Hmmm. Okay, now here's a small puzzle from the paper (http://www.stat.berkeley.edu/tech-reports/494.pdf) I'm currently stuck in. (It's equation 11).

If P(v) = I - vv^T / |v|²
can P(v) ever have any non-trivial value?
(v in this case is a vector in R³)

I'm sure I've missed something blindingly obvious, but ... well... I can't see it!

Originally posted by EI_Sparks
If P(v) = I - vv^T / |v|²


OK



can P(v) ever have any non-trivial value?


Sure, because vv^t is an outer product and |v|²=v^Tv is an inner product. They ain't the same thing. In fact, you should note carefully that the numerator is a 3x3 matrix, and the denominator is a 1x1 matrix (a scalar). That means that the "1" is to be understood as the identity 3x3 matrix.

how depressing, i totally don't understand any of that math. And what the heck is a "Geometrically Intrinsic Nonlinear Recursive Filter"? and what is it for. It looked quantum mechanicish, but I am not really sure.

Sure, because vvt is an outer product

If this in cartesian coordinates, then the transpose of a cvariant vector is a contravariant vector and vice versa. In cartesian coordinates, the metric is simply the Kronecker delta, thus its ommision here.

The outer product of a covariant and contravariant vector, is a second rank tensor, covariant of rank 1 and contravariant of rank one.

This equation is another way of stating the Gram-Schmidt orthogonalisation process from linear algebra, such that P(v) is orthogonal to v

Sure, because vvt is an outer product and |v|2=vTv is an inner product.
*blink*
*pause*
Oh *#@%$#@%!
*bangs head on desk*

Like I said, blindingly obvious :(
Thanks guys... back to the grindstone now...