# Kronecker product

## Primary tabs

Synonym:
tensor product (for matrices), direct product
Type of Math Object:
Definition
Major Section:
Reference

## Mathematics Subject Classification

### invariants of the tensor product

The entry for "Kronecker Product" or alternatively "Tensor Product"
shows formulas for the trace, rank, and determinant of the product
in terms of those for its factors.

Are there corresponding formulas for the other invariants, and in
particular, can the characteristic equation of the product be
related to the characteristic equations of its factors?

At worst, I suppose they could be deduced by knowing all the roots.

- hvm

### Re: invariants of the tensor product

this is possible, if not all that illuminating. recall that the $k$th coefficient of the characteristic polynomial of A is $(-1)^k {\rm tr}(\wedge^k A)$. Thus, for $A\otimes B$ we get ${\rm tr}(\wedge^n A\otimes B)={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A\otimes B)+\cdots={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A){\rm tr}(B)+\cdots$.

### Re: invariants of the tensor product

Allright, why not try to make it more illuminating? Those
wedgies are determinants, the trace takes sums, and the final
form looks like a convolution. But I'm suspicious of anything
that starts off with something depending only on A; the
determinant of the tensor product doesn't look like that,
although the trace does. Call those wedgies, which are the
symmatric functions of the roots, sigma-k. Then Sigma-2 (for
the tensor product) would be sigma-2-A + sigma-1-A * sigma-1-B
+ sigma-2-B. Is that correct?

Is it possible to run this in Mathematica(TM) and get a human-