# Eigenvalues of an $n \times n$ symmetric tridiagonal matrix

How to show that an $$n \times n$$ tridiagonal matrix only has positive eigenvalues?

$$\begin{pmatrix} 70 & -35 & 0 & \dots & 0 \\\ -35 & 120 & -35 & \ddots & \vdots \\\ 0 & -35 & 120 & \ddots & 0 \\\ \vdots & \ddots & \ddots & \ddots & -35 \\\ 0 & ... & 0 & -35 & 15 \end{pmatrix}$$

I want to show that all the eigenvalues of the tridiagonal matrix above are positive. My first attempt is to show that this matrix is positive definite, but it seems that it is not a good solution. So I just find the characteristic equation of the matrix:

$$(\lambda - 70)(\lambda - 120)(\lambda - 120)...(\lambda - 120)\left((120)(15) - ({-35})^2 \right) + (-35)(-35)(\lambda - 120)(\lambda -120)...(\lambda - 120)\left((120)(15) - ({-35})^2 \right)=0$$

Any hint on continuing this and do you think it would give me the result that I want?

• Is $a_{nn}=15?$ What are the diagonal elements? Is it $\{70,120,120,\ldots,120,15\}$? Commented Mar 29, 2023 at 5:19
• May be this will help? math.stackexchange.com/questions/2776864/… Commented Mar 29, 2023 at 5:24
• Sorry, I edited my post
– user1165024
Commented Mar 29, 2023 at 5:24
• It's still not clear what's on the diagonal. Commented Mar 29, 2023 at 5:54
• The main diagonal is $\{70,120,120,...120,15\}$ and the lower and upper diagonal have entries equal to $-35$
– user1165024
Commented Mar 29, 2023 at 5:59

The $$n \times n$$ matrix can be rewritten as a sum of $$n-2$$ positive semi-definite matrices (consists of a single $$2\times 2$$ non-zero block $$\left[\begin{smallmatrix}35 & -35\\-35 & 35\end{smallmatrix}\right]$$) and a close to diagonal matrix (diagonal entries $$35,50,\ldots,50,85,15$$ with a single pair of non-diagonal entries of $$-35$$). $${\tiny\begin{bmatrix} 35 & -35 & 0 & 0 & \dots & 0 \\ -35 & 35 & 0 & 0 & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ddots & \vdots \\ \vdots & \ddots &\ddots & \ddots & \ddots & \vdots \\ 0 & \dots & \dots & \dots & \dots & 0 & \end{bmatrix}} + {\tiny\begin{bmatrix} 0 & 0 & 0 & 0 & \dots & 0 \\ 0 & 35 & -35 & 0 & \ddots & \vdots \\ 0 & -35 & 35 & 0 & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ddots & \vdots \\ \vdots & \ddots &\ddots & \ddots & \ddots & \vdots \\ 0 & \dots & \dots & \dots & \dots & 0 & \end{bmatrix}} + \dots + {\tiny\begin{bmatrix} 35 & 0 & 0 & \dots & 0 & 0 & 0\\ 0 & 50 & 0 & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 50 & \ddots & \vdots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots & \vdots\\ 0 & ... & \dots & 0 & 50 & 0 & 0 \\ 0 & ... & \dots & \dots & 0 & 85 & -35 \\ 0 & ... & \dots & \dots & 0 & -35 & 15 \end{bmatrix}}$$
Since $$85\cdot 15 > 35^2$$, the last matrix is positive definite. Being a sum of a positive definite matrix with a bunch of positive semi-definite matrices. The matrix at hand is positive definite.