X is an sdpvar(n,n); [X>=0]
A is a known (n,n) Hermitian matrix.
and I have [XT-A]>=0
To write this in Schur Complement form can I decompose 'A' into B*B' and write [XT-BB']>=0
:= [X,B';B',T]>=0
Is this a correct way ?
do I have to take anything else into consideration while decomposing ?

kind regards,
Peter

So what is T? TX is most likely not a symmetric matrix, so it is not clear what you mean by your constraint. And no, the Schur complement is not correct (clearly, assuming scalars, X*T-1>0 is not the same as [X 1;1 T]>0, since the latter constraint forces X and T to be positive, while the first allows both to be negative)

Thank you very much for the reply and apologizes for missing 'T' in the post.

the problem actual is

P1:= minimize trace(A*inv(T));
constraint:= trace(R)<=P

where
T=[H*R*H' + C ]
R is an sdpvar(m,m) and R>=0 and Hermitian
C and A are (nxn) known Hermitian matrices.
H is an (nxm) matrix
so T is also an (nxn) Hermitian matrix
P is a know scalar.

I reformulate it as
P2:= minimize trace(X) ;
A*inv(T)<=X;
X>=0;

so A*inv(T)<=X becomes [X*T-A]>=0
and I decompose the hermitian matrix A as B'*B
=> [X*T - B'*B]>=0;
=> [X,B ; B',T]>=0;

ofcourse this SDP formulation of P1 into P2 gives a lower bound.
I solved it for a special case A=eye(nxn), so want to find if I can do it for a general Hermitian matix A

kind reagrds,
Peter.

A*inv(T)<=X becomes [X*T-A]>=0... is not correct (i.e. multiplying on left and right of an inequality does not generalize from positive scalars to positive semidefinite matrices

and your use of the Schur complement is not correct. It is X-Binv(T)B which yields the matrix you have. You cannot simply multiply with T from right and magically pull out the inv(T) from inside Binv(T)B, i.e B(inv(T))BT is not equal to B

Nice post thanks you posting