<s>[INST] Diagonalization of a Toeplitz matrix 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?
 [/INST] 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 $v$.
</s>