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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?

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

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user1165024
    Commented Mar 29, 2023 at 8:07
  • $\begingroup$ [+1] Very astute ! $\endgroup$
    – Jean Marie
    Commented Mar 29, 2023 at 8:10
  • $\begingroup$ @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. $\endgroup$ Commented Mar 29, 2023 at 8:19

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