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Let $A$ be a tridiagonal matrix as below: $${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}} {a}&{b_1}&{}&{}&{}\\ {c_1}&{a}&{b_2}&{}&{}\\ {}&{c_2}&{\ddots}&{\ddots}&{}\\ {}&{}&{\ddots}&{\ddots}&{b_{n-1}}\\ {}&{}&{}&{c_{n-1}}&{a} \end{array}} \right]$$ I want to show that for any eigenvalue $\lambda$ of $A$: $$|\lambda-a|\leq 2\sqrt{\max\limits_{j}|b_j|\max\limits_{j}|c_j|}$$ I think, this link can help us!

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  • $\begingroup$ Hint: Gershgorin's lemma. $\endgroup$ – TZakrevskiy Jan 5 '15 at 15:23
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    $\begingroup$ This is not true. Consider $n=3$, $a=0$ and $b_i=c_i=1$ for all $i$. The eigenvalues of $A$ are $0$ and $\pm\sqrt{2}$, but your RHS is $1$. $\endgroup$ – user1551 Jan 5 '15 at 15:30
  • $\begingroup$ please see inserted link above in post. $\endgroup$ – SKMohammadi Jan 5 '15 at 15:32
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    $\begingroup$ @MathMan Why? Your inequality is simply wrong. $\endgroup$ – user1551 Jan 5 '15 at 15:36
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    $\begingroup$ In fact, if $a=0$ and $b_i=b$ and $c_i=c$ (such that $bc>0$) for all $i$, the eigenvalues are given by $2\sqrt{bc}\cos\left(\frac{i\pi}{n+1}\right)$ for $i=1,\ldots,n$. So if $n$ is odd, there's a zero eigenvalue and if $n$ is even, the smallest eigenvalue is given by $2\sqrt{bc}\cos\left(\frac{n\pi}{2(n+1)}\right)$ which can be made arbitrarily close to zero for sufficiently large $n$. $\endgroup$ – Algebraic Pavel Jan 5 '15 at 16:08
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The best you can do is invoke Gershgorin's theorem. This gives you that

$$|\lambda_i-a|\leq |c_{i-1}|+|b_i|\leq \max_j |c_j|+\max_j |b_j|,$$

where $c_{i-1},b_i$ are zero outside their specification in $i$.

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Post updated. So as Alex said, we can use Gershgorin's theorem. we have $$|\lambda_i-a|\leq |c_{i-1}|+|b_i|\leq \max_j |c_j|+\max_j |b_j|.$$ I think there is a relation with Inequality of arithmetic and geometric mean because: $$\max_j |c_j|+\max_j|b_j|\geq 2\sqrt{\max_j|b_j|\max_j|c_j|}$$

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  • $\begingroup$ There is a relation, but you can't combine those inequalities because they go in the wrong sense. To make an easier example, if you know that $2<3$ and $1<3$ you can't conclude that $2<1$. $\endgroup$ – Federico Poloni Jan 6 '15 at 13:33

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