kronecker product property I'm confused about this concept. Is this right? If not what is the correct version
$$
    \begin{bmatrix}\mathbf{I}\otimes\mathbf{x}_{1}^{\prime}\\
    \vdots\\
    \mathbf{I}\otimes\mathbf{x}_{i}^{\prime}\\
    \vdots\\
    \mathbf{I}\otimes\mathbf{x}_{n}^{\prime}
    \end{bmatrix}=\mathbf{I}\otimes\begin{bmatrix}\mathbf{x}_{1}^{\prime}\\
    \vdots\\
    \mathbf{x}_{i}^{\prime}\\
    \vdots\\
    \mathbf{x}_{n}^{\prime}
    \end{bmatrix}
$$
Thanks in advance for your help.
 A: We have
$$\begin{bmatrix} \mathbf I \otimes \mathbf x_1' \\ \vdots \\ \mathbf I \otimes \mathbf x_n' \end{bmatrix} =
\begin{bmatrix}
\mathbf y_1 \\ \vdots \\ \mathbf y_n
\end{bmatrix}$$
where
$$\mathbf y_i = \mathbf I \otimes \mathbf x_i' =
\begin{bmatrix}
\mathbf x_i' & 0 & 0 & \cdots & 0 \\
0 & \mathbf x_i' & 0 & \cdots & 0 \\
0 & 0 & \mathbf x_i' & \cdots & 0 \\
\vdots & \vdots &  & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \mathbf x_i'
\end{bmatrix}$$
and 
$$\mathbf I \otimes 
\begin{bmatrix} \mathbf x_1' \\ \vdots \\ \mathbf x_n' \end{bmatrix} =
\mathbf I \otimes 
\mathbf X =
\begin{bmatrix}
\mathbf X & 0 & 0 & \cdots & 0 \\
0 & \mathbf X & 0 & \cdots & 0 \\
0 & 0 & \ddots & \cdots & \vdots \\
\vdots & \vdots & & \ddots\\
0 & 0 & 0 & \cdots & \mathbf X
\end{bmatrix}
=
\begin{bmatrix}
\mathbf x_1' & 0 & \cdots & 0 \\
\mathbf x_2' & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
\mathbf x_n' & 0 & \cdots & 0 \\
0 & \mathbf x_1' & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
0 & \mathbf x_n' & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \mathbf x_1' \\ 
\vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \mathbf x_n'
\end{bmatrix}
$$
So no, they are not the same thing.
