I simply wouldn't, out of spite.

>>17023152
And that's what no one did.

probably using tensor products

>>17023160
>Boiling down a 3rd rank inverse operation into products
This is why no one respects linear algebra.

If x and y are tensors of rank M and N and the result is intended to be a tensor of rank M+N:
[math]
z[i,j,...,\mu,\nu,...] = {y[\mu,\nu,...]}^{\frac{1}{x[i,j,...]}}
[/math]

>>17023151
Just take the power series for y^(1/x) for x,y real. The summation index is powers of the matrix x. At that point it's well known how to define A^B for matrices A and B.

>>17023171
Shouldn't we expect the result to end up with a rank of M*N?

>>17023204
For complete elementwise operations, like taking an outer product between tensors, your final rank is the sum of the ranks of the constituent tensors.
[math]x_{i,j} \otimes y_{k,l,m} = z_{i,j,k,l,m}[/math]

Ex. Consider the 1x2 matrix X (rank 2 tensor) and the 2x2x2 array Y (rank 3 tensor) below:
X = ((0,1))
Y = (((2,3),(4,5)),((6,7),(8,9)))
Taking the outer product yields
Z = ((((0,0),(0,0)),((0,0),(0,0))),(((2,3),(4,5)),((6,7),(8,9))))
A rank 5 tensor (2*3) with 16 elements (1*2*2*2*2)

So in OP's case, if your tensor arrays are some arbitrary ranks M and N, the rank of the resulting tensor is M+N. The total number of elements depend on the lengths of the various indices.

If X is a 4x7x10 tensor (rank 3, 280 elements) and Y is a 2x4x20x3 tensor (rank 4. 480 elements), then the resultant Z is a 4x7x10x2x4x20x3 tensor (rank 7, 134,400 elements).

Now, if summation over indices is implied that's a whole different ballgame.

>>17023227
>A rank 5 tensor (2+3)
Fuck I'm too tired for this shit.

>>17023151
Using eigenvalues or something

>>17023151
Write the natural definition of [math]\sqrt[X]{Y}[/math]

[eqn]\qquad \sqrt[X]{Y} = Y^{X^{-1}} = e^{X^{-1}\log(Y)}[/eqn]

Define the formal power series
[eqn]
\begin{aligned}
\qquad e^{A} &= \sum_{k = 0}^\infty\frac{A^k}{k!} = I + A + \frac{A^2}{2} + \frac{A^3}{3!} + \frac{A^4}{4!} +\cdots\\
\qquad \log(I + A) &= \sum_{k=1}^\infty (-1)^k\frac{A^k}{k} = A - \frac{A^2}{2} + \frac{A^3}{3} - \frac{A^4}{4} + \cdots
\end{aligned}
[/eqn]

To find [math]\log(Y)[/math], set [math]I + A = Y[/math] and find [math]A = Y - I[/math]. Plug [math]A[/math] in the series for logarithm above.

Find the inverse matrix [math]X^{-1}[/math] then multiply it with [math]\log(Y)[/math] to get [math]Z = X^{-1}\log(Y)[/math].

Plug [math]Z[/math] into the exponential series and you're done.

Of course, both [math]X[/math] and [math]Y[/math] have to be invertible square matrices i.e. members of the general linear group [math]GL(n,\mathbb{C})[/math]

>>17023983
>only works for special case
pass

>>17024433
retard

>>17023983
nice

>>17023983
Agree; in general matrix log and exp and stuff are easily handled by diagonalisation. But what if [eqn]e^{X^{-1}\log(Y)} ≠ e^{\log(Y)X^{-1}}[/eqn]?

>>17026047
pretty sure you can use BCH formula for this case and you'll atleast get a lot of the analytic properties. you could even evaluate some nasty stuff if the limits converge for your particular group of elements!