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How can one show that the following tridiagonal matrix

$$M_n= \begin{pmatrix} -1&3&0&\dots&\dots&\dots&0\\ 3&2&-1&0&&&\vdots\\ 0&-1&2&-1&\ddots&&\vdots\\ \vdots&0&-1&2&\ddots&0&\vdots\\ \vdots&&\ddots&\ddots&\ddots&-1&0\\ \vdots&&&0&-1&2&-1\\ 0&\dots&\dots&\dots&0&-1&2 \end{pmatrix}$$

has exactly one negative eigenvalue?

The initial problem was showing it has at least some number of real positive eigenvalues (not necessarily distinct) for sufficiently large $n$, which can be solved by using Gershgorin circle theorem:

Which implies $-4\le \lambda\le6$ if I'm not mistaken, since $M_n$ has real eigenvalues (it's hermitian),

Then since the trace $\operatorname{Tr}M_n =2n-3$ of a matrix is also the sum of its eigenvalues, we must have at least $\frac{2n-3}{6}$ real positive eigenvalues, otherwise $\operatorname{Tr}M_n < 2n-3$ would be a contradiction.


But I noticed $M_n$ seems to have exactly one negative real eigenvalue, and the rest positive real eigenvalues, for every $n$.

How can we prove this strict bound?

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  • $\begingroup$ seems a pretty good, provable, pattern in solving $P^T MP = D$ becoming diagonal, with $\det P = 1.$ By Sylvester's Law of Inertia, there is one negative eigenvalue, else positive $\endgroup$ – Will Jagy Jun 7 at 0:49
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THREE:

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ \frac{ 3 }{ 11 } & \frac{ 1 }{ 11 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} - 1 & 3 & 0 \\ 3 & 2 & - 1 \\ 0 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & \frac{ 3 }{ 11 } \\ 0 & 1 & \frac{ 1 }{ 11 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} - 1 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ 0 & - \frac{ 1 }{ 11 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} - 1 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & - \frac{ 1 }{ 11 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} - 1 & 3 & 0 \\ 3 & 2 & - 1 \\ 0 & - 1 & 2 \\ \end{array} \right) $$

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FOUR:

$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ \frac{ 3 }{ 11 } & \frac{ 1 }{ 11 } & 1 & 0 \\ \frac{ 1 }{ 7 } & \frac{ 1 }{ 21 } & \frac{ 11 }{ 21 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} - 1 & 3 & 0 & 0 \\ 3 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 3 & \frac{ 3 }{ 11 } & \frac{ 1 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 11 } & \frac{ 1 }{ 21 } \\ 0 & 0 & 1 & \frac{ 11 }{ 21 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} - 1 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 3 & 1 & 0 & 0 \\ 0 & - \frac{ 1 }{ 11 } & 1 & 0 \\ 0 & 0 & - \frac{ 11 }{ 21 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} - 1 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - 3 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 11 } & 0 \\ 0 & 0 & 1 & - \frac{ 11 }{ 21 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} - 1 & 3 & 0 & 0 \\ 3 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 2 \\ \end{array} \right) $$

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FIVE:

$$ P^T H P = D $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 11 } & \frac{ 1 }{ 11 } & 1 & 0 & 0 \\ \frac{ 1 }{ 7 } & \frac{ 1 }{ 21 } & \frac{ 11 }{ 21 } & 1 & 0 \\ \frac{ 3 }{ 31 } & \frac{ 1 }{ 31 } & \frac{ 11 }{ 31 } & \frac{ 21 }{ 31 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} - 1 & 3 & 0 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 3 & \frac{ 3 }{ 11 } & \frac{ 1 }{ 7 } & \frac{ 3 }{ 31 } \\ 0 & 1 & \frac{ 1 }{ 11 } & \frac{ 1 }{ 21 } & \frac{ 1 }{ 31 } \\ 0 & 0 & 1 & \frac{ 11 }{ 21 } & \frac{ 11 }{ 31 } \\ 0 & 0 & 0 & 1 & \frac{ 21 }{ 31 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} - 1 & 0 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } & 0 \\ 0 & 0 & 0 & 0 & \frac{ 41 }{ 31 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ - 3 & 1 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 11 } & 1 & 0 & 0 \\ 0 & 0 & - \frac{ 11 }{ 21 } & 1 & 0 \\ 0 & 0 & 0 & - \frac{ 21 }{ 31 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} - 1 & 0 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } & 0 \\ 0 & 0 & 0 & 0 & \frac{ 41 }{ 31 } \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & - 3 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 11 } & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 11 }{ 21 } & 0 \\ 0 & 0 & 0 & 1 & - \frac{ 21 }{ 31 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} - 1 & 3 & 0 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & - 1 & 2 \\ \end{array} \right) $$

