I am not sure how to prove this partial derivative. It seems to me that they are both vectors of integers. So when taking a derivative of the values we get 0 but then the total would be zero overall. Instead the value at one position seems to be 1 and it removes the transpose form the non derived value. I am not quite grasping this can anyone provide an explanation?
https://i.stack.imgur.com/fUTOt.jpg
\begin{equation} \frac{\partial}{\partial Vc} = Uo^T Vc \end{equation}
Thank you
$u_0$ and $v_c$ are vectors of the same dimension. The expression $\frac{\partial}{\partial v_c}(u_0^\top v_c)$ is not really a partial derivative, but rather the gradient of the function $f(v_c) = u_0^\top v_c = \sum_i (u_0)_i (v_c)_i$. In other words, $\frac{\partial}{\partial v_c}(u_0^\top v_c)$ is a vector of the partial derivatives $\frac{\partial}{\partial (v_c)_i} (u_0^\top v_c)$ for each $i$. Noting that $\frac{\partial}{\partial (v_c)_i} (u_0^\top v_c) = (u_0)_i$ shows that the gradient is indeed the vector $u_0$.