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Define an $(n \times m)$ matrix $E_{(i,j)}$ as the matrix which has one element, element $(i,j)$, that is equal to $1$ and the rest equal to $0$. Assume $A$ is an arbitrary matrix of shape $(m \times k)$. I am trying to write the following sum in einstein notation:

$$B = \sum_{i, j}E_{(i, j)} A$$

However I am getting confused. For arbitrary indices $(k, l, m, n)$ I know that element $(m, n)$ of $E_{(k, l)}$ equals $1$ if and only if $k=m$ and $l=n$.

From this I would assume that multiplication by $E_{(k, l)}$ would look something like $\delta^k_m\delta^l_n$ in Einstein notation, where $\delta$ is the kronecker delta.

However, in matrix form the kronecker delta is the identity matrix and $E_{(k, l)}$ is not a product of identity matrices. I am very confused, please could someone provide some intuition on how to represent $E_{(k, l)}$ in relation to einstein notation.

JDoe2
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1 Answers1

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You situation is similar to that of this problem. Let $C$ denote the matrix whose entries are all equal to $1$. We can then write your sum as $$ \left[\sum_{i,j} E_{(i,j)} A\right]_{p,q} = \delta_{pi}\delta_{ij}C_{ij} A_{jq}. $$ Alternatively, you might prefer to use $1_{ij}$ to denote the entries of the matrix whose entries are all equal to $1$.

Ben Grossmann
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  • Thank you for your answer. I guess my confusion also stems from the fact that $m\neq n \neq k$ so if I were to try and express my equation in matrix form I could end up with $\delta_{i, j} $ corresponding to a non-square matrix and thus might not be the identity. Can ricci calculus notation be used to matrix algebra involving non-square matrices? Looking at the definition of a tensor I'm thinking not... – JDoe2 Jan 25 '22 at 20:58
  • @JDoe2 It can! For example, if $A$ has size $m \times p$ and $B$ has size $p \times n$, then $$ [AB]{ij} = A{ik}B_{kj} $$ This works in the above case because the number of columns in $A$ (corresponding to the first $k$ index) matches the number of rows in $B$ (corresponding to the second). We just need to be careful about which indices are repeated – Ben Grossmann Jan 25 '22 at 21:37
  • I guess my confusion is if I have the below matrix equation:

    $$A=\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \end{bmatrix}B$$

    where $A$ is $(2 \times 4)$ in size and $B$ is $(4 \times 4)$ in size, then if I were to write it as a tensor it would look like:

    $$A^i_k=\delta^i_jB^j_k$$

    where $\delta^i_j$ does satisfy the definition of the kronecker delta (as it is $1$ when $i=j$ and $0$ otherwise) but doesnt correspond to the identity matrix.

    By the above, it must be true that either $\delta^i_j$ doesn't always correspond to the identity or tensors must be square. Which is true?

     – JDoe2 Jan 26 '22 at 18:26
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    $\delta^i_j$ does not always correspond to the identity. It only corresponds to the identity if the (implicit) bounds on $i$ and $j$ are the same. – Ben Grossmann Jan 26 '22 at 18:27
  • I see, thank you for explaining this to me! It is very appreciated! I don't suppose you know any good references with good discussion of these sort of nuances in ricci calculus notation? – JDoe2 Jan 26 '22 at 18:32
  • I don't, sorry. That might be a good question to ask with the [tag:reference-request] tag – Ben Grossmann Jan 26 '22 at 18:39
  • Okay, no worries! Thanks again! – JDoe2 Jan 26 '22 at 18:40