I've come across the identity ∂ Tr { F [ M ( x ) ] } ∂ M ( x ) = F ′ [ M ( x ) ] T where F' is the scalar derivative of F but I've never seen the proof of it. Does somebody know how to do it ?

Question

I&#039;ve come across the identity            ∂      Tr      {      F      [      M      (      x      )      ]      }              ∂      M      (      x      )        =      F    ′    [  M  (  x  )      ]    T  where F&#039; is the scalar derivative of F but I&#039;ve never seen the proof of it. Does somebody know how to do it ?

Spielgutq1 · Accepted Answer

Write the Taylor series of a function of a scalar variable x and its derivative                    F        (        x        )                    =                  ∑                      k            =            0                    ∞                          α          k                          x          k                ,                          F          ′                (        x        )                    =                  ∑                      k            =            0                    ∞                (        k                  α          k                )                  x                      k            −            1                              Apply the function to a matrix argument X and take the trace                    ϕ                    =        Tr        ⁡                  (          F          (          X          )          )                        =                I        :                  (                      ∑                          k              =              0                        ∞                                α            k                                X            k                    )                    Then calculate its differential and gradient                    d        ϕ                    =        I        :                  (                      ∑                          k              =              0                        ∞                                α            k                                          d                          X                              k                                              )                                          =        I        :                  (                      ∑                          k              =              0                        ∞                                α            k                                                        ∑                              j                =                                  1                                            k                                      X                              j                −                                  1                                                      d            X                                      X                              k                −                j                                              )                                          =                  (                      ∑                          k              =              0                        ∞                                α            k                                          ∑                          j              =                              1                                      k                                              (                              X                                  j                  −                                      1                                                              )                        T                    I                                              (                              X                                  k                  −                  j                                            )                        T                    )                :        d        X                                  =                              (                          ∑                              k                =                0                            ∞                                      α              k                                                  (              k                              X                                  k                  −                                      1                                                              )                        )                    T                :        d        X                                  =                  F          ′                (        X                  )          T                :        d        X                                                  ∂            ϕ                                ∂            X                                      =                  F          ′                (        X                  )          T                                  In the preceding, a colon is used as a convenient product notation for the trace, e.g.                    A        :        B                    =                  ∑                      i            =            1                    m                          ∑                      j            =            1                    n                          A                      i            j                                    B                      i            j                                  =                Tr        ⁡                  (                      A            T                    B          )                                    A        :        A                    =                              ‖                          A                                            ‖                                F          2                                    I        :        B                    =        Tr        ⁡                  (                      I            T                    B          )                        =                Tr        ⁡                  (          B          )                    The properties of the underlying trace function allow the terms in such a product to be rearranged in many different but equivalent ways, e.g.                    A        :        B                    =        B        :        A                            A        :        B                    =                  A          T                :                  B          T                                    C        :        A        B                    =        C                  B          T                :        A        =                  A          T                C        :        B

I've come across the identity (del Tr{F[M(x)]})/(delM(x))=F′[M(x)]^T where F' is the scalar derivative of F. Does somebody know how to do it ?

Answered question

Answer & Explanation

New Questions in Precalculus