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I am trying to calculate an upper bound for the determinant of a matrix of the form

$\begin{pmatrix}a_{n-1} & a_{n-2} & a_{n-3} & \cdots & a_{n-m+1} & a_{n-k+1}\\ a_{n} & a_{n-1} & a_{n-2} & \cdots & a_{n-m+2} & a_{n-k+2}\\ 0 & a_{n} & a_{n-1} & \cdots & a_{n-m+3} & a_{n-k+3}\\ \vdots & & \ddots & \ddots & & \vdots \\ 0 & 0 & \cdots & \cdots & a_n & a_{n-k+m}\end{pmatrix}$

which is an $m\times m$ upper-Hessenberg matrix, and a Toeplitz matrix but for a single column.

By Cramer's rule, the determinant of this matrix is the determinant of a Toeplitz-Hessenberg matrix (for which Trudi's formula gives a value) times one component of a solution of a linear system of equations. But I'm just not putting the pieces together. Please advise.

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  • $\begingroup$ Why do you suspect that there's a nice (easily computable) upper bound? What are you hoping to do with this upper bound? $\endgroup$ – Ben Grossmann Jan 24 at 19:29
  • $\begingroup$ One upper bound can be attained as follows: given any matrix $A$ and submultiplicative matrix-norm $\|\cdot\|$, we have $\det(A) \leq \|A\|^n$. $\endgroup$ – Ben Grossmann Jan 24 at 19:31
  • $\begingroup$ @Omnomnomnom Can't tell you, it's a state secret. However, have you any proof of that inequality you presented? $\endgroup$ – matty_k_walrus Jan 24 at 21:20
  • $\begingroup$ Yes, that's an easy one: such a norm is necessarily bigger than the largest absolute value of an eigenvalue. This is a pretty standard fact; you could use Horn and Johnson's Matrix Analysis as a reference for it, for instance. $\endgroup$ – Ben Grossmann Jan 25 at 3:04

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