# Determining eigenvalues of a matrix

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.

• 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.
– mvw
Mar 12, 2015 at 15:56

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 $$a_{n,m} = -\binom{\frac{n+m-1}2}{\frac{n-m-1}2} \qquad n\not\equiv m \pmod2$$

• The coefficients were something that I'd taken note of. For $n = 10$, the characteristic polynomial is $x^10 - C(9, 0)*x^9 + C(9, 1)*x^8 - C(8, 1)*x^7 + C(8, 2)*x^6 - C(7, 2)*x^5 + C(7, 3)*x^4 - C(6, 3)*x^3 + C(6, 4)*x^2 - C(5, 4)*x^1 + C(5, 5). However, attempting to show that the only valid combination of eigenvalues is requiring that they all have positive real part seems like it will be difficult for the general case. Mar 12, 2015 at 15:37 • @PistolPete Yeah, I was just editing the post to include a complete description of them (without proof). A closed form of$\chi_{A_n}\$ can now be written down by messing with indices. Mar 12, 2015 at 15:39