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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. $\endgroup$ – mvw Mar 12 '15 at 15:56
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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. $\endgroup$ – Pistol Pete Mar 12 '15 at 15:37
  • $\begingroup$ @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. $\endgroup$ – AlexR Mar 12 '15 at 15:39

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