1
$\begingroup$

In chapter 6.1 on 'Example: Learning XOR', the bottom of page 168 mentions:

The activation function $g$ is typically chosen to be a function that is applied element-wise, with $h_i = g(x^TW_{:,i}+c_i).$

Then we see equation 6.3 is defined as (assuming g as ReLU):

We can now specify our complete network as $f(x; W,c,w,b) = w^T$ max$\{0, W^Tx + c\} + b$

Wondering why the book uses $W^Tx$ in equation 6.3, while I expect it to be $x^TW$. Unlike XOR example in the book where $W$ is a $2\times2$ square matrix, we may have non-square $W$ as well, and in such cases, $x^TW$ is not same as $W^Tx$.

Please help me understand, if I'm missing something here.

$\endgroup$

1 Answer 1

2
$\begingroup$

Let $\mathbf{y} = \mathbf{W}^T \mathbf{x}$

Then, $\mathbf{y}^T =(\mathbf{W}^T \mathbf{x})^T =\mathbf{x}^{T}(W^T)^T = \mathbf{x}^{T}W $. Note that $\mathbf{W}$ does not have to be a square matrix.

Let $e^{(i)}_{j} = \delta_{i,j} $.

Then, $y_{i} = \mathbf{y}^{T}e^{(i)} = (\mathbf{x}^T W) e^{(i)} = \mathbf{x}^{T}(We^{(i)}) = \mathbf{x}^{T}W_{:,i}$ and thus

$h_{i} = g(\mathbf{x}^T W_{:,i}+c_{i}) = g(y_{i}+c_{i})$

On the other hand, $f(..) = w^{T} \max\{\mathbf{0},W^{T}\mathbf{x}+\mathbf{c}\}+b = w^{T} \max\{\mathbf{0},\mathbf{y}+\mathbf{c}\}+\mathbf{b}$.

Does that answer your question ?

$\endgroup$
1
  • $\begingroup$ Excellent explanations. Thanks a lot. May I ask some suggestions for books, especially for Mathematics that will help me grasp the concepts better. $\endgroup$
    – KGhatak
    Commented Sep 25, 2020 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.