some combination of this new set of categories of linear transformations?

    Is the set that mathematicians have chosen the best way to go about categorising linear/affine transformations, or is it arbitrary and there is no “best way”, or on the other hand, could there be improvements made in the way they are categorised?

  3. Applying what I suggested in 5. to 6., since many different bases can be selected for spaces of the form $\Bbb R^n$ (and I would imagine this to be true for most, if not all vector spaces), perhaps the same can apply for the space formed by linear transformations (if they do form a space).

  4. And lastly, could there be some way to select ‘better’ basis vectors for this space so that the space is commutative (and hence more ‘normal’)?

  • 2
    $\begingroup$ A pretty natural structure with non-commutativity is the algebra of square matrices, it is a vector space with a product $\endgroup$ – janmarqz Oct 26 '15 at 18:38
  • $\begingroup$ What do you mean by a "commutative space"? What do you mean "non-commutative vector space"? Without mentioning the binary operation, this doesn't have any meaning. $\endgroup$ – rschwieb Oct 26 '15 at 18:52
  • $\begingroup$ however the en.wikipedia.org/wiki/Affine_group already is at ours disposal $\endgroup$ – janmarqz Oct 26 '15 at 19:10
  • $\begingroup$ @rschwieb By "non-commutative vector space", I meant that the addition of vectors within this space wouldn't be commutative. (Sorry for bad use of terminology) $\endgroup$ – Shuri2060 Oct 27 '15 at 20:26

The question is better for invertible affine maps. Then you have the Lie group. But then the translations form a subgroup than you can handle later. In which case you work in the Lie group $GL(n,\Bbb R)$. This is not a vector space so it does not make sense to ask for a basis, but if you work infinitesimally you get the Lie algebra which has lots of convenient basis. These come exactly as stated from breaking up into rotations and the rest. The rotations are what is called the maximal compact. Then you put all these ingredients back together.


There are lots of ways to characterize linear transformations. You might look into Jordan canonical form, rational canonical form, polar decomposition, singular value decomposition, QR decomposition, LU decomposition, ...

If none of those fit your needs, you're welcome to define another one.


Choose a basis of $\mathbb R^2$. The basis $\{v_1 = (1,0), v_2 = (0,1)\}$ is good, but technically any basis will suffice as long as you specify what basis you have chosen.

Any linear transformation of $\mathbb R^2$ then can be uniquely specified by the pair of vectors $v_1', v_2'$ to which the transformation maps the pair of vectors $v_1, v_2$. Conversely, any pair of vectors $v_1', v_2'$ corresponds to a unique linear transformation that maps $v_1, v_2$ to $v_1', v_2'$.

As you can see from this description, two two-dimensional vectors are sufficient to identify any linear transformation of $\mathbb R^2$, so the linear transformations have four dimensions, not five.

Basically any approach to linear algebra is going to have to characterize all linear transformations somehow, and I do not think that beyond a very elementary level the primitives from which we choose to build all transformations will be any subset of the transformations you listed. The last time in my mathematical education that I recall those transformations being used as the "basis" for all linear transformations was in a "pre-calculus" type of curriculum during high school. (Of course in more advanced math it is still the case that we might remark that certain transformations fall into one of those categories, and sometimes we work with an interesting subset of transformations that can be characterized in such ways, for example, all the transformations are rotations of some kind.)

As for commutativity, you can get that only if you are willing to discard many possible transformations from the "space" that you are willing to consider. The reason for this is simply that some pairs of transformations do not commute. The only way to get a commutative algebra is to eliminate from your "space" at least one of the transformations in any such non-commuting pair. That's a lot of transformations you have to get rid of.

For example, the transformations consisting only of rotations and dilations (scaling uniformly in every direction) commute. But now you have lost stretching and shearing.


Rotation, stretching dilation, and translation are conformal, i.e. they do not change angles. Projection is singular, i.e. it maps regions of positive volume to regions of zero volume.

The linear transformation $(x,y) \mapsto (x, x+y)$, is a shear. It is non-singular, so it cannot have any projections among its factors. But it changes some angles (e.g. the angle between $(1,0)$ and $(0,1)$ becomes the angle between $(1,1)$ and $(0,1)$, which is different), so it cannot be made up only of transformations that never change angles.

Shearing happens whenever you do the elementary row operation of adding a multiple of one row of a matrix to another row.

POSTSCRIPT: It seems something other than multiplying every vector by the same scalar was meant by "stretching", so the question of the closure under composition, of the proposed set of transformations, is more than what the paragraphs above address.

  • $\begingroup$ I agree that rotation/translation do not change angles... but surely stretching in a particular direction does? Also, an answer here disagrees with that.... who is correct? math.stackexchange.com/questions/1498833/… $\endgroup$ – Shuri2060 Oct 27 '15 at 20:29
  • $\begingroup$ @MasterShuriken : It seems I misunderstood what was meant by "stretching". I took it to mean multiplication of every vector by the same scalar. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 27 '15 at 20:33
  • $\begingroup$ From my textbook, stretching occurs in the $x$ or $y$ directions by a certain factor (and hence my particular take on the meaning). $\endgroup$ – Shuri2060 Oct 27 '15 at 20:36

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