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It seems to me that any linear transformation in ${\Bbb R}^{n \times m}$ is just a series of applications of rotation — actually i think any rotation can be achieved by applying two reflections, but not sure — reflection, shear, scaling and projection transformations. One or more of each kind in some order. This is how I have been imagining it to myself, but I was unable to find proof of this on the internet. Is this true? And if this is true, is there a way to find such a decomposition?

EDIT: to make it clear, I am asking whether it is true that $\forall A \in {\Bbb R}^{n \times m} $,

$$ A = \prod_{i=1}^{k} P_i $$

where $P_i$ is rotation, reflection, shear, scaling, or projection matrix in ${\Bbb R}^{n_i\times m_i}$. Also, $n, m, k \in {\Bbb N}$, and $n_i, m_i \in {\Bbb N}$ for all $I$. And, if it is true, then how can we decompose it into that product?

Sunny88
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  • The generality in which "linear transformation" makes sense is not the generality in which the other terms you're using make sense. – Qiaochu Yuan Feb 14 '12 at 01:41
  • @QiaochuYuan Can you elaborate? – Sunny88 Feb 14 '12 at 02:01
  • "Linear transformation" makes sense for vector spaces over any field, as does "shear" and "scaling." However, to define a notion of "rotation" requires an inner product (so we should ideally be working over $\mathbb{R}$ or $\mathbb{C}$), and depending on what you want out of "projection" or "reflection" those notions may also require an inner product. – Qiaochu Yuan Feb 14 '12 at 02:05
  • @QiaochuYuan Sorry, I had in mind matrices of real numbers when I asked this question. I modified my question to reflect that. – Sunny88 Feb 14 '12 at 02:20
  • http://math.stackexchange.com/questions/13150/extracting-rotation-scale-values-from-2d-transformation-matrix ? – Inquest Feb 14 '12 at 10:30
  • Similar decomposition applies to Mobius Transformations – Maesumi Jan 18 '13 at 01:39
  • @QiaochuYuan, what is shear and scaling in finite fields for example, or non-zero characteristic? My understanding is that most generally projections are idempotents, and reflections are involutions (but perhaps don't exist in certain classes of fields). – alancalvitti Jan 18 '13 at 22:29
  • @alan: a shear in two dimensions is a Jordan block with eigenvalue $1$ and scaling is just multiplication by a scalar (multiple of the identity matrix). – Qiaochu Yuan Jan 19 '13 at 02:24

4 Answers4

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The question is not posed completely clearly, but I think that something close to what the questioner wants should follow quickly from the singular value decomposition, which states that any real matrix $A$ can be written in the form $$ A=UDV, $$ where $U$ and $V$ are square real orthogonal matrices and $D$ is a (possibly rectangular) diagonal matrix with nonnegative entries on the diagonal. Since $U$ and $V$ are orthogonal they are products of rotations and reflections, while $D$ can be thought of as a product of projections and scalings.

For example, if $$ A=\left(\begin{array}{cc}1&2x\\0&1\end{array}\right), $$ then $$ A= \left(\begin{array}{cc}\cos \phi&-\sin\phi\\\sin\phi&\cos\phi\end{array}\right) \left(\begin{array}{cc}\sqrt{x^2+1}-x&0\\0&\sqrt{x^2+1}+x\end{array}\right) \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right), $$ where $$ \phi=-\frac{\pi}{4}-\frac{1}{2}\arctan x, \qquad \theta=\frac{\pi}{4}-\frac{1}{2}\arctan x. $$

In reply to the comments below: Interpreting a diagonal matrix with positive entries along the diagonal as a scaling relies on allowing the scaling to be non-uniform, i.e., allowing it to scale different axes by different amounts. If the scaling matrices are restricted to be uniform, then, by using examples like the one above, you can write a square diagonal matrix with positive entries as a product of orthogonal matrices, shears, and a uniform scaling.

