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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?

Rodrigo de Azevedo
  • 23,223

1 Answers1

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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.

achille hui
  • 125,323
  • What result did you use to conclude that if the determinant of the lower right submatrix (of the last matrix) is positive, then the matrix is positive definite? –  Mar 29 '23 at 08:07
  • [+1] Very astute ! – Jean Marie Mar 29 '23 at 08:10
  • @JordanG Since it is a block real symmetric matrix. By Sylvester's criterion, it suffices to look at the leading principal minors of all the blocks. In our case, aside from the diagonal entries, there is one and only one determinant to check. – achille hui Mar 29 '23 at 08:19