SciForums.com > Science > Physics & Math > Hmmm. Or, projections onto orthogonal complements. PDA View Full Version : Hmmm. Or, projections onto orthogonal complements. Post ReplyCreate New Thread EI_Sparks07-11-03, 09:45 PMHmmm. 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 - vvT / |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! Tom207-11-03, 09:54 PMOriginally posted by EI_Sparks If P(v) = I - vvT / |v|2 OK can P(v) ever have any non-trivial value? Sure, because vvt is an outer product and |v|2=vTv 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. cephas101207-12-03, 02:45 AMhow 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. ryans07-12-03, 02:57 AMSure, 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 EI_Sparks07-12-03, 12:04 PMSure, 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... Post ReplyCreate New Thread