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I would like to check if the determinant of the following matrix is nonzero.

\begin{bmatrix} 1 & -2 & 1 & & & & & \\ 1 & 4 & 1 & & & & & \\ & 1 & 4 & 1 & & & & \\ & & 1 & 4 & \ddots & & & \\ & & & \ddots & \ddots & \ddots \\ & & & & \ddots & 4 & 1 & & \\ & & & & & 1 & 4 & 1 & \\ & & & & & & 1 & 4 & 1 \\ & & & & & & 1 & -2 & 1 \end{bmatrix}

I tried using Laplace expansion but I think this leads to nothing. Is there another way to see regularity of such a "tridiagonal" matrix. Unfortunately, the matrix is not strictly diagonal dominant either. That is why I tried using Laplace but the submatrices I get from doing that are looking almost the same. Maybe one has another idea.

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  • $\begingroup$ Where does this quasi-tridiagonal matrix come from? A book? $\endgroup$ Commented 9 hours ago

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I assume your $n\times n$ matrix, $M$, is sufficiently large -- in particular $n\geq 7$. You can directly calculate the determinant of $M$ when $n\leq 6$.

$\Sigma:=\begin{bmatrix} B & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & I_{n-6} & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & C \end{bmatrix}$ with

$B:=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \\ \end{bmatrix}$ and $C:=\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

$\implies \frac{1}{3^{4}}\cdot \det\big(M\big)=\det\big(M\Sigma\big)\neq 0$
where the right hand side holds by Taussky's refinement of Gerschgorin Discs since $(M\Sigma)$ is irreducible, weakly diagonally dominant and the dominance is strict in at least one row [row 4]. I gave a proof of Taussky's refinement under "Optional Second" here: Prove that this block matrix is positive definite

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  • $\begingroup$ This is pretty cool to be honest, but surley I'm not allowed to use this. $\endgroup$ Commented 9 hours ago
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    $\begingroup$ As is your question is "missing context" since it tells us nothing about where this problem came from or what constraints you have on techniques involved. Since you evidently have unstated constraints but have indicated strict diagonal dominance is ok, consider modifying the above: e.g. do a couple row operations on $M$ so the 1st and last rows have three 2's [and rest all zero] use $B:=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2+\epsilon} & 0 \\ 0 & 0 & \frac{1}{2} \\ \end{bmatrix}$ for small $\epsilon\gt 0$ and similar for $C$. Then $M\Sigma$ would be strictly diagonally dominant. $\endgroup$ Commented 8 hours ago

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