2011年8月23日星期二

LaTeX+MathJax+Blogger test

Consider the $n\times n$ matrix
$A_n = \begin{pmatrix} 0&1&\ldots&1\\-1&0&\ddots&\vdots\\\vdots&\ddots&\ddots&1\\-1&\ldots&-1&0\end{pmatrix}.$
It can be shown that the characteristic polynomial of $A_n$ is given by $p(\lambda)=\det(\lambda I-A_n)=\frac12 \left((\lambda+1)^n+(\lambda-1)^n\right)$. So, when $n$ is even, we have $\det(A_n)=(-1)^np(0)=1$ and the eigenvalues of $A_n$ appear in reciprocal pairs, i.e. if $\lambda$ is an eigenvalue, so is $1/\lambda$.