Facts: For matrices $A_i\in \mathbb{R}^{n\times n}$ with $i=1, 2, 3$, we have the following equation: $$ A_1\otimes A_2 \otimes A_3 = (A_1\otimes I_{n^2})(I_{n}\otimes A_2 \otimes I_n)(I_{n^2}\otimes A_3), $$ where $I_n$ represents an $n$ by $n$ identity matrix, and $\otimes$ denotes the Kronecker product.

My question is what would happen in the equation if matrices $A_i$ are not square matrices, i.e., $A_i\in \mathbb{R}^{m\times n}$, where $m,n$ are not necessarily equal. Is there a way to prove it?

Thank you very much!


  • $\begingroup$ Why do you need three As? For your non-square matrices, how do you write your equation for two As? with what "identity" matrices? $\endgroup$ – Cosmas Zachos Jun 26 '18 at 19:42
  • $\begingroup$ If n>m , for instance, you may embed your m×n matrices as blocks in n×n ones, and apply your formula unchanged. $\endgroup$ – Cosmas Zachos Jun 26 '18 at 19:46
  • $\begingroup$ @CosmasZachos For instance, if I have three covariance matrix in three different spaces $\mathcal{X}_1, \mathcal{X}_2, \mathcal{X}_3$, I will have three squared matrices. The kronecker product of these three matrices gives the covariance matrix for the product space $\mathcal{X}_1\times \mathcal{X}_2 \times \mathcal{X}_3$. These three matrices do not necessarily have same dimensions, since the number of points in each space can be different.There is a typo in my above question for the notation $A_i\in \mathbb{R}^{m\times n}$. It should be $A_i\in \mathbb{R}^{m_i\times n_i}$ $\endgroup$ – Bayes Jun 27 '18 at 15:00
  • $\begingroup$ @CosmasZachos Thanks for your comment. I am not sure how I can embed my $m_i\times n_i$ matrices as blocks in a one big matrice. For the identity matrix $I_{n}$, I mean its diagnoal is 1, and other entries are zeros. $\endgroup$ – Bayes Jun 27 '18 at 15:03
  • $\begingroup$ An m×n matrix has m rows and n columns (more than m). Consider the n×n matrix whose upper m rows are those of the m×n, and the rest zero. You are now dealing with square matrices. Appreciate the particular dimension of each in each tensor subspace is completely irrelevant. Now your Identity matrices are square and well-defined. My point is you needlessly confuse yourself by a triple tensor product, instead of a double one. $\endgroup$ – Cosmas Zachos Jun 27 '18 at 15:13

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