I'm attempting to show that the matrix $\left[ \begin{array}{ccccc} 1 & -1 & 0 & \cdots & 0\\ 1 & 0 & -1 & \cdots & \vdots\\ 0 & 1 & \ddots &\ddots &0\\ \vdots & 0 & \ddots & 0 &-1\\ 0 &\cdots &0 & 1 &0\end{array} \right]$, the matrix with a 1 in the top diagonal position, 1s on the subdiagonal, and -1s on the superdiagonal, has only eigenvalues with positive real part. Initially, I thought about using similarity transformations and Sylvester's Law of Inertia, but then I realized that SLoI requires the matrix to be symmetric.
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$\begingroup$ I tried several attempts to exploit the "slight" pertubation from the Toeplitz version, but that $a_{11} = 1$ entry seems to have a profound influence on the eigenvalues, in case of the instances I calculated with so far. $\endgroup$– mvwMar 12, 2015 at 15:56
1 Answer
This is not an answer, but too long for a comment. The coefficients of the characteristic polynomials of the above matrix $A_n \in \mathbb R^{n\times n}$ for $n=2..10$ are as follows:
x^10 x^9 x^8 x^7 x^6 x^5 x^4 x^3 x^2 x 1
1 -1 1
1 -1 2 -1
1 -1 3 -2 1
1 -1 4 -3 3 -1
1 -1 5 -4 6 -3 1
1 -1 6 -5 10 -6 4 -1
1 -1 7 -6 15 -10 10 -4 1
1 -1 8 -7 21 -15 20 -10 5 -1
1 -1 9 -8 28 -21 35 -20 15 -5 1
Maybe this will help in finding a closed form for $\chi_{A_n}$ and thus aid in proving this.
EDIT: The coefficients vaguely resemble pascal's triangle. Specifically $$a_{n+k,n-k} = \binom nk$$ Where $a_{n,m}$ is the coefficient of $x^m$ in $\chi_{A_n}$. Furhermore the "odd" indices are just the negative binomial coefficients: $$a_{n+k+1,n-k} = -\binom nk$$ Index renaming leads to $$a_{n,m} = \binom{\frac{n+m}2}{\frac{n-m}2} \qquad n\equiv m \pmod2$$ and $$a_{n,m} = -\binom{\frac{n+m-1}2}{\frac{n-m-1}2} \qquad n\not\equiv m \pmod2$$
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$\begingroup$ The coefficients were something that I'd taken note of. For $n = 10$, the characteristic polynomial is $x^10 - C(9, 0)*x^9 + C(9, 1)*x^8 - C(8, 1)*x^7 + C(8, 2)*x^6 - C(7, 2)*x^5 + C(7, 3)*x^4 - C(6, 3)*x^3 + C(6, 4)*x^2 - C(5, 4)*x^1 + C(5, 5). However, attempting to show that the only valid combination of eigenvalues is requiring that they all have positive real part seems like it will be difficult for the general case. $\endgroup$ Mar 12, 2015 at 15:37
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$\begingroup$ @PistolPete Yeah, I was just editing the post to include a complete description of them (without proof). A closed form of $\chi_{A_n}$ can now be written down by messing with indices. $\endgroup$– AlexRMar 12, 2015 at 15:39