ORTHOGONAL

Tell me things about O_n the group of Orthogonal Matrices.

HOT NAKED STRIPPING COEDS

Depends... Real or complex ?

Let's go with... real.

>>3
An orthogonal operator is real by definition.  A complex isometry is a unitary operator.

but they both take dicks up the ass. orthogonally. and with high dT/dt.

O(n) is not connex in the real case, but is compact. It has 2 connex parts, SO(n) and O-(n)= {M from O(n) / det M=-1}.

You have also the polar decomposition : Gl(n,R) homeomorph to S+ x O(n) where S+ are the symmetric matrix postively defined. Shit's not a trivial result.

Also  >>5 I think you're mixing U(n) and O(n). U(n) are the matrix M taht verifies (M*)M = Id and O(n) are those who verifies (tM)M=Id.

Orthogonal matrices can be defined over any ring as the invertible matrices that preserve some symmetric bilinear form on a free module M. The group is then called O(M), the orthogonal group of M. If the ring has an involution (like complex conjugation) one can also require that the matrices of O(M) should preserve sesquilinear forms. Then one can define U(M) as matrices satisfying (M*)M=I and if 2 is invertible in the ring U(M)=O(M). With these definitions O(C^n)=U(C^n)=U(n) for the standard hermitian inner product.

>>8
Looking at this definition, I agree 100%. You're totally gay :)



Ah, bad habit.

One of the problems with this board is that when you post something that actually contains math or science some high school kid doesn't understand what you're talking about and makes sure to inform everyone on the board of this fact.

By the way I'm not completely sure that 2 needs to be invertible in >>8 . I'll check that today.

Btw, if you define M*(M)=Id, doesn't it imply that M is invertible in U(n) ? M* = M^(-1) and since (M*)* = M...

>>11
If your module has finite rank, matrices satisfying M*(M)=Id are invertible. So in particular this holds for U(n). In the infinite dimensional case consider the matrix M for the unilateral shift operator. (note that the operator injective)
(a1,a2,a3,...) |-> (0,a1,a2,a3,...)
This matrix looks like the identity matrix with the diagonal shifted down one slot.
(a1,a2,a3, ...) |-> (a2,a3,...)
defines a left inverse to our first operator but is not injective. Its matrix N looks like the identity with the diagonal shifted up one slot.

The matrices for the two operators are conjugate transposes of each other, but only one of their compositions is the identity.
NM=Id but composing the other way MN does this to a vector:
(a1,a2,a3,...) |-> (0,a2,a3,...)
clearly not the identity or even invertible.

(this was an exercise a few weeks ago in a linear algebra class that I'm a TA in)

Fucking infinite dimension. I'm working so much in finite dimension that I forgot that there is often things that are true in finite case but not in infinite ones.

Do you have some sauce on this shift operator ? I was working on some problem recently where we used it to study hypercyclics vectors...

>>13
Yay for Hilbert spaces.  The quantum harmonic oscillator raising/lowering operators are quite similar to the shift operator.

>>13
http://en.wikipedia.org/wiki/Shift_operator