how to compute the vector derivative of this matrix equation 
How to compute the derivative of the latent factor where $u_n$, $v_n$ are column vectors. How did he take derivative and obtain the below equation?
I was reading probabilistic matrix factorization where I came across this equation. I don't know how to take vector derivative of another vector. To learn about vector derivatives in general can you point a source?
 A: Note that indexed vectors are really the columns of a matrix, e.g.
$$\eqalign{
 u_n &= Ue_n \cr
 v_m &= Ve_m\cr
}$$
where $e_k$ denotes the standard vector basis. 
Similarly, $X$ is also a matrix with columns and components given by
$$\eqalign{
 x_n &= Xe_n \cr
 X_{mn} &= e_m^Tx_n = e_m^TXe_n \cr
}$$
Now we can write the problem in matrix from
$$\eqalign{
 2{\cal L}_n &= \sigma^{-2}(Vu_n-x_n)^T(Vu_n-x_n) + \lambda u_n^Tu_n  \cr
}$$
at this point the $n$ subscripts are distracting, so let's drop them
$$\eqalign{
 2{\cal L} &= \sigma^{-2}(Vu-x)^T(Vu-x) + \lambda u^Tu  \cr
}$$
and take the differential
$$\eqalign{
 2\,d{\cal L} &= 2\sigma^{-2}(Vu-x)^T\,d(Vu-x) + 2\lambda u^T\,du  \cr
d{\cal L}
 &= \sigma^{-2}(Vu-x)^TV\,du + \lambda u^T\,du  \cr
 &= \big(\sigma^{-2}(Vu-x)^TV + \lambda u^T\big)\,du  \cr
}$$
Write down the gradient, set it equal to zero, and solve for the optimal vector $u$ 
$$\eqalign{
&\frac{\partial{\cal L}}{\partial u}
 = \sigma^{-2}(Vu-x)^TV + \lambda u^T = 0 \cr
&u^TV^TV + \lambda\sigma^2u^T = x^TV \cr
&V^TVu + \lambda\sigma^2u = V^Tx \cr
&(V^TV + \lambda\sigma^2I)\,u = V^Tx \cr
&u = (V^TV + \lambda\sigma^2I)^{-1}V^Tx \cr
\cr
}$$
