# kronecker product property

I'm confused about this concept. Is this right? If not what is the correct version

$$\begin{bmatrix}\mathbf{I}\otimes\mathbf{x}_{1}^{\prime}\\ \vdots\\ \mathbf{I}\otimes\mathbf{x}_{i}^{\prime}\\ \vdots\\ \mathbf{I}\otimes\mathbf{x}_{n}^{\prime} \end{bmatrix}=\mathbf{I}\otimes\begin{bmatrix}\mathbf{x}_{1}^{\prime}\\ \vdots\\ \mathbf{x}_{i}^{\prime}\\ \vdots\\ \mathbf{x}_{n}^{\prime} \end{bmatrix}$$

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 $$\mathbf I \otimes \begin{bmatrix} \mathbf x_1' \\ \vdots \\ \mathbf x_n' \end{bmatrix} = \mathbf I \otimes \mathbf X = \begin{bmatrix} \mathbf X & 0 & 0 & \cdots & 0 \\ 0 & \mathbf X & 0 & \cdots & 0 \\ 0 & 0 & \ddots & \cdots & \vdots \\ \vdots & \vdots & & \ddots\\ 0 & 0 & 0 & \cdots & \mathbf X \end{bmatrix} = \begin{bmatrix} \mathbf x_1' & 0 & \cdots & 0 \\ \mathbf x_2' & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ \mathbf x_n' & 0 & \cdots & 0 \\ 0 & \mathbf x_1' & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & \mathbf x_n' & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \mathbf x_1' \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & \mathbf x_n' \end{bmatrix}$$