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How would you show that a given tridiagonal matrix is nonsingular?

I have a tridiagonal linear system and I would like to show it's nonsingular, which means zero cannot be an eigenvalue.

The book I am using suggests that one use the Gershgorin circle theorem

$$D_i = \{z:|z-a_{ii}|\} \le \sum_{j\ne i} a_{ij}$$

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  • $\begingroup$ Is the matrix also Toeplitz? $\endgroup$ Oct 30 '17 at 21:42
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Do you have a specific triagonal matrix in mind? If so, just compute the Gersgorin Circles, and make sure they don't intersect the origin of the complex plane. If your matrix is $n \times n$ you have $n$ circles. The center of the $j$-th circle is the diagonal entry of the $j-$th column. The radius of the $j$-th circle is the sum of the absolute values of the off-diagonal entries of the $j$-th column. Since your matrix is triagonal, this will be extremely easy to compute. You could also define Gersgorin Circles in terms of the rows of your matrix as well.

EDIT: I made a slight correct, the radius of the $j$-th circles comes from adding the absolute values of the off-diagonal entries of the $j$-th column.

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  • $\begingroup$ Here is a nice applet that lets you play around with Gersgorin Circles to get a feel for them: geogebra.org/m/wDEj3Xg9. Your eigenvalues must lie in each of the circles, and the center of the circles are the diagonal entries, and the radius of the circles come from the off-diagonal entries. $\endgroup$
    – amarney
    Oct 30 '17 at 21:18

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