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Let $0<\lambda\leq1$ so that the $n \times n$ matrix $$\Sigma = \begin{pmatrix} 1&1-\lambda& \cdots &1-\lambda\\ 1-\lambda&\ddots&\ddots& \vdots\\ \vdots &\ddots&\ddots&1-\lambda\\ 1-\lambda&\cdots&1-\lambda&1\\ \end{pmatrix}$$ is positive definite. I believe we can orthogonally diagonalize $\Sigma$ as $$\Sigma = VDV^T$$ where $$ V = \begin{pmatrix} \frac{-1}{\sqrt{2 \cdot 1}} & \frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\cdots&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&\cdots&0&\sqrt{\frac{n-1}{n}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&0&\sqrt{\frac{n-2}{n-1}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \vdots&\iddots&\iddots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\vdots&\vdots\\ 0&\iddots&\iddots&\vdots&\vdots&\vdots\\ \sqrt{\frac{1}{2}}&\frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \end{pmatrix}$$

$$D = \begin{pmatrix} \lambda&0& \cdots &0\\ 0&\ddots&\ddots& \vdots\\ \vdots &\ddots&\lambda&0\\ 0&\cdots&0&n-(n-1)\lambda\\ \end{pmatrix}$$

I am having some trouble showing this result, can someone offer a suggestion for the proof?

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migrated from mathoverflow.net Jun 25 '14 at 1:15

This question came from our site for professional mathematicians.

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The matrix $\Sigma$ can be written as $$ \Sigma=\lambda I+(1-\lambda)E, $$ where $E=ee^T$, $e=[1,1,\ldots,1]^T$. If $V$ is an orthogonal matrix such that $V^TEV=D$ is diagonal, then $$\tag{1} V^T\Sigma V=\lambda I+(1-\lambda)D $$ is the diagonalization of $\Sigma$.

The only thing which remains to find are the matrices $D$ and $V$. Since $E$ has rank one, it has one nonzero eigenvalue and $n-1$ zero eigenvalues and we can chose an orthonormal set of eigenvectors since $E$ is symmetric.

We have $$ Ee=ne, $$ hence $n$ is the only nonzero eigenvalue of $E$ with the constant eigenvector $v_1=e/\sqrt{n}$ (normalized to $\|v_1\|=1$). The other eigenvectors $v_2,\ldots,v_n$ of $E$ corresponding to the zero eigenvalues can be any set of orthonormal eigenvectors orthogonal to $e$. Indeed, if $0\neq v\perp e$, then $$ Ev=(ee^T)v=e(e^Tv)=0. $$ Since the eigenvalues of $\Sigma$ are $\lambda$ plus $(1-\lambda)$ times the eigenvalues of $E$, the eigenvalues of $\Sigma$ are $\lambda$ with the multiplicity $n-1$ and the simple eigenvalue $\lambda+(1-\lambda)n$.

Note that there are infinitely many bases of the eigenspace of $E$ corresponding to the zero eigenvalues. You can chose any $n-1$ orthonormal vectors orthogonal to a constant vector.

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