Salam Mohammad,

Do you mean by A>B, the usual entry-wise partial ordering?

@ Mohammad Issa Sowaity,

Consider the following 2x2 Hermitiaon matrices:

A = [ 2 , 2

2 , 1]

Eigenvalues of A are (3 +sqrt(17) )/2 and (3 - sqrt(17))/2

E(A) =sqrt(17)

and B = [ -16, -1]

-1 , 0]

Eigenvalues of B are -8 +sqrt(65) and - 8 --sqrt(65)

E(B) = 2sqrt(65)

Now

A-- B = [ 18. 3

3, 1]

A-- B > 0 is positive definite ( check ) That means A > B.

But E(A)= sqrt(17) , E(B)= 2sqrt(65)

and then E(A) < E(B).

Consequently the answer is NO.

Best regards

Mohammad Adm,

Then A- B > 0 is positive definite, that means A > B.

Issam Kaddoura,

thanks a lot. But what about, if A>B>0?

Mohammad,

In the case A & B are pd we have all the evs of A & B are positive. For all i=1, ..., n, we have a_ii - b_ii>0 since A-B is pd. Hence

E(A) = sum_{i=1}^{n} lambda_i(A) = tr(A) > tr(B) = sum_{i=1}^{n} lambda_i(B) =E(B).

Therefore, E(A) > E(B).

@ Mohammad Issa Sowaity,

Under the new condition, A > B > 0, the question is simple,

the direct properties of traces and positive eigenvalues

show E(A-B) = E(A) - E(B) > 0

The answer is apparently yes.

Best regards

P.S

In this case E(A) = Tr(A), E(B)=Tr(B), E(A-B)=Tr(A-B)=Tr(A)-Tr(B)

Which agree with Mohammad Adm answer.