# LaTeX and Wordpress (in the wild)

Apparently, you can use LaTeX in wordpress. Alright, he is a practice problem I was working on for my exam in two weeks.

Let $\Sigma=(1-\rho)I_{k}+\rho J_{k}$ with $0 \le \rho \le 1$ where $I_{k}$ is a $k x k$ identity matrix and $J$ is a $k x k$ matrix of ones. Find the eigenvalues of $\Sigma$.

We seek $\lambda$ such that $det(\Sigma - \lambda I)=0$ where $det(A)$ is the determinant of a matrix $A$

$det((1-\rho)I_{k}+\rho J_{k}-\lambda I_{k})=0$
$det((1-\lambda-\rho)I_{k}+\rho J_{k})=0$
$det(\rho(J_{k}+\frac{1-\rho-\lambda}{\rho}I_{k}))=0$
$det(J_{k}+\frac{1-\rho-\lambda}{\rho}I_{k}))=0$
$det(J_{k}-(\frac{-(1-\rho-\lambda)}{\rho}I_{k}))=0$
$det(k(\frac{J_{k}}{k}-(\frac{-(1-\rho-\lambda)}{\rho k}I_{k})))=0$
$det(\frac{J_{k}}{k}-(\frac{-(1-\rho-\lambda)}{\rho k}I_{k}))=0$

Now this is merely the equation for determining the eigenvalues of $\frac{J_{k}}{k}$. Since, $\frac{J_{k}}{k}$ is idempotent the eigenvalues of $\frac{J_{k}}{k}$ must be either zero or one. In fact, since this matrix has rank one, $\frac{J_{k}}{k}$ has eigenvalues one with multiplicity one and zero with multiplicity $(k-1)$. Therefore the eigenvalues of $\Sigma$ can be found by setting $\frac{-(1-\rho-\lambda)}{\rho k}=0$ and $\frac{-(1-\rho-\lambda)}{\rho k}=1$. This yields $\lambda=1+(k-1)\rho$ with multiplicity $1$ and $(1-\rho)$ with multiplicity $(k-1)$ which are exactly the eigenvalues of $\Sigma$.

Cheers.

Posted on February 9, 2010, in Uncategorized. Bookmark the permalink. 1 Comment.