## New answers tagged determinants

0

Hancock seems to do not such a good job with the translation of Minkowski's work. I've posted the original here. See for example the double occurrence of $a_{mm}$ in the last inequality above versus what's written in the English version.

5

This is more of an extended comment. It is in fact possible to guess formulas for larger $k$ using the FriCAS guessing package.
At least for $k=3$ I obtain something reasonable:
$$
q^{\frac{n(n-1)(4n+13)}{6}}
\frac{(q^{n+3}+q^{n+1}-q-1)(q^{n+2}-1)(q^{n+1}-1)}
{(q;q)_3}
$$
The trick is to remove the power of $q$ first. Explicitly (I used FriCAS from within ...

2

If the $n\times n$ matrix $M$ is decomposed into submatrices,
$$M=\begin{pmatrix}A&B\\ C&D\end{pmatrix},$$
where $A$ has dimension $m\times m$, then the determinant of $M$ can be decomposed as
$$\det M=\det A\det D+X.$$
The multinomial $X$ in the matrix elements of $M$ contains $n!-m!(n-m)!$ terms, for a general matrix $M$. If the matrix is symmetric,...

answered Jan 5 at 17:47

Carlo Beenakker

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