# Deep NNs, backpropagation and error calculation

I was following the backpropagation tutorial by Michael Nielsen on http://neuralnetworksanddeeplearning.com/chap2.html, one of the very few places where the backprop algorithm is nicely explained both in theory and practice with an actual implementation in python.

Now, something is not clear when he explains how to calculate the error on a specific layer in terms of error in the next layer (I will provide a screenshot because I don't want to mess with notation):

Here, delta^l is the error in layer l, the circled dot is the elementwise product betweeen vectors, w is a weight matrix, and sigma primed indicates the derivative of an activation function.

Now, observe he points out that by applying the transpose operator to the next-layer-error vector (which is the previous layer during backprop), one is "sending back" the error across the network, and then, multiplying by the derivative of sigma, one applies the chain rule which completes the backprop step.

I have troubles understanding what does he really mean by saying that the traspose operator "sends back". Perhaps he wants to apply the inverse trasformation? But unless the matrix happens to be orthogonal, which is not the case AFAIK, the transpose does NOT equal the inverse.. In any case, I would have thought of applying some kind of differential operator to matrix w, and not the transpose one.

Opinions?

NOTE: Please let me know whether this post is correctly suited for stackoverflow, or should it be posted on stats or math stackexchange.

"When we apply the transpose weight matrix, $(w^{l+1})^T$, we can think intuitively of this as moving the error backward through the network"

Is that the sentence you're confused about? Note that in this sentence he's not trying to provide intuition for ''applying the transpose operator'', he's trying to provide intuition for ''aplplying the transpose weight matrix'', which basically means ''multiplying by the transpose weight matrix''. That entire act of multiplying the error vector with the transposed weight matrix, that is what can be understood intuitively as moving the error backward through the network. Not just the transpose operator on its own.

The reason for doing the transpose is simply to make the dimensions work out, that matrix needs to be transposed in order to guarantee that multiplying it with the error vector afterwards results in the correct dimensionality.

The reason why this application (~= multiplication) with the transposed weight matrix can be understood as moving the error backward through the network is kind of exactly the same intuition you probably already have for the forwards pass. In the forwards pass, a high/strong weight means a node in one layer has a large influence on the connected node in the next layer. Exactly the same intuition holds here for the backwards pass, because, just like in the forwards pass, we're multiplying something with a weight. In the forwards pass, that something is the activation level of a node in the previous layer. In the backwards pass, that something is the error observed in a node in the later layer. If a particular connection is ''strong'' (weight has a high value), and we also have a large error, we should put a large amount of ''blame'' for the error on that large weight. This is what we get by multiplying by the (transposed) weight matrix; large weights take a large portion of the ''blame'' for large errors.

More Elaborate explanation for transpose:

Let's consider the forwards pass, following the notation from your link. For simplicity, let's ignore activation functions and biases for now, they're not really important for our intuition. Then, equation 25 in your link tells us that the activation vector $a^l$ of layer $l$ is defined as $a^l = w^l * a^{l-1}$. To make the notation a bit more consistent with the equation for the backwards pass, we'll rewrite this as $a^{l+1} = w^{l+1} * a^l$ (simply added $1$ to all layer indices).

$w^{l+1}$ is a matrix, and the $a$ things are vectors, so it's convenient to consider again what the matrix-vector multiplication looks like. Take a look at the image near the top here for example: http://mathinsight.org/matrix_vector_multiplication , copied at the bottom of this answer.

Let's consider the activation level of the very first node (at the top) of layer $l+1$. This would be the top element of the vector in the right-hand side of the equation on the page I linked. This one activation level is determined by the complete vector of activation levels of the entire previous layer, and the very first row of weights in the weight matrix.

