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\section*{Chapter 4 Solutions}
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4.1 Consider the homogeneous recurrence

$$y(k)+a_{n-1}y(k-1)+...+a_0y(k-n)=0,$$

We know there are $n$ linerly independent solutions $\{z_i(k)\} i=1,..,n$
and any solution can be written as

$$y(k)=\sum_{i=0}^nc_iz_i(k)$$
where the constants $\{c_i\}$ are determined by the initial conditions.
A virtually identical result holds for linear differential equations.
If can regard the sequences $\{z_i(k)\} i=1,..,n$ as basis vectors
and thus the solution of the homogeneous equation is a point in the
vector space spanned by this basis.

For the inhomogeneous equation
$$y(k)+a_{n-1}y(k-1)+...+a_0y(k-n)=x(k),$$
where $\{x(k)\}$ is a known sequence, if $\{u(k)\}$ is {\em any} 
solution of this resurrence, any other solution can be written as 
$$y(k)=u(k)+\sum_{i=0}^nc_iz_i(k).$$
Thus $\{u(k)\}$ acts as a {\em point} and any solution is another
point that is obtained by adding a vector to it.

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4.5 There are 12 degrees of freedom in the three--dimensional affine
transformation. Suppose we consider a point ${\bf p}=[x,y,z,1]^T$ that
is transformed to ${\bf p}'=[x'y',z'1]^T$ by the matrix ${\bf M}$.
Hence we have the relationship ${\bf p}'={\bf Mp}$ where ${\bf M}$ has
12 unknown coefficients but ${\bf p}$ and ${\bf p}|'$ are known. Thus we
have 3 equations in 12 unknowns (the fourth equation is simply the
identity 1=1). If we have 4 such pairs of points we will have 12 equations
in 12 unknowns which could be solved for the elements of ${\bf M}$.
Thus if we know how a quadrilateral is transformed we can determine the
affine transformation.

In two dimensions, there are 6 degrees of freedom in ${\bf M}$ but 
${\bf p}$ and ${\bf p}'$ have only $x$ and $y$ components. Hence
if we know 3 points b oth before and after transformation, we will have
6 equations in 6 unknowns and thus in two dimensions if we know how
a triangle is tranformed we can determine the affine transformation.

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4.7 It is easy to show by simply multiplying the matrices that the 
concatenation of two rotations yields a rotation and that the concatenation
of two translations yields a translation. If we look at the product
of a rotation and a translation, we find left three columns of ${\bf RT}$ 
are the left three columns of ${\bf R}$ and the right column of ${\bf RT}$
is the right right column of the translation matrix. If we now consider
${\bf RTR}'$ where ${\bf R}'$ is a rotation matrix, the left three columns
are exactly the same as the left three columns of ${\bf RR}'$ and the 
and right column still has 1 as its bottom element. Thus, the form is
the same as ${\bf RT}$ with an altered rotation (which is the concatenation
of the two rotations)  and an altered translation. Inductively, we can
see that any further concatrenations with rotations and translations do no 
alter this form.

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4.9 If we do a translation by -h we convert the problem to reflection about
a line passing through the origin. From m we can find an angle by which
we can rotate so the line is aligned with either the x or y axis. Now
reflect about the x or y axis. Finally we undo the rotation and translation
so the sequence is of the form ${\bf T}^{-1}{\bf R}^{-1}{\bf SRT}$.

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4.11 The most sensible place to put the shear is second so that the
instance transformation becomes ${\bf I}={\bf TRHS}$. We can see that
this order makes sense if we consider cube centered at the origin
whose sides are aligned with the axes. The scale gives us the desired
size and proportions. The shear then converts the right parallelpiped
to a general parallepiped. Finally we can orient this parallelepiped with
a rotation and place it where desired with a translation. Note that the 
order ${\bf I}={\bf TRSH}$ will work too.


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4.13 
$${\bf R}={\bf R}_z(\theta_z){\bf R}_y(\theta_y){\bf R}_x(\theta_x)=
\left[\begin{array}{cccc}
\cos (\theta_y)\cos (\theta_z)&
   \cos \theta_z)\sin \theta_x \sin \theta_y - \cos \theta_x \sin \theta_z&
   \cos \theta_x \cos \theta_z \sin \theta_y + \sin \theta_x \sin \theta_z&0\\
  \cos \theta_y \sin \theta_z &\cos \theta_x) \cos \theta_z + 
    \sin \theta_x \sin \theta_y \sin  \theta_z &
   -\cos \theta_z) \sin \theta_x + \cos \theta_x) \sin \theta_y \sin \theta_z&0\\
  -\sin \theta_y &\cos \theta_y \sin \theta_x &\cos \theta_x \cos \theta_y&0 \\
0&0&0&1\end{array}\right]$$

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4.15 $${\bf R}={\bf R}_x(45){\bf R}_y(45)=\frac{1}{2}\left[\begin{array}{cccc}
\sqrt{2}&0&\sqrt{2}&0\\1&\sqrt{2}&-1&0\\-1&\sqrt{2}&1\\0&0&0&1
\end{array}\right].$$

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4.17 If they are collinear, one vertex is a linear combination of the
other two and the determinent of the matrix
$$\left[\begin{array}{ccc}\\x_1&x_2&x_3\\y_1&y_2&y_3\\
z_1&z_2&z_3\end{array}\right]$$
will be zero.

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