Is it true that any matrix can be decomposed into product of rotation, reflection, shear, scaling and projection matrices?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.