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