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 $a^{ji}$?
[Math] Einstein Notation Of An Inverse Matrix
index-notation
Related Solutions
Mainly, the Kronecker delta makes sums collapse, making the two indexes equal everywhere else in the expression. For example: $$\delta_j^i \delta^j_i = \delta_i^i = n,$$and $$\delta^{\color{red}{a}}_{\color{blue}{b}}g_{c\color{red}{a}}g^{bd}\delta^{c}_{d} = g_{c\color{blue}{b}}g^{bd}\delta^c_d.$$I'll use colors again to ilustrate how this computation proceeds: $$g_{\color{red}{c}b}g^{bd}\delta_{\color{blue}{d}}^{\color{red}{c}} = g_{\color{blue}{d}b}g^{bd} \stackrel{(\ast)}{=} \delta_d^d = n,$$where in $(\ast)$ I used the definition of the inverse metric tensor.
I was able to answer the question for myself well enough. The answer is that yes, for a given tensor $R_{ijkl}$ It is possible to find another tensor $L_{ijmn}$ with the property that
$$ L_{ijmn}R_{ijkl} = \delta_{mk}\delta_{nl} $$
The point is that $L_{ijmn}$ has $N^4$ components and the equation above stands for $N^4$ equations where $N$ is the length of the indices. We should then be able to solve the above $N^4$ equations for the $N^4$ components of $L_{ijmn}$ to find the desired tensor.
I implemented some code in Mathematica to convince myself which I'll post here for those curious.. I don't know the best way to input code here so apologies if it doesn't look great.
R = Table[RandomReal[], {i, 1, 6}, {j, 1, 6}, {k, 1, 6}, {l, 1, 6}];
L = Table[a[i, j, k, l], {i, 1, 6}, {j, 1, 6}, {k, 1, 6}, {l, 1, 6}];
K = Table[
KroneckerDelta[m, k]*KroneckerDelta[n, l], {k, 1, 6}, {l, 1,
6}, {m, 1, 6}, {n, 1, 6}];
Flatten[
Table[
Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}] ==
K[[k, l, m, n]]
, {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}]];
s = Solve[Flatten[
Table[
Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}] ==
K[[k, l, m, n]]
, {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}]]];
Chop[Table[
Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}]
, {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}] /. s[[1]]]
The final line shows the result of multiplying the tensors $L$ and $R$ when the components of $L$ are replaced with the components found in the above line. The result is the double Kronecker delta tensor $K$ as desired.
Fortunately for my application the code can run in a few seconds for me (on my laptop) with tensors with $6^4 = 1296$ components.
I'm not sure what to call $L_{ijmn}$. I could call it $R^{-1}_{ijmn}$, but I can also pose another problem:
Find a tensor $G_{ikmj}$ with the property that
$$G_{ikmj}R_{ijkl} = \delta_{ml}$$
This new tensor could similarly be called $R^{-1}_{ikmj}$ and so could a number of other tensors with different combinations and permutations of indices. Therefore, for the time being I'll just continue to use unique names for these inverse tensors and explicitly state the relevant property in terms of the Kronecker deltas.
Best Answer
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): $$ [A^{-1}]_\nu^\eta= n\left[ \varepsilon^{i_1\ldots i_n} \varepsilon_{j_1\ldots j_n} A^{j_1}_{i_1} \ldots A^{j_n}_{i_n} \right]^{-1} \varepsilon^{\eta i_2\ldots i_n}\varepsilon_{\nu j_2\ldots j_n} A^{j_2}_{i_2} \ldots A^{j_n}_{i_n} $$