I'm reading a book about neural network. In the section about back-propagation of an Affine-layer of the network, the author provides a formula and omits the details.

Say $$\mathbf{X}\cdot\mathbf{W}=\mathbf{Y},$$

which $\mathbf{X}$ is of dimension $(N, 2)$, $\mathbf{W}$ of $(2, 3)$ respectively. The loss function, say $L$, will do some modification to the $\mathbf{Y}$. Then the provided formula is:

$$\begin{align}\\ \frac{\partial L}{\partial\mathbf{X}} &= \frac{\partial L}{\partial\mathbf{Y}}\cdot\mathbf{W}^\mathrm{T},\\ \frac{\partial L}{\partial\mathbf{W}} &= \mathbf{X}^\mathrm{T}\cdot\frac{\partial L}{\partial\mathbf{X}}.\\ \end{align}$$

What I really want to know first is that what's the dimension of $\Large\frac{\partial\mathbf{Y}}{\partial\mathbf{X}}$? Since

$$\large y_{ij} = \sum^{2}x_{ik}\cdot w_{kj},$$

so for each $y_{ij}$ there are two related $x_{ik}$ to with respect to? I'm confused at this point. What's the definition of the partial derivative of a matrix multiplication/product? And I don't know why there come up with the transposes $\mathbf{W}^\mathrm{T}$ and $\mathbf{X}^\mathrm{T}$?

I draw a picture about this process(and I omit the $\mathbf{B}$, which means bias but since it's out of concern here I assume it be zero matrix):

enter image description here


1 Answer 1


Suppose we are given a loss function $L=L(Y)\,$ and its gradient $\,G=\frac{\partial L}{\partial Y}$

Then suppose we are told that $Y$ in turn, depends on two other variables $$\eqalign{ Y &= XW \cr dY &= dX\,W+X\,dW \cr }$$ Let's find the differential of the loss function with respect to these two variables. $$\eqalign{ dL &= G:dY \cr &= G:dX\,W + G:X\,dW \cr &= GW^T:dX + X^TG:dW \cr }$$ Holding $W$ constant means that $dW=0$, and yields the gradient $$\eqalign{ \frac{\partial L}{\partial X} &= GW^T \cr }$$ Similarly, holding $X$ constant yields the gradient wrt $W$ $$\eqalign{ \frac{\partial L}{\partial W} &= X^TG \cr\cr }$$ In some of the steps above, a colon was used to denote the trace/Frobenius product $$A:B = {\rm tr}(A^TB)$$

  • $\begingroup$ Am I right to say that $\large G=\frac{\partial L}{\partial\mathbf{Y}}$ implies $\large dL = G:d\mathbf{Y}$? Is $\partial\mathbf{Y}$ equivalent to $d\mathbf{Y}$? $\endgroup$ Commented Apr 27, 2018 at 3:54
  • $\begingroup$ If the gradient of the loss function is $G=\nabla_YL,\,$ then the differential is $$dL=G:dY=\sum_{ij}G_{ij}\,dY_{ij}$$ And vice versa. Sometimes the notation $\frac{\partial L}{\partial Y}$ is used instead of $\nabla_YL$. $\endgroup$
    – greg
    Commented Apr 27, 2018 at 4:28
  • $\begingroup$ In your similar question, the quantity you are asking about $\frac{\partial Y}{\partial X}$ isn't a matrix, it is a 4th order tensor with components $$T_{ijkl}=\frac{\partial Y_{ij}}{\partial X_{kl}}$$ It is difficult to utilize the chain rule in matrix calculus because it often requires intermediate quantities which are 3rd and 4th order tensors. The best approach is to use differentials, since the differential of a matrix is also a matrix. $\endgroup$
    – greg
    Commented Apr 27, 2018 at 13:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .