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sweetymoeyz 2022-07-08 Answered
Consider the problem for some vectors v , m R n :

f ( v ) = ( v T m ) 2

w.r.t v 2 = 1

I want to maximize f

If I consider the lagrangian, I get:

L ( v ) = ( v T m ) 2 + λ ( 1 v 2 )

Taking derivative, I get: 2 m m T v λ 2 v = 0 Therefore m m T v = λ v ( )

Multiplying by v T from left, I end up with

( v T m ) 2 = λ

If I put that in(*), I do cannot simplfy that.

Is there a trick I can apply?
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Answers (1)

Freddy Doyle
Answered 2022-07-09 Author has 20 answers
You do not need much optimization here. Just note that ( v T m ) 2 = v T m m T v and that m m T is symmetric positive semidefinite with rank 1. The only nonnezero eigenvalue is given by m T m. Therefore, we have that
0 v T m m T v m T m
for all v R n . The maximum is attained for v = m / | | m | | which is normalized eigenvector associated with the unique positive eigenvalue. Note that it is also attained for v = m / | | m | | .

The difficulty here with using Lagrange multipliers is that m m T is not invertible and one has to restrict v to the subspace where m m T v 0. That is you can solve for the problem where v = α m and pick α such that the norm of v is one.
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