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Deriving Back Propagation in Matrix Form

Roy Varon Weinryb

varonroy@gmail.com
Introduction
The forward propagation process of neural networks can be very concisely written using a series of Matrix-Vector multiplications. The same can be said about the back propagation process. In this article I will show step by step how derivatives propagate through Matrix-Vector multiplications and how to use these results in a neural network.
Forward Propagation
Lets consider a portion of a neural network: two layers and the weights between, and an activation function \(f\). Let the first layer have \(n\) neurons and the second one \(m\) neurons. Since these two layers are nothing more than arrays of numbers, they can be represented using vectors: \[first \; layer\: v^1 \in \mathbb{R}^n \] \[second \; layer\: v^2 \in \mathbb{R}^m \] Let \(W_{i,j}\) be the weight connecting the \(ith\) neuron of the second layer and the \(jth\) neuron of the first layer. This way, the forward propagation for each of the individual neurons can be written as: \[v^2_i = f\Big(\sum_{k = 1}^{n}W_{i, k} {v^1_k}\Big)\] Or more compactly: \[v^2 = f(W v^1) \] To simplify things, lets split the above step into two: \[s^2 = W v^1 \] \[v^2 = f(s^2) \] So far we have: \[v^1 \in \mathbb{R}^n \] \[s^2 \in \mathbb{R}^m \] \[v^2 \in \mathbb{R}^m \] \[W \in \mathbb{R}^{m \times n} \] \[ \] \[s^2 = W v^1 \] \[v^2 = f(s^2) \]
Back Propagation
Let the derivative of the second layer with respect to the loss function be given: \[\frac{\partial L}{\partial v^2}\] So the partial derivative of the loss with respect to the \(ith\) neuron in the second layer will be the \(ith\) element of the above vector: \[\frac{\partial L}{\partial v^2_i} = \Big(\frac{\partial L}{\partial v^2}\Big)_i\] Going a step back, let's write how the \(ith\) element of \(s^2\) is propagation: \[ v^2_i = f(s^2_i) \] Then by taking the derivative: \[ \frac{\partial v^2_i}{\partial s^2_i} = f'(s^2_i) \] We can write the following vector: \[ \frac{\partial v^2}{\partial s^2} = f'(s^2) \] And apply the chain rule: \[ \frac{\partial L}{\partial s^2_i} = \frac{\partial L}{\partial v^2_i} \frac{\partial v^2_i}{\partial s^2_i} = \frac{\partial L}{\partial v^2_i} f'(s^2) \] And in vector form: \[ \frac{\partial L}{\partial s^2} = \frac{\partial L}{\partial v^2} \circ f'(s^2) \] (where \(\circ\) represents element wise multiplication). Next, the propagation towards \(v^1\) will be more complicated, because unlike there is no longer a one to one relation between elements in \(v^1\) and \(s^2\). Each element of \(v^1\) depends on each elment of \(s^2\) in the following way: \[s^2_i = \sum_{k = 1}^{n} W_{i, k} {v^1_k} \] So by taking the derivative of the \(ith\) element of \(s^2\) with respect to the \(jth\) element of \(v^1\) we get: \[\frac{\partial s^2_i}{\partial v^1_j} = W_{i, j}\] This means that when we apply the chain rule to get \(\partial L / \partial v^1_i\), we need to consider how every element of \(s_2\) will impact the total derivative: \[ \frac{\partial L}{\partial v^1_i} = \sum_{j = 0}^{m}\frac{\partial L}{\partial s^2_j} \frac{\partial s^2_j}{\partial v^1_i} = \sum_{j = 0}^{m}\frac{\partial L}{\partial s^2_j} W_{j, i} \] Notice how the indices in \(W\) are flipped. We can flip the indices back by transposing the matrix \[ \frac{\partial L}{\partial v^1_i} = \sum_{j = 0}^{m}\frac{\partial L}{\partial s^2_j} (W^T)_{i, j} = \sum_{j = 0}^{m} (W^T)_{i, j} \frac{\partial L}{\partial s^2_j}\] So the \(ith\) element of \(\partial L / \partial v^1_i\) equals to the dot product of the \(ith\) row of matrix \(W^T\) with the vector \(\partial L / \partial s^2_j\). This is exactly what matrix multiplication does, so we can write: \[ \frac{\partial L}{\partial v^1} = W^T \frac{\partial L}{\partial s^2}\] Finally, we need to calculate the derivatives with respect to each of the weights in the matrix \(W\). Using the above forward propagation formula: \[s^2_i = \sum_{k = 1}^{n}W_{i, k} {v^1_k} \] We can find the partial derivative: \[ \frac{\partial s^2_i}{\partial W_{i, j}} = v^1_j\] Notice that the index of the element of \(s^2\) has to match the index of the row of \(W\). Since only the \(ith\) row has an effect on that element. And now by applying the chain rule: \[ \frac{\partial L}{\partial W_{i, j}} = \frac{\partial L}{\partial s^2_i} \frac{\partial s^2_i}{\partial W_{i, j}} = \frac{\partial L}{\partial s^2_i} v^1_j \] It is a bit hard to see at first, but we can rewrite this expression as a multiplication of a column vector by a row vector: \[ \frac{\partial L}{\partial W} = \frac{\partial L}{\partial s^2} (v^1)^T \] Here is a small example of how that multiplication would work: \[ \begin{bmatrix} a_1\\ a_2 \end{bmatrix} \begin{bmatrix} b_1 & b_2 & b_3 \end{bmatrix} = \begin{bmatrix} a_1b_1 & a_1b_2 & a_1b_3\\ a_2b_1 & a_2b_2 & a_2b_3\\ \end{bmatrix} \]
Summary
To Summarize, given the following structure of a neural network: \[v^1 \in \mathbb{R}^n \] \[s^2 \in \mathbb{R}^m \] \[v^2 \in \mathbb{R}^m \] \[W \in \mathbb{R}^{m \times n} \] \[ \] \[s^2 = W v^1 \] \[v^2 = f(s^2) \] And the derivative of the loss function with respect to the last layer: \[\frac{\partial L}{\partial v^2}\] We can back propagate the derivative with the following formulas: \[ \frac{\partial L}{\partial s^2} = \frac{\partial L}{\partial v^2} \circ f'(s^2) \] \[ \frac{\partial L}{\partial v^1} = W^T \frac{\partial L}{\partial s^2}\] \[ \frac{\partial L}{\partial W} = \frac{\partial L}{\partial s^2} (v^1)^T \]