Why can any affine transformaton be constructed from a sequence of rotations, translations, and scalings? A book on CG says:

... we can construct any affine transformation from a sequence of rotations, translations, and scalings.

But I don't know how to prove it.
Even in a particular case, I found it still hard. For example, how to construct 
a shear transformation from a sequence of rotations, translations, and scalings?
Can you please help? Thank you.
EDIT:  
Axis scalings may use different scaling factors for the axes.
Is there a matrix representation or proof for this?
For example, to show that a two-dimensional rotation can be decomposed into three shear transformation, we can write
$$
\begin{pmatrix}
\cos\alpha & \sin\alpha\\
-\sin\alpha & \cos\alpha
\end{pmatrix}
=
\begin{pmatrix}
1 & \tan\frac{\alpha}{2}\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
-\sin\alpha & 1
\end{pmatrix}
\begin{pmatrix}
1 & \tan\frac{\alpha}{2}\\
0 & 1
\end{pmatrix}
$$
 A: You can write any affine transformation
$$
\vec{x}'=A\vec{x}+\vec{t}\;,
$$
where $A$ is any non-singular matrix, as follows:
$$
\left(
\begin{array}{c}
\vec{x}'\\
1
\end{array}
\right)
=
\left(
\begin{array}{cc}
A&\vec{t}\\
0&1
\end{array}
\right)
\left(
\begin{array}{c}
\vec{x}\\
1
\end{array}
\right)
\;.
$$
This allows you to compose affine transformations by composing the corresponding matrices. In this approach, rotations, translations and axis scalings can respectively be written like this:
$$
\left(
\begin{array}{cc}
\Omega&0\\
0&1
\end{array}
\right)
\;,
$$
$$
\left(
\begin{array}{cc}
I&\vec{t}\\
0&1
\end{array}
\right)
\;,
$$
$$
\left(
\begin{array}{cc}
S&0\\
0&1
\end{array}
\right)
\;,
$$
where $\Omega$ is a rotation matrix, $I$ is the identity matrix and $S$ is a diagonal matrix with the scaling factors on the diagonal.
Given any affine transformation specified by $A$ and $\vec{t}$, you can split it into a translation and a linear part:
$$
\left(
\begin{array}{cc}
A&\vec{t}\\
0&1
\end{array}
\right)
=
\left(
\begin{array}{cc}
I&\vec{t}\\
0&1
\end{array}
\right)
\left(
\begin{array}{cc}
A&0\\
0&1
\end{array}
\right)
\;.
$$
So now we just need to be able to write any non-singular matrix as a product of rotations and axis scalings. This is possible due to the singular value decomposition.
A: Perhaps using the singular value decomposition?
For the homogeneus case (linear transformation), we can always write
$y = A x = U D V^t x$ 
for any square matrix $A$ with positive determinant, were U and V are orthogonal and D is diagonal with positive real entries. U and V would the be the rotations and D the scaling.
Some (trivial?) details to polish: what if A has negative determinant, what is U and V are not pure rotations but also involve axis reflections.
It only remains add the indepent term to get the affine transformation ($y = Ax +b$) and that would be the translation.
