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Is it possible to take a tensor to the other side of the equation, and the tensor becomes its inverse(i.e contravariant becomes covariant and vice versa)? It is a stupid question, but It confuses me.

For example, if

$A_{ij} = B_{ij},$

Can I write

$A_{ij}B^{ij} = \delta_{j}^{i}$, (though I think it should be $A^2$)

Or is it only valid for the metric tensor?

Also, is there a difference in the matrix representations of a tensor's contravariant and covariant form, or only the transformation rules differ?

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Your equation seems a bit off since you are summing over both $i$ & $j$ subscripts and the result would be a scalar and not the Kronecker delta. – Constandinos Damalas May 25 '14 at 10:53
I agree @PhotonicBoom, It should be the delta with indices summed over – GRrocks May 25 '14 at 10:57

Note that the matrix equation is: \begin{equation} A = B \implies A B^{-1} = I \end{equation} Taking the components of the above equation gives: \begin{equation} A_{ij} (B^{-1})_{jk} = \delta_{ik} \end{equation} In case it is a tensor rather than a matrix, then you would write: \begin{equation} A_{ij} (B^{-1})^{jk} = \delta_i^k \end{equation} What you have written is only valid for the metric tensor because it can be used to raise/lower indices.

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So we have to first find $B^{-1}$ ? – GRrocks May 25 '14 at 10:46
@GRrocks yeah, like I've done above. – Hunter May 25 '14 at 10:48
THANX @Hunter.. – GRrocks May 25 '14 at 10:49

Any non-degenerate bilinear form $g$ induces an isomorphism between a vector space and its dual. In components, this is done by lowering an index via contraction. The inverse map then raises an index by contraction with the inverse of the matrix representing our bilinear form.

Assuming $g$ is symmetric because I'm too lazy to figure out the index placements, we then have $$ g^{ij}\equiv (g^{-1})^{ki}(g^{-1})^{lj}g_{kl}=(g^{-1})^{ki}\delta^j_k=(g^{-1})^{ij} $$ So $g^{ij}=(g^{-1})^{ij}$ is actually a theorem and holds for the bilinear form that induces the isomorphism, but not in general as can be easily seen via counterexamples (eg using diagonal matrices for simplicity).

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