I am seeking to compute the following:

$\nabla ( \mathbf{1}^T f(X) )$

Where:

$\mathbf{1}^T = (1 , \ldots , 1)$ an $1 \times n$ vector.

$f(X)= (A*X) \circ (B*X)$

$A$ and $B$: $n \times n$ matrices

$X$: $n \times 1$ vector

$\circ$ the Hadamard or element wise product

$*$ matrix product.

I read the following rule $d(X \circ Y) = d(X) \circ Y + d(Y) \circ X$

But in the above I have a mixture of matrix and element wise product (dimensions are consistent at the end $\mathbf{1}^T f(X)$ is of size $1 \times 1$)

• Prove that your function equals $x^TA^TBx$ and take it from there. – Hans Engler Jun 2 '17 at 15:13
• thank you very much I worked it out! – Sha Jun 8 '17 at 14:49

Let's denote the elementwise/Hadamard and inner/Frobenius products respectively as \eqalign{ A &= B\circ C \cr \alpha &= B:C = {\rm tr}(B^TC) \cr } Recall that these products are commutative, and mutually commutative \eqalign{ A\circ B &= B\circ A \cr A:B &= B:A \cr A:B\circ C &= A\circ B:C \cr } and that the matrix of all ones is the identity element for the Hadamard product. Note that the matrices $(A,B,C)$ must have the same shape for these products to make sense.
Your scalar function can be written as \eqalign{ y &= 1:f \cr &= 1:(Ax)\circ(Bx) \cr &= Ax:Bx \cr } Whosee differential and gradient are \eqalign{ dy &= A\,dx:Bx + Ax:B\,dx \cr &= Bx:A\,dx + Ax:B\,dx \cr &= (A^TB + B^TA)\,x:dx \cr \cr \frac{\partial y}{\partial x} &= (A^TB + B^TA)\,x \cr\cr } (Note that I've used juxtaposition rather than $*$ for the ordinary matrix product.)