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
