# Tag Info

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 $... 1 There are two issues here. First, you are in a regression setting, where accuracy is meaningless - see What function defines accuracy in Keras when the loss is mean squared error (MSE)? Although the linked thread is not directly applicable here (since you use your own custom calc_accu function), the essence of the argument remains: assessing K.equal(K.round(... 1 Given the information you provided, the most honest answer is: You have to test it by yourself, there is no general answer for it. Still, it has been shown empirically in research that a neural network may benefit from having multiple outputs. So let's say we have a neural network that has multiple outputs. Further, let us group them into specific tasks: For ... 1 Apart from the correct answers which you find in the comment section, you mention "the square [..] prevent any negative loss". In principle you can also have a negative loss. The point is without the square, you have$(x-y) \neq (y-x)$for$x \neq y$. In particular, the loss would not be symmetric and for$x = 0$, you have$(x-y) = -y\$. So by ...