Inverse matrix notation question

[SamanthaYellow]

I'm hoping that you can help me settle an argument. For a matrix \textbf{M} with elements m_{ij}, is there any sitaution where the notation (M_{ij})^{-1} could be correctly interpreted as a matrix with elements 1/m_{ij}?

Personally I interpret (M_{ij})^{-1} in the usual sense of an inverse matrix, where it would have the property \textbf M \textbf M^{-1} = I, but perhaps there are other interpretations that I don't know about. Thanks!

 

[disregardthat]

The only interpretation I can think of is when you define matrix multiplication component-wise. In this case, the invertible matrices would be the ones with non-zero values, and the matrix you describe would be the inverse. Note that the identity matrix would be the one with 1's as values. I don't think this is commonly used.

 

[Fredrik]

I'm not a fan of the notation $(M_{ij})^{-1}$, mainly because $M_{ij}$ should refer to the component on row i, column j, not the matrix itself. I'm also not a fan of using a lowercase m for the components, because that prevents us from writing the definition of matrix multiplication in what I consider the obviously best way: $(AB)_{ij}=\sum_k A_{ik}B_{kj}$. I find it very puzzling that some authors go out of their way to avoid this notation, by writing things like "if $C=AB$, then $c_{ij}=\sum_k a_{ik}b_{kj}$".

If M is a diagonal matrix, for example
\begin{pmatrix}2 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 4\end{pmatrix} then its inverse is simply
\begin{pmatrix}\frac 1 2 & 0 & 0\\ 0 & \frac 1 3 & 0\\ 0 & 0 & \frac 1 4\end{pmatrix} But even if M is diagonal, and we use horrible notation, we still don't quite have $(M_{ij})^{-1}=1/m_{ij}$ because the off-diagonal elements of $M^{-1}$ aren't 1/0.

 

[SamanthaYellow]

This is helpful. The matrix in question isn't diagonal, and that's a good point about 1/0. Hopefully I can convince this other person to change their notation!