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Consider two polynomials P and Q and their companion matrices. It seems that char polynomial of tensor product of said matrices would be a polynomial with roots that are all possible pairs product of roots of P,Q.

I guess its coefficients could be expressed through coefficients of P and Q. But I don't know the explicit formula and I cannot find it. I also failed to find it out myself -- I tried different approaches. Maybe it should be that characteristic polynomial, maybe resultant of some form, but..

I hope this is done by someone already.

I'll try to give an example $c_0 = a_0^nb_0^m$, or maybe $c_0 = (-1)^{n+m}a_0^nb_0^m$ ($m,n$ are the degrees of $P and Q$.

$c_1 = a_0^{n-1}a_1b_0^{m-1}b_1$, this is one is probably incorrect, but I obviously cannot provide the correct answer if I don't know it yet. It looks like $c_i$ is a sum of some sort.

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closed as off-topic by user44191, Steven Landsburg, David Handelman, Igor Khavkine, abx Jun 17 at 8:56

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    $\begingroup$ Funny, how this is discussed here mathoverflow.net/questions/188814/… (all this I know) but no explicit formula given( $\endgroup$ – Andrew S. Jun 16 at 20:42
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    $\begingroup$ I don't see what more you want than what's given in the answer: if $F(x)$ is the characteristic polynomial of $P$, and $G(x)$ is the characteristic polynomial of $G$, then the characteristic polynomial of the tensor product should be (if I haven't made a mistake) the resultant with respect to $y$ of $G(y), F(\frac{x}{y})$, multiplied by a power of $y$. $\endgroup$ – user44191 Jun 16 at 21:16
  • $\begingroup$ I want the answer of a form: Let's say a_i, b_j are coeffs of P and Q,and c_k are coeffs of what I want) The answer would look like c_k = f(a_i,b_j). I want the explicit f. $\endgroup$ – Andrew S. Jun 16 at 21:19
  • $\begingroup$ A resultant can be expressed as the determinant of the Sylvester matrix; the Sylvester matrix is explicit in the $a_i, b_j$, and the formula for a determinant is well-known. Does that answer the question? If not, I don't see what would; the formulae depend on the degrees of $P$ and $Q$. $\endgroup$ – user44191 Jun 16 at 23:24
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    $\begingroup$ Possible duplicate of Characteristic polynomial of Kronecker/tensor product $\endgroup$ – Igor Khavkine Jun 17 at 8:39