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When we look at a matrix $A$ as a linear map, we write the element of the matrix as $a^{i}_{j}$ so the inverse matrix will be?

In the case of bilinear form for $a_{ij}$ the inverse is $a^{ij}$ or $a^{ji}$?

gbox
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    If the elements of $A$ are denoted as $a^{i}_j$, then the elements of the transpose $A^T$ are simply $a^{j}_i$. However, the elements of the matrix inverse $A^{-1}$ are in general not linear functions of the elements of $A$. – Semiclassical Jul 10 '17 at 16:56

1 Answers1

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In general, inversion has nothing to do with raising or lowering indices. E.g. see the inverse of the Ricci tensor here.

One exception to this is the metric tensor $g$, where: $g^{ij}=[g^{-1}]^{ij}$ from $g_{ij}$.

For $a_{ij}$, we have $a^{\alpha\beta} = g^{\alpha i}g^{\beta j}a_{ij}$. For a matrix to be an inverse, we need $M_{ik} [M^{-1}]^{kj} = \delta_j^i$. So $a^{ik}a_{kj}=g^{i \alpha}g^{k \beta}a_{\alpha\beta} a_{kj}$. Only in rather special cases would this equal $\delta_j^i$.

There is a nice answer for inverses in indicial notation here. It says (in $n$D): $$ [A^{-1}]_\nu^\eta= n\left[ \varepsilon^{i_1\ldots i_n} \varepsilon_{j_1\ldots j_n} A^{j_1}_{i_1} \ldots A^{j_n}_{i_n} \right]^{-1} \varepsilon^{\eta i_2\ldots i_n}\varepsilon_{\nu j_2\ldots j_n} A^{j_2}_{i_2} \ldots A^{j_n}_{i_n} $$

user3658307
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