Now suppose we have a vector $\delta^{l+1}$ of errors in layer $l+1$. Let's again consider only the top node of this layer. The activation level of this particular node was determined by the top row of the weight matrix $w^{l+1}$, and the entire vector of activation levels $a^l$. We'll want to "punish" each of the weights in that top row proportional to the magnitude of that weight and our error. So, to do that, we'll want some kind of multiplication between $w^{l+1}$ and $\delta^{l+1}$ (note that this is different from the forwards pass, in which we had a multiplication between the same weight matrix, but an activation vector from a different layer). This should again result in a vector with the same shape as $a^l$ (this is already one clue that we have to do the transpose, otherwise the shape simply won't be correct).

To figure out how this multiplication should look exactly, we'll have to investigate which weights exactly are to blame for which errors. The top row of weights in the matrix $w^{l+1}$ had (during the forwards pass) influence on only the top element of $a^{l+1}$, and is therefore now only to blame for the top element of the error vector $\delta^{l+1}$ and that entire top row of weights should therefore now also only be multiplied by the top element of $\delta^{l+1}$ in order to compute a part of $\delta^l$. In the notation of the image near the top of that page I linked, this means that we would like our top row of weights to be only multiplied $x_1$. But that's not what the picture of matrix-vector multiplication says, that picture says that elements of the top row of the matrix are multiplied with all $x$. If you look carefully, there is a vector in that matrix which is solely multiplied by the top element $x_1$ though, which is the very first column of the (weight matrix). By transposing the weight matrix, we interchange rows and columns, and we essentially get exactly the multiplications we desire.

Linked matrix-vector multiplication \begin{align*} A{x}= \left[ \begin{array}{cccc} a_{11}& a_{12}& \ldots& a_{1n}\\ a_{21}& a_{22}& \ldots& a_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ a_{m1}& a_{m2}& \ldots& a_{mn} \end{array} \right] \left[ \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right] = \left[ \begin{array}{c} a_{11}x_1+a_{12}x_2 + \cdots + a_{1n} x_n\\ a_{21}x_1+a_{22}x_2 + \cdots + a_{2n} x_n\\ \vdots\\ a_{m1}x_1+a_{m2}x_2 + \cdots + a_{mn} x_n\\ \end{array} \right] \end{align*}

• I understand your qualitative considerations about sensitivity and errors. What I still miss, though, is why we use the transposed matrix in order to "move the error backwards". It couldn't be just a matter of dimensions.. Remember that the matrix-vector product is like applying a linear transformation to that vector (and the entire space it lies within). – MadHatter Nov 28 '17 at 2:35
• @MadHatter I've edited a large block of extra text into my answer. Does this help? – Dennis Soemers Nov 28 '17 at 10:09
• Although, to be honest, there is still something I'm missing. The top row of W is to blame for errors in the top node. Fine, but: 1. What is the connection between the error vector (which comes backwards from the output layer) and the activation vector a_(l+1) which is directly produced by Wa_l? 2. If the top row is the only to blame, for errors in the top node, by trasposing we somewhat lose track of this, since as you duly noticed, we multiply that error by the first COLUMN, which has little to to, aside from its first element with how that error has been produced. – MadHatter Nov 28 '17 at 21:52
• 1. The connection between error vector and activation vector is that high activation levels (in an absolute sense, so also very negative numbers) should be blamed more for high errors than low activation levels. 2. The transposing is necessary because we change the equation a bit. In forwards pass, W*a^l = a^{l+1}. In backwards pass, W*delta^{l+1} = delta^l (ignoring activation functions etc). Note how the l+1 moved from right side of the equation to the left side of the equation. To make sure that the correct weights still get the correct amount of ''blame'', the transpose is necessary. – Dennis Soemers Nov 29 '17 at 9:20
• Good, thanks. Still, the question about row/column exchange remains. That is: " If the top row is the only to blame, for errors in the top node, by trasposing we somewhat lose track of this, since as you duly noticed, we multiply that error by the first COLUMN, which has little to to, aside from its first element with how that error has been produced." – MadHatter Nov 29 '17 at 17:39