# Index notation for inverse matrices

I have a question: There is an standard way to write the inverse of a matrix in index notation?.

The reason is that I don't want to write $(A^{-1})_{ij}$ or $(A^{-1})_i^j$ or $(A^{-1})^{ij}$ using the exponent $^{-1}$.

I working with a one-covariariant one-contravariant tensor with components $A_i^j$ and I only need to write the expression $y^i (A^{-1})^j_i=x^j$ without the exponent $^{-1}$.

• If there isn't a standard notation, note that you're free to define $B=A^{-1}$ if you want to avoid repeating that exponent over and over again. – Semiclassical Aug 23 '14 at 14:20
• Funnily, no answer so far addresses the fact that you are dealing with tensors, thus the up/down distinction of indices is actually important. – Jean-Claude Arbaut Aug 25 '14 at 9:10

There are two standard ways to index an element in a matrix.

First. If you note the matrix with uppercase letters from the beginning of the english alphabet, then you can use the lowercase version of the letter while indexing. For example the matrix is $A$ and the element in the $i$-th row and $j$-th column is $a_{ij}$. Or for $B$ you use $b_{ij}$. Etc. More at wikipedia about it.

Second. If the matrix is a most difficult expression, for example the Hadamard product if $A$ and $B$ matrices, and you note the Hadamard product with $\circ$. Then I prefer $[A \circ B]_{ij}$ notation. Some author also use the $\sideset{_i}{_j}{[A \circ B]}$ notation. It is also works for inverses of course: $[A^{-1}]_{ij}$.

Each way work with comma between indexes. As @Semiclassical wrote that, it is also a solution, that if the expression is very complex define an alternate variable.

I have seen the following notation, for the $i,j$th entry of $A$ use $A_{i,j}$, then for the $i,j$th entry of $A^{-1}$ use $A^{i,j}$.

Instead of specifying $A$ and then working with its inverse $A^{-1}$, do the opposite. Say calculate $B^{-1} = ..$ and the index with $B_{ij}$.

I have done this often when I include the $\mbox{ }^{-1}$ in my matrix definition/calculation.

Example:

$$\Lambda^{-1} = s^\top T I^{-1}$$ $$\Lambda_{11} = m_1+m_2$$