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I have been going through the description of the backpropagation algorithm found here. and I am having a bit of trouble getting my head around some of the linear algebra.

Say I have a final output layer $L$ consisting of two visible units, and layer $L-1$ consists of four hidden units. (this is just an example to illustrate my problem)

my understanding is that the weight matrix for this final layer ($w^L$) should be a 4 x 2 matrix.

the reference says to calculate the output error $\delta^{x,L}$ given by:

$\delta^{x,L} = \nabla_a C_x \odot \sigma'(z^{x,L})$

where:

$z^{x,L} = w^La^{x,L-1} + b^L$,

$a^{x,L} = \sigma(z^{x,L})$, and

$\odot$ is the hadamard product.

evaluating $\delta^{x,L}$ gives a 1 x 2 vector, as it should given there are two output units.

my problem is when calculating the hidden layer gradients (eg. $L-1$) given by:

$\delta^{x,L-1} = ((w^L)^T\delta^{x,L})\odot\sigma'(z^{x,L-1})$

now if $w^L$ is a 4x2 matrix and $\delta^{x,L}$ is a 1x2 vector then wouldnt $(w^L)^T\delta^{x,L}$ be a multiplication of a 2x4 matrix and a 1x2 matrix, which is impossible?

i feel like i have missed something vital in my understanding, but i cant work out what it is.

is it just as simple as making it $\delta^{x,L}(w^L)^T$? this would be a 1x2 matrix multiplied by a 2x4 matrix, which is perfectly legal. but the formula has it around the other way.

can anyone see where my understanding is flawed? any help would be greatly appreciated

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You've transposed the sizes of both $w^L$ and $\delta^{x,L}$. $w^L$ should be 2x4 and $\delta^{x,L}$ should be 2x1. $(w^L)^T$ is a 4x2 matrix that will be multipled by a 2x1 matrix yielding a 4x1 matrix suitable for the next step of backpropagation. In general for neural nets the activation units are represented as column vectors and the weights are matrices of dimension |L+1| x |L|, where L is the current layer and L+1 is the next layer (in the forward direction).

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  • $\begingroup$ thanks for clearing that up, i was under the impression that input weights were the rows and outputs the columns and you perform the multiplications as input * weight matrix, not the other way around.. i will have to go back and update my whole code now! thanks for letting me know $\endgroup$ – guskenny83 Oct 17 '14 at 1:11
  • $\begingroup$ actually, on second thoughts, using the inputs as row vectors and performing the multiplications as input * weights gives the same result as using column vectors and weights * input, so i dont think i do have to change my code, just the order of the equation (so long as i specify the convention used). thanks for clearing that up for me though.. $\endgroup$ – guskenny83 Oct 17 '14 at 1:16

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