In eq:decomp, is uniquely determined. However, is not. Indeed, if satisfies eq:decomp and is any invertible matrix which commutes with , we have :

So, satisfies eq:decomp too. The converse is also true. That is, if and both satisfy eq:decomp, then is a matrix which commutes with . It can be easily shown that such a matrix can be written :

Let be an orthogonal matrix such that . eq:ginf gives a real Jordan decomposition for :

Furthermore, if has been calculated by a real Jordan decomposition of (we will see later how to obtain it), we have, with respect to what has been shown previously, the following relationship : eq:ambiguity a,b,c =

Then,

And so, by ,

Let be and . We have finally :

eq:kkt

Mon Dec 7 13:48:06 GMT 1998