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I want to prove by virtue of Gershgorin's theorem that this matrix cannot have the numbers 0 and 4 as its eigenvalues: $$A= \begin{pmatrix} 2 & -1 & 0 & 0 &\ldots \\ -1 & 2 & -1 & 0 &\ldots \\ 0 & -1 & 2 & -1 & \ldots\\ \ldots\\ \end{pmatrix} $$

To this end, I've looked up this Gershgorin's theorem in the book by Ralston, but I do not follow the proof there, see the snippets below.

It would be enough for me to understand the solution of Exercise 30 in the section 9.6 the 2nd snippet below. The assumption for this is that $\det B=0$ but our matrix $A$ seems to be regular (I'm not sure about this). Moreover the solution on the page 477 introduces this notation which is never used in the sequel: $$x_t^{(i)}$$ and also $x_t$ which should be a solution of $Bx=0$ which again is not used any more.

Btw, how can I obtain from all this that $0$ is also not the eigenvalue of $A$ ?

Last but not least I do not follow why for $$B=A-\lambda I$$ holds this $$|a_{ii}-\lambda|\leq\sum_{k\neq i}a_{ik},$$ please see the first sinppet out of the total of three snippets, 30 (b).

1st SNIPPET

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2nd SNIPPET

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3rd SNIPPET

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user122424
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  • @MishaLavrov The matrix is naturally assumed to continue in the manner similar to its beginning so $[0 \ 0 \ldots 0\ 4]$ will never be the last row. I was unable to enter the continuation into it, but I should have mention it at least. – user122424 Nov 09 '22 at 18:05
  • @MishaLavrov Yes, of course! – user122424 Nov 09 '22 at 18:10
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    This seems to essentially be a duplicate of https://math.stackexchange.com/questions/3903369/prove-that-this-block-matrix-is-positive-definite/ . I believe your matrix $=B-2I$ from that link. Taussky's refinement of Gerschgorin Discs gives the desired result. – user8675309 Nov 09 '22 at 18:10
  • @user8675309 Your link is similar but too general for my purpose when I'm struggling with the very beginning of the machinery. – user122424 Nov 09 '22 at 18:13
  • In the case of this problem, we can also explicitly write down the quadratic forms $x^{\mathsf T}!Ax$ and $x^{\mathsf T}(4I-A)x$ as sums of squares and show that they are positive for all $x \ne 0$. – Misha Lavrov Nov 09 '22 at 18:13
  • @MishaLavrov In this comment you want to show that $A$ is non-singular only, right ? – user122424 Nov 09 '22 at 18:14
  • Yes, but proving that $A$ is positive definite is easier :) – Misha Lavrov Nov 09 '22 at 18:15
  • @MishaLavrov OK what about the rest of my OQ and how can I write down $x^{\top}Ax$ as positive-definite quadratic form for our matrix $A$ ? My purpose for my OQ is to fully understand at least one portion of the two books together with the notation there . – user122424 Nov 09 '22 at 18:16

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You're right that the matrix $A$ is regular. Exercice 30 does not apply.

An alternative approach (without Gerschgorin)
${A}$ is a tridiagonal $n×n$ matrix with constant diagonal and by-diagonals.
Let us refer to this article. With $a=2, b=c=-1,$ our matrix satisfies $A=D_n$ , see p.$130,$ and formula $(19)$ therein applies.
As $a^2=4bc,$ the determinant is $$\det A=n+1\neq 0,$$ therefore $0$ is not an eigenvalue.
Using $(19)$ for the matrix $B=A-4I$ we obtain $$\det B=(n+1)\left(-1\right)^n\neq 0.$$ This proves that $0$ is not an eigenvalue of $B,$ or equivalently, $4$ is not an eigenvalue of $A.$

Viera Čerňanová
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