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SIX:

$$ P^T H P = D $$ $$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 & 0 & 0 \\ \frac{ 3 }{ 11 } & \frac{ 1 }{ 11 } & 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 7 } & \frac{ 1 }{ 21 } & \frac{ 11 }{ 21 } & 1 & 0 & 0 \\ \frac{ 3 }{ 31 } & \frac{ 1 }{ 31 } & \frac{ 11 }{ 31 } & \frac{ 21 }{ 31 } & 1 & 0 \\ \frac{ 3 }{ 41 } & \frac{ 1 }{ 41 } & \frac{ 11 }{ 41 } & \frac{ 21 }{ 41 } & \frac{ 31 }{ 41 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} - 1 & 3 & 0 & 0 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & 0 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 3 & \frac{ 3 }{ 11 } & \frac{ 1 }{ 7 } & \frac{ 3 }{ 31 } & \frac{ 3 }{ 41 } \\ 0 & 1 & \frac{ 1 }{ 11 } & \frac{ 1 }{ 21 } & \frac{ 1 }{ 31 } & \frac{ 1 }{ 41 } \\ 0 & 0 & 1 & \frac{ 11 }{ 21 } & \frac{ 11 }{ 31 } & \frac{ 11 }{ 41 } \\ 0 & 0 & 0 & 1 & \frac{ 21 }{ 31 } & \frac{ 21 }{ 41 } \\ 0 & 0 & 0 & 0 & 1 & \frac{ 31 }{ 41 } \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 41 }{ 31 } & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{ 51 }{ 41 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ - 3 & 1 & 0 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 11 } & 1 & 0 & 0 & 0 \\ 0 & 0 & - \frac{ 11 }{ 21 } & 1 & 0 & 0 \\ 0 & 0 & 0 & - \frac{ 21 }{ 31 } & 1 & 0 \\ 0 & 0 & 0 & 0 & - \frac{ 31 }{ 41 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 21 }{ 11 } & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{ 31 }{ 21 } & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 41 }{ 31 } & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{ 51 }{ 41 } \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & - 3 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 11 } & 0 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 11 }{ 21 } & 0 & 0 \\ 0 & 0 & 0 & 1 & - \frac{ 21 }{ 31 } & 0 \\ 0 & 0 & 0 & 0 & 1 & - \frac{ 31 }{ 41 } \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} - 1 & 3 & 0 & 0 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & 0 & - 1 & 2 \\ \end{array} \right) $$

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For convenience, let us drop the subscript in $M_n$. The trailing principal $(n-1)\times(n-1)$ submatrix $P$ of $M$ is a symmetric tridiagonal Toeplitz matrix whose diagonal entries are $a=2$ and whose subdiagonal entries are $b=-1$. The eigenvalues of $P$ are therefore $$ \lambda_k(P)=a+2b\cos\left(\frac{k\pi}{n}\right)=2\left(1-\cos\left(\frac{k\pi}{n}\right)\right) $$ for $k=1,2,\ldots,n-1$. Hence $\lambda_k(P)>0$ for every $k$ and $P$ is positive definite. However, by Cauchy's interlacing inequality for bordered matrix, $$ \lambda_1(M)\le\lambda_1(P)\le\lambda_2(M)\le\lambda_2(P)\le \cdots\le\lambda_{n-1}(M)\le\lambda_{n-1}(P)\le\lambda_n(M). $$ Thus $M_n$ has at most one non-positive eigenvalue. This eigenvalue must be negative, otherwise $M$ would be positive semidefinite, which is a contradiction to the fact that the first entry of $M$ is negative.

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