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.