Dual of MVE

Hi,

Consider the problem (well known Minimum Volume Ellipsoid):

min -log det M
s.t. x_i^t M x_i <= 1
M is symmetric and positive semidefinite

Numerous papers on web suggest that the dual of this problem is

max \sum_{i} u_i x_i x_i^t

s.t. u >= 0
\sum_{i} u_i = 1

However, what is confusing is how the constraint \sum_{i} u_i = 1 is
obtained in the dual.

starting from the Lagrangean,
L = -log det M + \sum_i u_i ( x_i^t M x_i - 1 ), I am not able to see
how \sum_i u_i = 1 is obtained.
Any explanation would be greatly appreciated.

Thanks,
--HWG

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