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
$$ V = \begin{pmatrix}
\frac{-1}{\sqrt{2 \cdot 1}} & \frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\cdots&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
0&\cdots&\cdots&0&\sqrt{\frac{n-1}{n}}&\frac{1}{\sqrt{n}}\\
0&\cdots&0&\sqrt{\frac{n-2}{n-1}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
\vdots&\iddots&\iddots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\vdots&\vdots\\
0&\iddots&\iddots&\vdots&\vdots&\vdots\\
\sqrt{\frac{1}{2}}&\frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
\end{pmatrix}$$
$$D = \begin{pmatrix}
\lambda&0& \cdots &0\\
0&\ddots&\ddots& \vdots\\
\vdots &\ddots&\lambda&0\\
0&\cdots&0&n-(n-1)\lambda\\
\end{pmatrix}$$
I am having some trouble showing this result, can someone offer a suggestion for the proof?
 A: 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.
