Differentials to derivatives involving trace of matrices

Suppose $$P$$ is a real-valued function of the $$p\times m$$ (real) matrix $$\mathbf{Q}$$. After taking its differential, one arrives with the following:

$$d(P(\mathbf{Q})) = \operatorname{trace}\left\{\mathbf{1}^\top_p\left[ d\mathbf{Q}\odot \mathbf{W} \right]\left(\mathbf{w}\odot \mathbf{G}\right)\right\}$$ where $$\mathbf{1}_p$$ is the $$p\times 1$$ vector of $$1$$'s with $$\mathbf{W}$$ is $$p\times m$$ while $$\mathbf{w}$$ and $$\mathbf{G}$$ are both $$m \times 1$$. $$\mathbf{W}$$, $$\mathbf{w}$$ and $$\mathbf{G}$$ are matrices involving $$\mathbf{Q}$$.

Question: What is $$\dfrac{dP}{d\mathbf{Q}}$$ ?

Attempt: $$\dfrac{dP}{d\mathbf{Q}} = \mathbf{W} \left(\mathbf{w}\odot \mathbf{G}\right) \mathbf{1}_p$$

But I think it's wrong. So my problem really is that Hadamard product of $$d\mathbf{Q}$$ and $$\mathbf{W}$$.

Some identities I have found online are these:

$$\dfrac{d(\mathbf{a}^\top\mathbf{X}\mathbf{b})} {d \mathbf{X}} = \mathbf{a} \mathbf{b}^\top$$

$$\operatorname{trace} (\mathbf{A}\odot \mathbf{B})\mathbf{C} = \operatorname{trace} \mathbf{A} (\mathbf{B}^\top \odot \mathbf{C})$$

UPDATE: To make it simpler, a general problem would be

$$\frac{\mathbf{a}^\top\left[d\mathbf{Q}\odot f(\mathbf{Q}) \right]g(\mathbf{Q})}{d\mathbf{Q}}$$ where $$\mathbf{a}\in \mathbb{R}^{p}$$, $$f:\mathbb{R}^{p\times m}\rightarrow \mathbb{R}^{p\times m}$$ and $$g:\mathbb{R}^{p\times m}\rightarrow \mathbb{R}^{m}$$.

The available identity I have encountered similar to this is

$$\frac{\operatorname{trace}(\mathbf{A}d\mathbf{X})}{d\mathbf{X}} = \mathbf{A}$$

from page 2 of this link.

• Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
– greg
Commented Mar 22, 2019 at 14:41
• Thank you @greg for point out. I correted it already together with an attempt. Commented Mar 22, 2019 at 21:28

For convenience, let \eqalign{ b &= w\odot G, \quad a &= 1_p \cr } Rearrange the given differential to isolate the gradient wrt $$Q$$. \eqalign{ dP &= {\rm Tr}\Big(a^T\,(dQ\odot W)\,b\Big) \cr &= a^T\,(dQ\odot W)\,b \quad {\rm \{trace\,does\,not\,affect\,scalar\,values\}} \cr &= ab^T:(dQ\odot W) \cr &= (ab^T\odot W):dQ \cr \frac{\partial P}{\partial Q} &= ab^T\odot W \cr &= \Big(1_p(w\odot G)^T\Big)\odot W \cr } where a colon is used to write the trace in product form, i.e. $$A:B = {\rm Tr}\big(A^TB\big)$$

UPDATE
The updated question uses $$(f,g)$$ in place of $$(W,b)\,$$ so the gradient becomes \eqalign{ \frac{\partial P}{\partial Q} &= ag^T\odot f \cr }

• I apologize. Prior to my edit, I have written that $\mathbf{W}$ also involves $\mathbf{Q}$, i.e., it is not constant with respect to $\mathbf{Q}$. I have edited it now. Commented Mar 23, 2019 at 1:10
• Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
– greg
Commented Mar 23, 2019 at 1:27
• But I can't seem to get pass with the third line of yours : $X = Q\odot W \implies dX = dQ\odot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$? Commented Mar 23, 2019 at 1:34
• I changed the name of my variable from $P$ to $\phi$. The important point is that their differentials are equal.
– greg
Commented Mar 23, 2019 at 1:37
• The $X$ variable was a distraction. It has been removed.
– greg
Commented Mar 23, 2019 at 18:58