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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 $$ V = \begin{pmatrix} \frac{-1}{\sqrt{2 \cdot 1}} & \frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\cdots&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&\cdots&0&\sqrt{\frac{n-1}{n}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&0&\sqrt{\frac{n-2}{n-1}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \vdots&\cdots&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\vdots&\vdots\\ 0&\cdots&\cdots&\vdots&\vdots&\vdots\\ \sqrt{\frac{1}{2}}&\frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \end{pmatrix}$$

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

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

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We can rewrite $\Sigma$ as $(1 - \lambda)M + \lambda I$, where $M$ is given by $$ M = \pmatrix{ 1&\cdots & 1\\ \vdots &\ddots&\vdots\\ 1&\cdots &1 } $$ It follows that $V^T\Sigma V$ will be diagonal iff $V^T M V$ will be diagonal. And so, it suffices to show that $M = VDV^T$ with $$ D = \pmatrix{ 0&&&\\\ &\ddots&&\\ &&0&\\ &&&n } $$ Which is to say that $MV = VD$.

In fact, all we need to say in order to show this is the case is that $M$ is symmetric with rank $1$. From there, it suffices to show that $v$, the last column of $V$, is length one and satisfies $Mv = n\cdot v$, and that all other columns are length one, mutually orthogonal, and orthogonal to $v$.

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