Inverting a Tensor - How to Find Out?

[tulip]

Thank you for reading.

If I have an object (is it correct to call it a tensor?) whose components are defined by:

$$X_{mikl}=(R^{-1})_{mi}R_{kl}-(S^{-1})_{mi}S_{kl},$$

where R and S are invertible matrices. I want to find the "inverse" of X, i.e. to find (X^{-1}) such that,

$$(X^{-1})_{qkpm}X_{mikl}=\delta_{lq}\delta_{pi}$$.

Is there a way to find out whether (X^{-1}) exists? A matrix doesn't have an inverse if its determinant is zero - is there a similar rule here? I've tried some trial functions for X^{-1} but nothing works, and I want to know whether there's a better way of tackling this problem.

 

[Fernsanz]

With no further constraints the tensor you have proposed is, in general, not invertible.

Consider for example the case $$R=S$$. Then $$X$$ will be the null tensor. Or even the less trivial case $$R=S^{-1}$$ will lead also to $$X=0$$ if $$R^2=1$$ (idempotent).

Therefore, in general, it is not invertible.

 

[tulip]

R is not equal to S, or to the inverse of S. It is clearly true that for special cases where X=0 there is no inverse, but I don't see how this tells us that X is generally non-invertible. Can you explain?

 

[Fernsanz]

If for the problem stated there are cases where X is not invertible (even if there is just one of such cases) then it can not exist a theorem, a result, or an algorithm which allows us to invert the given expression with generelity, i.e. regardless of the values of R and S, (i.e., operating with "letters" in full generality insted of with "numbers"). Hence, the invertibility or not of X will depend on the particular values of R and S, and so we say that "in general" X is not invertible.

So, the meaning of "in general X is not invertible" is not "X is non-invertible more often than not"; the meaning is actually "the invertibility of X has to be determined on a particular case basis, it can not be determined generally (no matter the values of R and S)"