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Let $\mathbf A$ be an $n\times n$ matrix, and $\mathbf B$ be an $nr\times nm$, block diagonal matrix, with each block being a matrix of size $r\times m$. Also, let $\otimes$ denote the Kronecker product.

How do I conduct the following operation

$$ \mathbf B (\mathbf A \otimes \mathbf I_m ) \mathbf B ^\top?$$

Hopefully, I would like to rewrite this term in the following format

$$ \sum_{i=1}^n\sum_{j=1}^n F(\mathbf A)G(\mathbf B_i \mathbf B_j^\top),$$

where $\mathbf B_i$ corresponds to the $i$-th block matrix in the diagonal of $\mathbf B$, and $F$ and $G$ are linear operators.

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If $\{E_{kj}\}$ are the $n\times n$ matrix units, we have $$ B=\sum_{k,j=1}^nB_{kj}\otimes E_{kj},\ \ \ A\otimes I_m=\sum_{k,j=1}^na_{kj}I_m\otimes E_{kj}. $$ Then \begin{align} B(A\otimes I_m)B^T &=\sum_{k,j,s,t,x,y}B_{kj}a_{st}B_{yx}\otimes E_{kj}E_{st}E_{xy} =\sum_{k,j,t,y}a_{jt}B_{kj}B_{yt}^T\otimes E_{ky}\\ \ \\ &=\sum_{k,y}\left( \sum_{j,t}a_{jt}B_{kj}B_{yt}^T\right)\otimes E_{ky}. \end{align} I'm not sure how you expect to ignore the non-diagonal blocks.

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  • $\begingroup$ This is exactly what I needed. Thank you! $\endgroup$ – mzp Oct 11 '16 at 23:37
  • $\begingroup$ I think the equation for $A \otimes I_m$ should be $\sum_{k,j}a_{kj}E_{kj} \otimes I_m$. $\endgroup$ – mzp Oct 21 '16 at 6:29
  • $\begingroup$ Now I see that your $B $ is block diagonal, which is not what I did in my answer. If you also interpret $A\otimes I_m $ as block diagonal, then you are trying to do products $(B\otimes I_n)(A\otimes I_m) $; in that case my feeling is that there is little structure to play with (try an example with $n=2$ and $m=3$). $\endgroup$ – Martin Argerami Oct 21 '16 at 12:39

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