# generic rule matrix differentiation (Hadamard Product, element-wise)

I struggle with taking the derivative of the Hadamard-Product?

Let us consider $f(x)=x^TAx=x^T(Ax)$. We know

$$\frac{\partial}{\partial x} x^TAx = (A+A^T)x.$$

The Matrix-Cookbook claimed $d(XY)=d(X)Y+Xd(Y)$ and $$\frac{\partial}{\partial x} x^Ta = \frac{\partial}{\partial x}a^Tx = a.$$

Setting $X:=x^T$ and $Y:=Ax$ we have \begin{align*} X &= x^TE & d_x(X) = E\\ Y &= Ax & d_x(Y) = A\\ \end{align*} This gives $$d(XY)=d(X)Y+Xd(Y)= 1^TAx + x^TA= Ax + x^TA$$ This format (dimension) is incorrect. A generic rule seems to be $d(XY)=d(X)Y+(Xd(Y))^T$ instead? What does a generic rule look like for the Hadamard product $[a\odot b]_i=[a]_i\cdot [b]_i$? The Matrix-Cookbook states: $$d(X\odot Y)=d(X)\odot Y+X\odot d(Y).$$

For example, the derivative of $$(x\odot y)^TA(x\odot y)$$ we have \begin{align*} X &= (x\odot y)^TE & d_x(X) &= \left(d(x\odot y)\right)^TE\\ & & &= \left(1\odot y + x\odot 0\right)^TE\\ & & &= y^TE\\ Y &= A(x\odot y) & d_x(Y) &= A\left(d(x\odot y)\right)\\ & & &=A\left(1\odot y + x\odot 0\right)\\ & & &=Ay\\ \end{align*}

Which implies \begin{align*} d_x((x\odot y)^TA(x\odot y))&=y^T\odot A(x\odot y) + \left( (x\odot y)^T Ay \right)^T\\ &=y^T\odot A(x\odot y) + y^TA^T(x\odot y) \end{align*} as a derivative. But the correct derivative should be

$$2y\odot A(x\odot y)$$

What is missing?

• The cookbook is in principle correct: this is how we deal with differentiation of bilinear maps. The problem lies in the way you interpret it. The differential $df$ of a multivariable function $f$ is usually represented by a row vector instead of a column vector (in accordance with the convention of the Jacobian matrix of a multi-valued map). – Troy Woo Feb 2 '15 at 13:28
• But this does not solve the format-issue with $d(XY)=d(X)Y+Xd(Y)= 1^TAx + x^TA= Ax + x^TA$ – John Doe Feb 2 '15 at 14:09
• The cookbook rule is for differentials, for which it works perfectly. $d(XY) = (dX)Y+X(dY) = (dx^T)(Ax)+(x^T)(Adx) = (x^TA^T)(dx)+(x^T)(Adx) = x^T(A^T+A)dx$ – lynn Feb 3 '15 at 1:43
• @lynn What am I missing since i don't see $x^{\mathsf{T}}\right)\text{Ax} \left(d =\text{dx} \left(A^{\mathsf{T}} x^{\mathsf{T}}\right)$ – ALEXANDER Feb 8 '18 at 12:08

New episode of the misdeeds of the Matrix-Cookbook.

If $f(x)=x^\top Ax$, the derivative is the linear application \begin{align*} Df_x: h\in\mathbb{R}^n\rightarrow h^\top Ax+x^\top Ah=(x^\top A^\top+x^\top A)h \end{align*} and the gradient $\nabla f(x)$ is the vector defined, for every $h$, by the relation \begin{align*}Df_x(h)=\langle \nabla f(x),h\rangle={\nabla f(x)}^\top h. \end{align*} Thus $\nabla f(x)=(A+A^\top)x$.

