# Derivative of row-wise softmax matrix w.r.t. matrix itself

Define matrix $$\mathbf{M}' \in \mathbb{R}^{n \times k}$$ as the result of the row-wise softmax operation on matrix $$\mathbf{M} \in \mathbb{R}^{n \times k}$$. Hence,

$$\mathbf{M}'_{ij} = \frac{\exp{\mathbf{M}_{ij}}}{\sum\limits_{b=1}^k \exp{\mathbf{M}_{ib}}}.$$

Now, I look at the derivative of a scalar function, e.g., the Frobenius norm, with respect to $$\textbf{M}$$, namely

$$\frac{\partial E}{\partial \textbf{M}} = \frac{\partial \left\Vert \textbf{X} - \textbf{M}'\textbf{H}\right\Vert_F}{\partial \textbf{M}}.$$

I don't have any problem calulating the derivative of the above function w.r.t. $$\textbf{M}'$$. However, I am interested in finding the derivative w.r.t. $$\textbf{M}$$, which means that I somehow have to deal with the row-wise softmax operation. Since softmax is a vector function, but I am interested in finding the derivative w.r.t. the whole matrix $$\textbf{M}$$ at once, I don't know how to deal with it best. Do I need to calculate the derivative w.r.t. each vector $$\textbf{M}_{i:}$$ seperately? Also, the derivative of the softmax would yield a Jacobian matrix of dimensionality $$k \times k$$. Getting one Jacobian for each row vector $$\textbf{M}_{i:}$$ seems to mess up the dimensionality, assuming I would need to concatenate all those Jacobians... I am not sure where my mistake is. However, it feels like I am stuck.

It would be great if you could help me out :)

Thanks in advance and best regards.

Denote elementwise/Hadamard multiplication and division by the symbols $$(\odot,\oslash)$$ respectively.
Define an all-ones matrix $$J\in{\mathbb R}^{k\times k},\,$$ as well as the following matrices \eqalign{ P &= \exp(M) \quad&\implies dP &= P\odot dM \\ Q &= PJ &\implies dQ &= dP\,J = (P\odot dM)\,J \\ R &= P\oslash Q &\implies dR &= dP\oslash Q - P\odot dQ\oslash Q\oslash Q \\ &&&= R\odot dM - S\odot\Big((P\odot dM)\,J\Big) \\ S &= R\oslash Q \\ Y &= RH-X \\ } where the differentials of all the new matrices have been expressed in terms of that of $$M$$.
Also note that in your naming convention \eqalign{ R &= M' \\ Q_{ij} &= \sum_{b=1}^k \Big(\exp M_{ib}\Big) J_{bj}\\ } Define the scalar $${\cal E}$$ function in terms of the new matrices and calculate its gradient. \eqalign{ {\cal E}^2 &= \|Y\|_F^2 \;=\; Y:Y \\ 2{\cal E}\,d{\cal E} &= 2Y:dY \\ &= 2(RH-X):dR\,H \\ &= 2(RH-X)H^T:dR \\ d{\cal E} &= A:dR \\ &= A:\bigg(R\odot dM - S\odot\Big((P\odot dM)\,J\Big)\bigg) \\ &= (R\odot A):dM - P\odot\Big((S\odot A)J\Big):dM \\ \frac{\partial{\cal E}}{\partial M} &= R\odot A - P\odot\Big((S\odot A)J\Big) \\ \\ }
In the above steps, the implicit definitions \eqalign{ A &= \left(\frac{RHH^T-XH^T}{\|X-RH\|_F}\right) \\ A:B &= {\rm Tr}(A^TB) \\ } were utilized; the latter being the trace/Frobenius product.
• Hi greg, thanks so much for the detailed answer. I just have one question. To get from line 5 to 6 in the derivative calculation it looks like you are using the property $A \colon B \odot C = A \odot B \colon C$. I saw you used this and similar trace product properties in other answers to matrix calculus questions. Could you maybe elaborate on why this property holds, or point me to some book/paper, where this property is further explained? Also, in the particular example of getting from line 5 to 6 I am not sure how to treat the brackets. Can I just ignore them using the above property? Commented Apr 20, 2020 at 8:01
• Write the product definitions as summations \eqalign{ B:C &= \sum_{i,j} B_{ij}C_{ij} \quad&\big({\rm Frobenius\,Product}\big) \\ (A\odot B):C &= \sum_{i,j} (A_{ij}B_{ij})C_{ij} \quad&\big({\rm Triple\,Product}\big) \\ } On the RHS, the matrices $(A,B,C)$ can be freely rearranged, which means they can also be rearranged on the LHS.