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I think it means TRANSPOSE, but I can't figure out the need to perform a transpose operation:

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Similarly, the set of weights that go with the inputs is written with a small T as well:

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There is also some algebraic operation with this T that I don't understand (why is the T interchangeable?):

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  • $\begingroup$ The T operator is not generally interchangeable, but as Michael's answer notes, it is in this particular case because the dot product is commutative. This is basically because the transpose of a matrix is equal to itself for $1\times 1$ matrices, but not for larger matrices. $\endgroup$ – Erick Wong Jan 1 '16 at 16:51
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The $T$ is indeed transpose. The reason it used is to define a column vector using less space.

The equality for $y$ is a dot product between two vectors defined as $$u\cdot v =u^T v$$ which coincides with a matrix vector product (the matrix is $n\times 1$ in this case and a vector $1\times n$). Particularly you have $$u\cdot v =u^T v= \sum_i u_i v_i =\sum_i v_i u_i = v^T u =v\cdot u$$

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