Unitary and pseudounitary transformation

A so-called unitary transformation ${\bf U}$ conserves energy. In other words, if ${\bf v=Ux}$, then ${\bf x'x= v'v}$,which requires ${\bf U' U = I}$.Imagine an application where the transformation seems as if it should not destroy information. Can we arrange it to conserve energy? The conventional inversion

$$\begin{eqnarray} \bold y &=& {\bf B x} \ \bold x &=& {\bf (B'B)^{-1} B'y }\end{eqnarray}$$ (27) (28)

can be verified by direct substitution. Seeking a more symmetrical transformation between $\bold y$ and $\bold x$ than the one above, we define  

$$\bold U \eq \bold B (\bold B'\bold B)^{-1/2} \$$ (29)

and the transformation pair

$$\begin{eqnarray} \bold v &=& \bold U \bold x \ \bold x &=& \bold U'\bold v\end{eqnarray}$$ (30) (31)

where we can easily verify that $\bold x' \bold x = \bold v' \bold v$ by direct substitution. In practice, it would often be found that $\bold v$ is a satisfactory substitute for $\bold y$,and further that the unitary property is often a significant advantage.

Is the operator $\bold U$ unitary? It would not be unitary for NMO, because equation (19) is not invertible. Remember that we lost (x₁,x₂,x₃,x₄, and x₅) in (17). $\bold U$ is unitary, however, except for lost points, so we call it ``pseudounitary." A trip into and back from the space of a pseudounitary operator is like a pass through a bandpass filter. Something is lost the first time, but no more is lost if we do it again. Thus, $\bold x \neq \bold U'\bold U \bold x$,but $\bold U'\bold U \bold x = \bold U'\bold U (\bold U'\bold U \bold x)$for any $\bold x$.Furthermore, $(\bold U'\bold U)^2 = \bold U'\bold U$,but $\bold U'\bold U \neq \bold I$.In mathematics the operators $\bold U'\bold U$and $\bold U\bold U'$are called ``idempotent" operators. Another example of an idempotent operator is that of subroutine advance() 1