h(x)'*P*h(x)

Dear Debopam,

if I understand well, (eigenvalue-max(P)) is the greatest eigenvalue of P, (eigenvalue-min(P)) is the smallest one.

If yes, I agree with the upper bound you propose (I recommend), but have doubts about the lower bound.

thanks Viera,

Yes the lower bound i present is only a possible one, it can further be tightened.

Substitute y = h(x) and assume that $y \neq 0$. Since P is a Hermitian matrix, we can use the min-max theorem for its eigenvalues. Then:

(eigenvalue-min(P)) = min_{z \neq 0} (z' Pz)/(z'z) = (eigenvalue-min(P)) (y'y) (lower bound)

Similarly:

(eigenvalue-max(P)) = max_{z \neq 0} (z' Pz)/(z'z) >= (y'Py)/(y'y)

Hence:

y'Py

Right, Gabriel. I saw (wrongly!) in Debopam's lower bound

(eigenvalue-min(P))*X'*H'H*X

:)