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Let $A \in \mathbb{R}^{n_2 \times n_1}, B \in \mathbb{R}^{n_2 \times n_3}, C \in \mathbb{R}^{n_1 \times n_3}$ be matrices. Define:

$$S_j = \sum_{k=j}^{n_3} \sum_{i=1}^{n_1} C_{ik} (A_i - B_k)$$

for $j=1, \dots, n_3$, where $C_{ik}$ is the $ik$-th element of $C$, $A_i$ is the $i$-th column of $A$ and $B_k$ is the $k$-th column of $B$. I am wondering how to write the matrix

$$S = (S_1 \dots S_{n_3})$$

as a product of matrices, rather than the double summation above. I am asking because I need to compute $S$ in a program. For optimization purposes, it would be best to write it as a matrix product.

My attempts so far have involved the following transformation matrix:

$$\Lambda = \begin{pmatrix} 1 & 0 & 0 & \dots & 0 \\ 1 & 1 & 0 & \dots & 0 \\ 1 & 1 & 1 & \dots & 0 \\ & && \ddots \\ 1 & 1 & 1 & \dots & 1 \end{pmatrix}.$$

I am able to write

$$\left(\sum_{k=1}^{n_3} \sum_{i=1}^{n_1} C_{ik} A_i , \dots, \sum_{k=n_3}^{n_3} \sum_{i=1}^{n_1} C_{ik} A_i \right) = A C \Lambda^T$$

which solves half the problem. However, I do not see a way to write

$$\left(\sum_{k=1}^{n_3} \sum_{i=1}^{n_1} C_{ik} B_k , \dots, \sum_{k=n_3}^{n_3} \sum_{i=1}^{n_1} C_{ik} B_k \right).$$

Thank you in advance for any suggestions.

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1 Answer 1

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Note that $(\sum_{i=1}^{n_1} C_{i1}B_1,\dots,\sum_{i=1}^{n_1} C_{in_3}B_{n_3}) = BC^T$. So, your matrix can be written as $$ \left(\sum_{k=1}^{n_3}[BC^T]_k, \dots, \sum_{k=n_3}^{n_3}[BC^T]_k\right) $$ Conclude that the desired matrix is in fact $(BC^T)\Lambda_{n_1}$. All together, your matrix is $S = AC\Lambda_{n_3}^T - BC^T\Lambda_{n_1}$.

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  • $\begingroup$ Thanks for your answer. If I'm not mistaken, there is a dimension mismatch: $B C^T \Lambda_{n_1}$ has size $n_2 \times n_1$ but $AC\Lambda_{n_3}^T$ has size $n_2 \times n_3$. $\endgroup$
    – atzol
    May 23, 2017 at 17:21
  • $\begingroup$ Thank you also for correcting the first term. $A$ indeed does not require a transpose. I edited my post accordingly. $\endgroup$
    – atzol
    May 23, 2017 at 17:22

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