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
2nd SNIPPET
3rd SNIPPET
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 $A.$