Hi everyone,

This is a newbie question. Just going through the initial chapter on
vectors and I think I understand the orthogonalization process but I am
having trouble visualizing it geometrically...

Geometrically, this method proceeds as follows: to compute ui, it
projects vi orthogonally onto the subspace U generated by u1, …,
ui−1, which is the same as the subspace generated by v1, …, vi−1.
The vector ui is then defined to be the difference between vi and this
projection, guaranteed to be orthogonal to all of the vectors in the
subspace U.

Say we are in working in 3 dimensional space... So there exist 3
vectors in the orthogonal basis..

So the first vector does not change...

The second vector is calculated by projecting the second vector on the
first vector and subtracting it from the second vector so whatever s
left over must be orthogonal...

Anja ha scritto:

Anja,

the Gram-Schmidt orthogonalization can be written
somewhat simplified by using normalized base vectors.
A sketch here illustrates the creation of the third base
vector:
http://www.fho-emden.de/~hoffmann/gram-schmidt.pdf

The doc refers as well to Sigurd Falk, Matrizen, Springer.
Here we can find an elegant orthogonalization, merely
by matrix operations (without the somewhat tedious
Gram-Schmidt decomposition). The book is more than
20 years old, the method dates back to 1951 or so.

Best regards --Gernot Hoffmann

"Anja" googlemail.com> wrote in message
news:1164323109.592032.296380@l39g2000cwd.googlegroups.com...
Hi everyone,

This is a newbie question. Just going through the initial chapter on
vectors and I think I understand the orthogonalization process but I am
having trouble visualizing it geometrically...

Geometrically, this method proceeds as follows: to compute ui, it
projects vi orthogonally onto the subspace U generated by u1, .,
ui?1, which is the same as the subspace generated by v1, ., vi?1.
The vector ui is then defined to be the difference between vi and this
projection, guaranteed to be orthogonal to all of the vectors in the
subspace U.

Say we are in working in 3 dimensional space... So there exist 3
vectors in the orthogonal basis..

So the first vector does not change...

The second vector is calculated by projecting the second vector on the
first vector and subtracting it from the second vector so whatever s
left over must be orthogonal...

Jon Slaughter wrote: