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I'm a bit confused by using Einstein summation over more than two matrices being multiplied together. I want to write $CD^T=A(I-BB^T)C^T$ in Einstein summation notation, where $D^T$ is the transpose of the matrix $D$. How do I write the indices that we're summing over?

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  • $\begingroup$ I don't understand the question. You have two matrices -- one on the left and one on the right. The left would be $(CD^T)_{ij} = C_{ik}D_{jk}$. $\endgroup$
    – amsmath
    Oct 8, 2019 at 13:36
  • $\begingroup$ Yes so I need to write CD^T in Einstein summation, but I thought that it would be c_ij d_ji Then I need to write A(I-BB^T)C^T in Einstein summation notation but I am unsure of how to write this. $\endgroup$
    – Hector
    Oct 8, 2019 at 14:16
  • $\begingroup$ No. $C_{ij}D_{ji}$ is $(CD)_{ii}$. $\endgroup$
    – amsmath
    Oct 8, 2019 at 14:18
  • $\begingroup$ OK thanks, and do you know how to write the second part? $\endgroup$
    – Hector
    Oct 8, 2019 at 14:25
  • $\begingroup$ Sure. You can start writing it as $AC^T-ABB^TC^T$. $\endgroup$
    – amsmath
    Oct 8, 2019 at 14:43

1 Answer 1

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In Einstein's summation notation, $$\begin{align} (CD^T)_{ij} &= C_{ia}(D^T)_{aj} = C_{ia}D_{ja}\\ (A(I-BB^T)C^T)_{ij} &= A_{ia}(I - BB^T)_{ab}(C^T)_{bj}\\ &= A_{ia}(\delta_{ab} - (BB^T)_{ab})C_{jb}\\ &= A_{ia}(\delta_{ab} - B_{ac}(B^T)_{cb})C_{jb}\\ &= A_{ia}(\delta_{ab} - B_{ac} B_{bc})C_{jb} \end{align} $$ where $\delta_{ab}$ is the Kronecker delta.

The equality $CD^T=A(I−BB^T)C^T$ becomes

$$C_{ia}D_{ja} = A_{ia}(\delta_{ab} - B_{ac} B_{bc})C_{jb} = A_{ia}C_{ja} - A_{ia}B_{ac}B_{bc}C_{jb} $$

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