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Results of invertibility of a matrix involving the Szego kernel

Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=...
Christian Remling's user avatar
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Results of invertibility of a matrix involving the Szego kernel

Sorry for my previous answer. This is a partial answer for the $2\times 2$ case. Notice first that we can assume without loss of generality that $z_1=0$. Otherwise we can apply the Moebius ...
an_ordinary_mathematician's user avatar
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Determinant of Jacobian and directional derivatives

As you point out, there are some general matrix facts in here. Consider the eigendecomposition of a matrix: $A = S \Lambda S^{-1}$, where $S$ and $S^{-1}$ are orthonormal (implying their determinant ...
Juan Martinez's user avatar
6 votes

Maximum determinant of binary matrices with special properties

I'll give a partial answer to your Question #2. If you know $k$ and also how many entries are equal to $1$ in each column, you can actually compute the absolute value of the determinant exactly. By ...
Nathaniel Johnston's user avatar
2 votes

Maximum determinant of binary matrices with special properties

In case 1) a cheap bound is $|{\rm det}(A)|\le k^n$. That's because $A/k$ is a stochastic matrix and therefore has all its eigenvalues in the closed unit circle. See https://en.wikipedia.org/wiki/...
Abdelmalek Abdesselam's user avatar
2 votes

Testing for equal characteristic polynomials using a single determinant calculation

The coefficients of the characteristic polynomial are bounded by $n!$ so the coefficients of the difference are bounded by $2 n!$. The leading coefficient of the difference is integer, so its absolute ...
Command Master's user avatar

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