# Eigenvalues of a tridiagonal matrix [duplicate]

I am currently studying this type of Jacobi matrix for all n

Example of the type of matrix $n=4$ $$\mathbf{X}=\left[\begin{array}{*{20}{c}} {0}&{1}&{0}&{0}\\ {1}&{0}&{1}&{0}\\ {0}&{1}& {0} &{1}\\ {0}&{0}&{1}&{0} \end{array}\right]$$

I am struggling to prove that all eigenvalues for all n will be; $\lambda \leq |2|$

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If all you need is to show that the spectral radius $\rho(X)$ is bounded above by 2, there are many ways to do it:

1. As pointed out in a user's comment to your question, this follows directly from Geršgorin disc theorem.
2. The result also follows immediately from the inequality $\rho(X)\le\|X\|$ if you take the induced 1-norm (i.e. the maximum absolute column sum norm $\|X\|=\max_{j}\sum_i|x_{ij}|$).
3. Another approach is to add 1 to the top right and bottom left corners of $X$, and call the resulting matrix $Y$. Since $X$ and $Y$ are irreducible nonnegative matrices and $X$ is entrywise bounded above by $Y$, we get $\rho(X)\le\rho(Y)$. However, as $\frac12Y$ is a stochastic matrix, we have $\rho(\frac12 Y)=1$.
4. Alternatively, in (3) above, as $Y$ is also a circulant matrix, you can express its eigenvalues explicitly: they are $\{2\operatorname{Re}(\omega^k): k=0,1,2,\ldots,n-1\}$, where $\omega$ is the $n$-th root of unity. Hence $\rho(Y)=2$.

A computational way:

$$\det (tI-X)=\begin{vmatrix}t&-1&0&0\\ -1&t&-1&0\\ 0&-1&t&-1\\ 0&0&-1&t\end{vmatrix}=t\begin{vmatrix}t&-1&0\\ -1&t&-1\\ 0&-1&t\end{vmatrix}+\begin{vmatrix}-1&0&0\\ -1&t&-1\\ 0&-1&t\end{vmatrix}=$$

$$=t\left(t^3-2t\right)+\left(-t^2+1\right)=t^4-3t^2+1$$

Using the roots formula, we get that the eigenvalues are:

$$\left|\pm\sqrt{\frac{3\pm\sqrt5}2}\right|\le 2$$

• sorry my mistake, i didnt make it clear, I am trying to prove that the eigenvalues are $\leq 2$ for all n, not just $n=4$ – mike Nov 9 '14 at 19:49
• @Zhang, ok: do you know something about tridiagonal Toeplitz matrices? In particular, about their eigenvalues? – Timbuc Nov 9 '14 at 20:08
• I would assume that their eigenvalues are always real?? not 100% sure though – mike Nov 9 '14 at 21:50
• Well, in this case they sure are as the matrix is also symmetric. – Timbuc Nov 9 '14 at 22:04
• how would i prove this, in this case? – mike Nov 9 '14 at 22:51