Let w , a ∈ R n , and B ∈ R + + n × n (the set of n × n positive definite matrices).We know that the following function (which is a specific form of the Rayleigh quotient) has a unique maximum, and a closed-form solution in w: f ( w ) = w T a a T w w T B w It's maximum is achieved at w ∗ = B − 1 a and its value is f ( w ∗ ) = a T B − 1 a. (by using the generalized eigenvalue decomposition)Now here is my question:If I have w , a 1 , a 2 ∈ R n , and B 1 , B 2 ∈ R + + n × n , and the following function: g ( w ) = w T a 1 a 1 T w w T B 1 w + w T a 2 a 2 T w w T B 2 w Then, what can we say about the maximum of g ( w ), can we still solve for w in closed-form?

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Let   w  ,  a  ∈      R    n  , and   B  ∈      R          +      +              n      ×      n       (the set of   n  ×  n positive definite matrices).We know that the following function (which is a specific form of the Rayleigh quotient) has a unique maximum, and a closed-form solution in   w:  f  (  w  )  =                    w        T            a              a        T            w                      w        T            B      w      It&#039;s maximum is achieved at       w    ∗    =      B          −      1        a and its value is   f  (      w    ∗    )  =      a    T        B          −      1        a. (by using the generalized eigenvalue decomposition)Now here is my question:If I have   w  ,      a    1    ,      a    2    ∈      R    n  , and       B    1    ,      B    2    ∈      R          +      +              n      ×      n      , and the following function:  g  (  w  )  =                    w        T                    a        1                    a        1        T            w                      w        T                    B        1            w        +                    w        T                    a        2                    a        2        T            w                      w        T                    B        2            w      Then, what can we say about the maximum of   g  (  w  ), can we still solve for   w in closed-form?

Marco Meyer · Accepted Answer

The maximum is reached in points   w s.t.                    (                  a          1          T                w        )                              w          T                          B          1                w                  a    1    +                    (                  a          2          T                w        )                              w          T                          B          2                w                  a    2    =                    (                  a          1          T                w                  )          2                            (                  w          T                          B          1                w                  )          2                          B    1    w  +                    (                  a          2          T                w                  )          2                            (                  w          T                          B          2                w                  )          2                          B    2    wIn the unknown   w, it is an homogeneous equation of degree 7, and consequently, can only be numerically solved.EDIT 1. Multiplying on the left the previous equation by       w    T  , we obtain an identity.EDIT 2. You can use the Newton&#039;s method, adding the conditions       w    T    w  =  1  ,      w    1    &amp;gt;  0. Yet, you must give a good initial estimate, otherwise the sequence does not converge! This is difficult, especially when the system has many solutions. Unfortunately, this happens here. Indeed, generically, when   n=2, there are 2 real solutions and, when   n=3, 5 real solutions. If   n is great, then it may exist many candidates. I see only one case where the choice is easy; that is when       B    1          −      1            a    1   and       B    2          −      1            a    2   are close.

poklanima5lqp3 · Accepted Answer

some clarifications:we got                    (                  a          1          T                w        )                              w          T                          B          1                w                  a    1    +                    (                  a          2          T                w        )                              w          T                          B          2                w                  a    2    =                    (                  a          1          T                w                  )          2                            (                  w          T                          B          1                w                  )          2                          B    1    w  +                    (                  a          2          T                w                  )          2                            (                  w          T                          B          2                w                  )          2                          B    2    wbecause  ∇  f  (  w  )  =                    (                  a          T                w        )                              w          T                B        w              a  −                    (                  a          T                w                  )          2                            (                  w          T                B        w                  )          2                      B  w

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