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