David Moews
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    So it means that shear is also a product of reflections, projections and scaling? Can that be true? – Sunny88 Jan 18 '13 at 16:34
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    I added an example of the SVD of a shear above. – David Moews Jan 18 '13 at 22:22
  • How can a diagonal matrix be thought of as a product of projections and scalings? [Also, the middle matrix in your decomposition of the shear isn't diagonal.] – user108903 Jan 18 '13 at 22:27
  • I made a typographical error; thank you for spotting that. A diagonal matrix with nonnegative entries projects onto the set of axes given by its non-zero entries, and then scales these axes. The scaling may be non-uniform as the diagonal matrix may use a different scale factor for each axis. – David Moews Jan 18 '13 at 22:35
  • So are you saying that every positive semidefinite diagonalizable matrix is a scaling? I had assumed that "scaling" means a scalar multiple of the identity, and certainly no invertible diagonal matrix is a product of (orthogonal) projections and scalar multiples of $I$. – user108903 Jan 18 '13 at 22:39
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    In some lines of work, it's common to call a diagonal matrix with positive entries a "scaling matrix", although, as you point out, it's not a scalar multiple of the identity. – David Moews Jan 18 '13 at 22:53
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    This interpretation makes me wonder -- why do we need two rotations (since both $U$ and $V$ are orthogonal)? Why not just one? Isn't one enough to express any rotation in the final composite transformation? – Joseph Garvin Mar 21 '20 at 23:13
  • "square real orthogonal matrices": if it's an orthogonal matrix, it's a square and real matrix, so it's redundant to write "square real orthogonal matrices", isn't it? https://en.wikipedia.org/wiki/Orthogonal_matrix – ThePhi Jan 18 '22 at 09:35
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There are several things that are true and known for a long time, related to the question. First, we can consider non-singular matrices without much loss of generality, since any singular matrix can be written (in many ways) as non-singular composed with (orthogonal) projection.

As in another answer, the "singular value decomposition" (an example of a Cartan decomposition, valid in many other scenarios) expresses an arbitrary non-singular (real, for specificness) matrix as a product of rotation, diagonal matrix, and another rotation. Yes, this does express "shearing" as such a composition.

Another decomposition of non-singular real matrices is as a product of shear (upper triangular with 1's on the diagonal), diagonal, and then rotation. Among its other names, this is a special case of "Iwasawa decomposition".

Yet another is as product of upper triangular, permutation matrix (=exactly one non-zero entry, a "1", in each row and column), then another shear. This is a special case of "Bruhat decomposition".

Yes, each of these applies to the components of the others.

paul garrett
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The claim "any linear transformation is a series of applications projection transformations" is not correct. J. A. Erdos showed that every noninvertible $n\times n$ matrix is a finite product of projection matrices. An elmentary proof can be found here [An Elementary Proof That Every Singular Matrix Is a Product of Idempotent Matrices by J. Araújo and J. D. Mitchell, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), pp. 641-645] And it is not the case for invertible matrices generally.

Sunni
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  • If "any linear transformation is a series of applications projection transformations" is not true, I don't see how it says anything about "any linear transformation is a series of applications of projection, shear,reflection,rotation and scaling transformations". Now if it was true, then that is different matter :) – Sunny88 Feb 14 '12 at 01:27
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For every real $n\times n$ matrix $\mathbf{A}$, we have singular value decomposition $\mathbf{A}=\mathbf{U}\Sigma\mathbf{V}$, where both $\mathbf{U},\mathbf{V}$ are orthogonal matrix, and $\Sigma$ is diagonal matrix. Note that this decomposition is not unique, and we can choose $\mathbf{V}, \mathbf{U}$ with determinant $+1$.

Consider the following facts:

  • Every orthogonal matrix with determinant $+1$ could be decomposed to products of Givens rotation. related question
  • Givens rotation correponds to rotation in a subplane spanned by 2 orthogonal basis with one-hot coordinate.
  • Therefore, every orthogonal matrix corresponds to a series of rotation in 2-dimensional subspaces.
  • Real diagonal matrix corresponds to scaling at all directions.
  • Operations like reflection and shearing could be decomposed to rotation and scaling.

We can see that linear map represented real matrix could always be decomposed into a series of rotations, one scaling at each direction, and another series of rotations.