Why does a quaternion rotation matrix simplify to this? I'm reading Ken Shoemake's explanation of quaternions in David Eberly's book Game Physics.  In it, he defines the rotation matrix for a quaternion $q = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} + w\mathbf{1}$ to be the product of two matrices:
$\begin{align}
\mathbf{Q\overleftarrow{Q}}^T &=
\begin{bmatrix}
 w & -z &  y & x \\
 z &  w & -x & y \\
-y &  x &  w & z \\
-x & -y & -z & w
\end{bmatrix}
\begin{bmatrix}
 w & -z &  y & -x \\
 z &  w & -x & -y \\
-y &  x &  w & -z \\
 x &  y &  z &  w
\end{bmatrix} \\
&=
\begin{bmatrix}
\mathbf{R} & 0 \\
0 & xx + yy + zz + ww
\end{bmatrix}
\end{align}
$
where
$
\mathbf{R} = \begin{bmatrix}
ww - zz - yy + xx & -wz -zw + yx + xy & wy + zx + yw + xz \\
zw + wz + xy + yx & -zz + ww - xx + yy & zy - wx - xw + yz \\
-yw + xz - wy + zx & yz + xw + wx + zy & -yy - xx + ww + zz
\end{bmatrix}
$
I can follow this so far.  $\mathbf{R}$ is easy to see from the rules of matrix multiplication.  But then the book goes on to say that because the bottom right corner of the block matrix (ie, $xx + yy + zz + ww$) is $\textrm{N}(q)$, and because we're concerned with normalized rotations $\textrm{N}(q) = 1$, $\mathbf{R}$ can be simplified to
$
\begin{bmatrix}
1 - 2(z^2 + y^2) & 2(xy - wz) & 2(zx + wy) \\
2(xy + wz) & 1-2(x^2 + z^2) & 2(yz - wx) \\
2(zx - wy) & 2(yz + wx) & 1 - 2(x^2 + y^2)
\end{bmatrix}
$
I don't understand why the elements of the main diagonal can be simplified this way.  How does $x^2 + w^2 - y^2 - z^2 = 1 - 2(z^2 + y^2)$?  Does this only work because $x^2 + y^2 + z^2 + w^2 = 1$?

Exactly, you can easily see that
$\begin{align}x^2 + w^2 - y^2 - z^2 &= x^2+w^2+(y^2-2y^2)+(z^2-2z^2) \\
&= \underbrace{x^2+w^2+y^2+z^2}_{=1} - 2(z^2+y^2) \\
&= 1 - 2(z^2 + y^2)\end{align}$
It works exactly the same way for all the entries of the diagonal.
