Taking the derivative of the loss function of a neural network can be quite cumbersome. Even taking the derivative of a single layer in a neural network often results in expressions cluttered with indices. In this post I’d like to show an index-free way to do it.

Consider the map ${\sigma(Wx+b)}$ where ${W\in{\mathbb R}^{m\times n}}$ is the weight matrix, ${b\in{\mathbb R}^{m}}$ is the bias, ${x\in{\mathbb R}^{n}}$ is the input, and ${\sigma}$ is the activation function. Usually ${\sigma}$ represents both a scalar function (i.e. mapping ${{\mathbb R}\mapsto {\mathbb R}}$) and the function mapping ${{\mathbb R}^{m}\rightarrow{\mathbb R}^{m}}$ which applies ${\sigma}$ in each coordinate. In training neural networks, we would try to optimize for best parameters ${W}$ and ${b}$. So we need to take the derivative with respect to ${W}$ and ${b}$. So we consider the map $\displaystyle \begin{array}{rcl} G(W,b) = \sigma(Wx+b). \end{array}$

This map ${G}$ is a concatenation of the map ${(W,b)\mapsto Wx+b}$ and ${\sigma}$ and since the former map is linear in the joint variable ${(W,b)}$, the derivative of ${G}$ should be pretty simple. What makes the computation a little less straightforward is the fact the we are usually not used to view matrix-vector products ${Wx}$ as linear maps in ${W}$ but in ${x}$. So let’s rewrite the thing:

There are two particular notions which come in handy here: The Kronecker product of matrices and the vectorization of matrices. Vectorization takes some ${W\in{\mathbb R}^{m\times n}}$ given columnwise ${W = [w_{1}\ \cdots\ w_{n}]}$ and maps it by $\displaystyle \begin{array}{rcl} \mathrm{Vec}:{\mathbb R}^{m\times n}\rightarrow{\mathbb R}^{mn},\quad \mathrm{Vec}(W) = \begin{bmatrix} w_{1}\\\vdots\\w_{n} \end{bmatrix}. \end{array}$

The Kronecker product of matrices ${A\in{\mathbb R}^{m\times n}}$ and ${B\in{\mathbb R}^{k\times l}}$ is a matrix in ${{\mathbb R}^{mk\times nl}}$ $\displaystyle \begin{array}{rcl} A\otimes B = \begin{bmatrix} a_{11}B & \cdots &a_{1n}B\\ \vdots & & \vdots\\ a_{m1}B & \cdots & a_{mn}B \end{bmatrix}. \end{array}$

We will build on the following marvelous identity: For matrices ${A}$, ${B}$, ${C}$ of compatible size we have that $\displaystyle \begin{array}{rcl} \mathrm{Vec}(ABC) = (C^{T}\otimes A)\mathrm{Vec}(B). \end{array}$

Why is this helpful? It allows us to rewrite $\displaystyle \begin{array}{rcl} Wx & = & \mathrm{Vec}(Wx)\\ & = & \mathrm{Vec}(I_{m}Wx)\\ & = & \underbrace{(x^{T}\otimes I_{m})}_{\in{\mathbb R}^{m\times mn}}\underbrace{\mathrm{Vec}(W)}_{\in{\mathbb R}^{mn}}. \end{array}$

So we can also rewrite $\displaystyle \begin{array}{rcl} Wx +b & = & \mathrm{Vec}(Wx+b )\\ & = & \mathrm{Vec}(I_{m}Wx + b)\\ & = & \underbrace{ \begin{bmatrix} x^{T}\otimes I_{m} & I_{m} \end{bmatrix} }_{\in{\mathbb R}^{m\times (mn+m)}}\underbrace{ \begin{bmatrix} \mathrm{Vec}(W)\\b \end{bmatrix} }_{\in{\mathbb R}^{mn+m}}\\ &=& ( \underbrace{\begin{bmatrix} x^{T} & 1 \end{bmatrix}}_{\in{\mathbb R}^{1\times(n+1)}}\otimes I_{m}) \begin{bmatrix} \mathrm{Vec}(W)\\b \end{bmatrix}. \end{array}$

