Critical Points of Quadratic FormsJust a question about finding critical points (points where the differential is not surjective). I have the equation f ( x ) = x t A xwhere A is a symmetric n by n matrix and x is an element of R n . Now, I've found the derivative of the function to be d f ( x ) = 2 x t A = 2 A xby symmetry of A. I'm having trouble determining values in R n where the derivative is not surjective. It's clear that for x = ( 0 , . . . , 0 ), the derivative is not surjective, but my intuition is that for any other value of x, and for any value in R n , we can choose a corresponding A. Am I thinking about this correctly? Is my derivative correct for the quadratic form f ( x ) = x t A x?

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Critical Points of Quadratic FormsJust a question about finding critical points (points where the differential is not surjective). I have the equation  f  (  x  )  =      x    t    A  xwhere A is a symmetric n by n matrix and x is an element of             R        n  . Now, I&#039;ve found the derivative of the function to be  d  f  (  x  )  =  2      x    t    A  =  2  A  xby symmetry of A. I&#039;m having trouble determining values in             R        n   where the derivative is not surjective. It&#039;s clear that for   x  =  (  0  ,  .  .  .  ,  0  ), the derivative is not surjective, but my intuition is that for any other value of x, and for any value in             R        n  , we can choose a corresponding A. Am I thinking about this correctly? Is my derivative correct for the quadratic form   f  (  x  )  =      x    t    A  x?

Envetenib8ne · Accepted Answer

Step 1As I said in my comment, because f is a map from             R        n   to       R  , its derivative at a point   x  ∈            R        n   will be a linear map   d      f    x    :            R        n    →      R  . You can compute the value of this linear map applied to a vector   y  ∈            R        n   by taking the ordinary derivative (with respect to s) of the smooth function   f  (  x  +  s  y  ), and setting   s  =  0. Thus                    d                  f          x                (        y        )                            =                                                        d                              d                s                                      |                                s            =            0                                                (                          (        x        +        s        y                  )          t                A        (        x        +        s        y        )                              )                                                            =                                                        d                              d                s                                      |                                s            =            0                                                (                                    x          t                A        x        +        s                  y          t                A        x        +        s                  x          t                A        y        +                  s          2                          y          t                A        y                              )                                                            =                  y          t                A        x        +                  x          t                A        y        .            Step 2Because each term above is a scalar quantity (i.e., a   1  ×  1 matrix, which is equal to its own transpose) and A is symmetric, this can be written as                    d                  f          x                (        y        )                            =        (                  y          t                A        x                  )          t                +                  x          t                A        y        =                  x          t                          A          t                y        +                  x          t                A        y        =        2                  x          t                A        y        .            Therefore,   d      f    x   is the linear map represented by the row matrix   2      x    t    A. (But this is not equal to 2Ax, because the latter is a column matrix, and therefore doesn&#039;t represent a linear map from             R        n   to       R  .)

Critical Points of Quadratic Forms. I have the equation f(x)=x^t Ax where A is a symmetric n by n matrix and x is an element of R^n.

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