The Hadamard product $a\odot b$ is bilinear and the derivative satisfies \begin{align*}\mathrm{d}(a\odot b)=\mathrm{d}a\odot b+a \odot \mathrm{d}b \end{align*} like a standard product of matrices. For instance if $f(x)=(x\odot y)^\top A(x\odot y)$, then \begin{align*} Df_x:h\rightarrow (h\odot y)^\top & A(x\odot y)+(x\odot y)^\top A(h\odot y)\\&=[(x\odot y)^\top A+(x\odot y)^\top A^\top](y\odot h) \\&=([(A+A^\top)(x\odot y)]\odot y)^\top h \\&=\langle [(A+A^\top)(x\odot y)]\odot y,h \rangle \end{align*} because $\langle u, v\odot w \rangle=\langle u\odot v, w \rangle$ and $\nabla f(x)=[(A+A^\top)(x\odot y)]\odot y$.

Concerning the derivative or the gradient, the standard notation is as follows. Let \begin{align*}f: X=(x,y)\in \Omega\subset \mathbb{R}^p\times\mathbb{R}^q\rightarrow f(X)\in \mathbb{R}^n. \end{align*} Note that $\mathrm{d}f_X, DF_X, {\mathrm{d}f(X)}/{\mathrm{d}X}$ refer to the same concept: the total differential or the total derivative in $X$; it is a linear application $(h,k)\in\mathbb{R}^p\times\mathbb{R}^q\rightarrow \mathbb{R}^n$. In particular, in the formula \begin{align*}\frac{\mathrm{d}f}{\mathrm{d}X}=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y, \end{align*} the linear applications $\mathrm{d}x, \mathrm{d}y$, are defined as $\mathrm{d}x:(h,k)\rightarrow h$ and $\mathrm{d}y:(h,k)\rightarrow k$. The "partial derivative" $\partial f(X)/{\partial x}:\mathbb{R}^P\rightarrow\mathbb{R}^n$ is also a linear application.

For the case $n=1$, we can define the gradient of $f$ by duality, using the scalar product $\langle H, K \rangle=H^\top K$ for vectors or $\langle H, K \rangle=\mathrm{trace}(H^\top K)$ for square matrices (cf. the beginning of the post).

With our notation, \begin{align*}Df_X(h,k)=\dfrac{\partial f(X)}{\partial x}h+\dfrac{\partial f(X)}{\partial y}k=\left[\dfrac{\partial f(X)}{\partial x},\dfrac{\partial f(X)}{\partial y}\right][h,k]^\top. \end{align*} Thus, \begin{align*}\nabla(f)(X)=\left[\frac{\partial f(X)}{\partial x}, \frac{\partial f(X)}{\partial y}\right],\end{align*} that is, the transpose of the Jacobian matrix of $f$.

Of course, the calculation in your post is correct but you calculate the gradient, not the differential nor the derivative.

Your "known" result in the second example (with the Hadamard products) is wrong.

Define the diagonal matrix \eqalign{ M &= {\rm Diag}(y) = M^T \cr Mx &= (x\circ y) \cr x^TM &= (x\circ y)^T \cr }

Let's assume $y$ is constant and find the differential of your proposed function \eqalign{ f &= (x\circ y)^TA(x\circ y) \cr &= x^TMAMx \cr\ &= x^TBx \cr } As you state in your first example, the derivative of this function is well-known \eqalign{ \frac{\partial f}{\partial x} &= (B^T+B)\,x \cr &= M(A^T+A)\,Mx \cr &= y\circ(A^T+A)\,(x\circ y) \cr } Also, you seem to be confused about the difference between a differential and a derivative. If $y$ is a vector, then $dy$ is also a vector, but $(\frac{dy}{dx})$ is a matrix. It is incorrect to replace differentials with derivatives in an equation; you must properly account for the tensorial character of each term in the equation.

• I just read your post. I think that your first line refers to my post. If you want that I am aware of your remarks, then write "arobase my pseudo" at the beginning of your post. I have no time to discuss about the notations; yet I'll write an edit for you in my post above. – user91684 Jan 6 '16 at 10:48