Finding the eigenvalues and eigenvectors of tridiagonal matrix

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!

• Hint: Gershgorin's lemma. – TZakrevskiy Jan 5 '15 at 15:23
• 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 $1$. – user1551 Jan 5 '15 at 15:30
• @MathMan Why? Your inequality is simply wrong. – user1551 Jan 5 '15 at 15:36
• 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 $n$. – Algebraic Pavel Jan 5 '15 at 16:08

$$|\lambda_i-a|\leq |c_{i-1}|+|b_i|\leq \max_j |c_j|+\max_j |b_j|,$$
where $c_{i-1},b_i$ are zero outside their specification in $i$.
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: $$\max_j |c_j|+\max_j|b_j|\geq 2\sqrt{\max_j|b_j|\max_j|c_j|}$$
• 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 $2<1$. – Federico Poloni Jan 6 '15 at 13:33