Dear all,
suppose that a nonlinear vector function h(x) is bounded as below:
h(x)'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
:)
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