I apologize if my question is duplicate, but I really need to know this. I tried to verify equation $\frac{\partial vech(X)}{\partial vech(C)}$ suggested by Lynn in this post. As a simple case, I consider a 2x2 symmetric p.s.d matrix $X=CC^T$. To this end, I computed every element of $\frac{\partial vech(X)}{\partial vech(C)}$ one by one as follows (equation 1). Assuming matrix $C$ to be: \begin{equation} C=\begin{bmatrix} c_{11} & 0 \\ c_{21} & c_{22} \end{bmatrix} \end{equation} then matrix X will be: \begin{equation} X=\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} = \begin{bmatrix} c_{11}^2 & c_{11}c_{12} \\ c_{11}c_{21} & c_{21}^2+c_{22}^2 \end{bmatrix} \end{equation} $=>$
\begin{equation} \frac{\partial vech(X)}{\partial vech(C)}= \begin{bmatrix} \frac{\partial c_{11}^2}{\partial c_{11}} & \frac{\partial c_{11}^2}{\partial c_{21}} & \frac{\partial c_{11}^2}{\partial c_{22}} \\ \frac{\partial c_{11}c_{21}}{\partial c_{11}} & \frac{\partial c_{11}c_{21}}{\partial c_{21}} & \frac{\partial c_{11}c_{21}}{\partial c_{22}} \\ \frac{\partial (c_{21}^2+c_{22}^2)}{\partial c_{11}} & \frac{\partial (c_{21}^2+c_{22}^2)}{\partial c_{21}} & \frac{\partial (c_{21}^2+c_{22}^2)}{\partial c_{22}} \\ \end{bmatrix} =\begin{bmatrix} 2c_{11} & 0 & 0 \\ c_{21} & c_{11} & 0 \\ 0 & 2c_{21} & 2c_{22} \end{bmatrix} \quad\quad\quad\quad (1) \end{equation}
Now using equation $\frac{\partial vech(X)}{\partial vech(C)}=L[(C \otimes I)+(I \otimes C)K]D$ from this post, the following result is obtained:
\begin{equation} \frac{\partial vech(X)}{\partial vech(C)}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \left\{ \begin{bmatrix} c_{11} & 0 & 0 & 0 \\ 0 & c_{11} & 0 & 0 \\ c_{21} & 0 & c_{22} & 0 \\ 0 & c_{21} & 0 & c_{22} \end{bmatrix} + \\ \begin{bmatrix} c_{11} & 0 & 0 & 0 \\ c_{21} & c_{22} & 0 & 0 \\ 0 & 0 & c_{11} & 0 \\ 0 & 0 & c_{21} & c_{22} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \right\} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} => \end{equation}
\begin{equation} \frac{\partial vech(X)}{\partial vech(C)}= \begin{bmatrix} 2c_{11} & 0 & 0 \\ c_{21} & c_{11}+c_{22} & 0 \\ 0 & 2c_{21} & 2c_{22} \end{bmatrix} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (2) \end{equation} As can be seen, value at cell (2,2) of the Jacobian matrix in equation 2 is different from the value at the same cell in equation 1. Please correct me if I am making a mistake here. But if I'm right, then does it mean that $\frac{\partial vech(X)}{\partial vech(C)}$ equation suggested by Lynn is wrong? If that's the case, can anyone give the correct answer, please? thank you.
I've found the solution myself. The problem with the approach suggested by Lynn is that we can't use duplication matrix to convert $vec(dC)$ to $D_k vech(dC)$ because matrix $dC$ is lower triangular and not symmetric. Recall that, for any $n \times n$ matrix $A$, $vec(A)=D_n vech(A)$ only when $A$ is symmetric [see here].
The correct answer for $\frac{\partial vech(X)}{\partial vech(C)}$ is available in section 10.5.4 of book Handbook of Matrices: \begin{equation} \frac{\partial vech(CC^T)}{\partial vech(C)^T} = 2D_k^+(C \otimes I_k)L^T_k \quad (1) \end{equation}
In the equation above $D_k^+ = (D_k^T D_k)^{-1} D_k^T$ is Moore-Penrose pseudoinverse of duplication matrix [see here].
In the mentioned book, it is said that proof for equation (1) is available in book Linear Structures. Unfortunately, I don't have access to that book and I have failed to write the proof for equation (1) myself. So I appreciate if anyone can come up with the proof for equation (1).