Using diagonality in Einstein notation

index-notationlinear algebra

Given a diagonal matrix $D$, with diagonal elements given by vector $\mathbf{d}$. Representing this in Einstein notation gives

$$
D_{ij} = \delta_{ijk} d_k
$$

where

$$
\delta_{ijk} =
\begin{cases}
1 & \text{if } i = j = k \\
0 & \text{ otherwise}
\end{cases}
$$

If I now apply this in a matrix multiplication, e.g.

$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} \delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} \delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$

The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.

This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:

  1. Where do I go wrong in my reasoning?
  2. Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?

Best Answer

What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this

$$ D_{ij} = \delta_{ijk}d_k \color{red}{\stackrel{!!}{=}} d_i $$

which clearly shows the problem much earlier than you noticed: expanding the symbol $\delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option

\begin{eqnarray} (A D)_{lj} &=& \sum_{i}A_{li}\color{blue}{D_{ij}} = \sum_iA_{li}\color{blue}{\delta_{ij}d_j} = A_{lj}d_j ~~~\mbox{(sum not implied)} \end{eqnarray}

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