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.

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

  1. This article was very useful for a paper I am writing for my thesis.

    Thanks

    Bernice Franklin
    UGG Purses
    UGG Bags
    Classic Tall Chestnut

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