Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I re-tagged this "linear algebra". I'm not sure that's right (but it's better than the original tag, "education"!). –  Gerry Myerson Feb 22 '12 at 23:32

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.