Is it true that any matrix can be decomposed into product of rotation,reflection,shear,scaling and projection matrices? It seems to me that any linear transformation in $R^{n\times m}$ 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 R^{n \times m}  $,$$A=\prod_{i=1}^{k}P_i$$
Where $P_i$ is rotation, reflection, shear, scaling, or projection matrix in $R^{n_i\times m_i}$. Also  $n,m,k\in N$,and $n_i,m_i\in N$ for all i.
And if it is true then how can we decompose it into that product.
 A: 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.
A: 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.
A: 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.
