Consider the problem for some vectors v , m ∈ R n : f ( v ) = ( v T m ) 2 w.r.t ‖ v ‖ 2 = 1I want to maximize fIf 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 = λIf I put that in(*), I do cannot simplfy that.Is there a trick I can apply?

Question

Consider the problem for some vectors   v  ,  m  ∈            R        n    :  f  (  v  )  =  (      v    T    m      )    2  w.r.t   ‖  v      ‖    2    =  1I want to maximize fIf 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    =  λIf I put that in(*), I do cannot simplfy that.Is there a trick I can apply?

Freddy Doyle · Accepted Answer

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    mfor 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.

Consider the problem for some vectors v , m ∈ <mrow class="MJ

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