So our map ${G(W,b) = \sigma(Wx+b)}$ mapping ${{\mathbb R}^{m\times n}\times {\mathbb R}^{m}\rightarrow{\mathbb R}^{m}}$ can be rewritten as $\displaystyle \begin{array}{rcl} \bar G( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}) = \sigma( ( \begin{bmatrix} x^{T} & 1 \end{bmatrix}\otimes I_{M}) \begin{bmatrix} \mathrm{Vec}(W)\\b \end{bmatrix}) \end{array}$

mapping ${{\mathbb R}^{mn+m}\rightarrow{\mathbb R}^{m}}$. Since ${\bar G}$ is just a concatenation of ${\sigma}$ applied coordinate wise and a linear map, now given as a matrix, the derivative of ${\bar G}$ (i.e. the Jacobian, a matrix in ${{\mathbb R}^{m\times (mn+m)}}$) is calculated simply as $\displaystyle \begin{array}{rcl} D\bar G( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}) & = & D\sigma(Wx+b)( \begin{bmatrix} x^{T} & 1 \end{bmatrix}\otimes I_{M})\\ &=& \underbrace{\mathrm{diag}(\sigma'(Wx+b))}_{\in{\mathbb R}^{m\times m}}\underbrace{( \begin{bmatrix} x^{T} & 1 \end{bmatrix}\otimes I_{M})}_{\in{\mathbb R}^{m\times(mn+m)}}\in{\mathbb R}^{m\times(mn+m)}. \end{array}$

While this representation of the derivative of a single layer of a neural network with respect to its parameters is not particularly simple, it is still index free and moreover, straightforward to implement in languages which provide functions for the Kronecker product and vectorization. If you do this, make sure to take advantage of sparse matrices for the identity matrix and the diagonal matrix as otherwise the memory of your computer will be flooded with zeros.

Now let’s add a scalar function ${L}$ (e.g. to produce a scalar loss that we can minimize), i.e. we consider the map $\displaystyle \begin{array}{rcl} F( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}) = L(G(Wx+b)) = L(\bar G( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}). \end{array}$

The derivative is obtained by just another application of the chain rule: $\displaystyle \begin{array}{rcl} DF( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}) = DL(G(Wx+b))D\bar G( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}). \end{array}$

If we want to take gradients, we just transpose the expression and get $\displaystyle \begin{array}{rcl} \nabla F( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix}) &=& D\bar G( \begin{pmatrix} \mathrm{Vec}(W)\\b \end{pmatrix})^{T} DL(G(Wx+b))^{T}\\ &=& ([x^{T}\ 1]\otimes I_{m})^{T}\mathrm{diag}(\sigma'(Wx+b))\nabla L(G(Wx+b))\\ &=& \underbrace{( \begin{bmatrix} x\\ 1 \end{bmatrix} \otimes I_{m})}_{\in{\mathbb R}^{(mn+m)\times m}}\underbrace{\mathrm{diag}(\sigma'(Wx+b))}_{\in{\mathbb R}^{m\times m}}\underbrace{\nabla L(G(Wx+b))}_{\in{\mathbb R}^{m}}. \end{array}$

Note that the right hand side is indeed vector in ${{\mathbb R}^{mn+m}}$ and hence, can be reshaped to a tupel ${(W,b)}$ of an ${m\times n}$ matrix and an ${m}$ vector.

A final remark: the Kronecker product is related to tensor products. If ${A}$ and ${B}$ represent linear maps ${X_{1}\rightarrow Y_{1}}$ and ${X_{2}\rightarrow Y_{2}}$, respectively, then ${A\otimes B}$ represents the tensor product of the maps, ${X_{1}\otimes X_{2}\rightarrow Y_{1}\otimes Y_{2}}$. This relation to tensor products and tensors explains where the tensor in TensorFlow comes from.