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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|$


marked as duplicate by user1551 linear-algebra Nov 10 '14 at 11:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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:

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


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

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

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

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