Eigenvalues of correlation matrix $\mathbf{\rho}$ with constants I'm tasked with finding the eigenvalues of $M = \begin{bmatrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{bmatrix}$. My difficulty here is in finding roots of the characteristic polynomial, since it's a cubic.
Attempt. Let's call the characteristic polynomail by $chp$, and let's use $x$ instead of $\lambda$ to have a more familiar-looking expression. We have
$$
\operatorname{chp}(x) = \det\left(\mathbf{\rho} - xI\right) = \det\begin{bmatrix} 1-x & \rho & \rho \\ \rho & 1-x & \rho \\ \rho & \rho & 1-x \end{bmatrix}
$$
Expanding on the first line, we obtain
$$
chp(x) = (1-x) \det \begin{bmatrix} 1-x & \rho \\ \rho & 1-x \end{bmatrix} - \rho \det \begin{bmatrix} \rho & \rho \\ \rho & 1-x \end{bmatrix} + \rho \det \begin{bmatrix} \rho & 1-x \\ \rho & \rho \end{bmatrix}.
$$
After simplification, you'd get
$$
chp(x) = 1 - 3\rho^2 + 2\rho^3 - 3x + 3\rho^2 x + 3x^2 - x^3.
$$
I couldn't solve $chp(x)=0$ analitically. Using Wolfram though, and coming back to the usual $\lambda$ notation, I got
$$
\lambda_1 = \lambda_2 = 1-\rho \qquad \text{and} \qquad \lambda_3 = 1+2\rho.
$$
Thanks for any help.
 A: Your question is actually no easier than the general case $n$, so I will treat the general case directly. Note that we could avoid evaluating the eigen-polynomial.
Let $e$ be the $n$-column vector consisting of all ones, then it is easy to verify that
\begin{align*}
A = (1 - \rho)I_{(n)} + \rho ee^T.
\end{align*}
By this answer, all $n$ eigenvalues of the matrix $ee^T$ are $n, 0, \ldots, 0$, which implies all eigenvalues of $A$ are
$(1 - \rho) + n\rho, 1 - \rho, \ldots, 1 - \rho$, where we used the fact that if $\lambda_1, \ldots, \lambda_n$ are all eigenvalues of the matrix $B$, and $f$ is a polynomial, then $f(\lambda_1), \ldots, f(\lambda_n)$ are all eigenvalues of the matrix $f(B)$.
You can verify that when $n = 3$, it matches the answer given by Wolfram.
A: This matrix is
\begin{align}
& (1+2\rho)\left[ \begin{array}{ccc} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{array} \right] + (1-\rho) \left[ \begin{array}{rrr} 2/3 & -1/3 & -1/3 \\ -1/3 & 2/3 & -1/3 \\ -1/3 & -1/3 & 2/3 \end{array} \right] \\[8pt]
={} & (1+2\rho)P+ (1-\rho) Q.
\end{align}
You can check that $P^2=P =P^T$ and $Q^2=Q=Q^T$ and $PQ=QP=0.$
Observe that if all components of a $3\times1$ column vector $\mathbf x$ are equal to each other, then $P\mathbf x=\mathbf x$ and $Q\mathbf x= \mathbf 0,$ and if the sum of the components of a column vector $\mathbf y$ is $0$ (which is the same as saying $\mathbf y\perp\mathbf x$) then $P\mathbf y=\mathbf 0$ and $Q\mathbf y = \mathbf y.$ That means the $2$-dimensional column space of $Q$ is an eigenspace with of $Q$ with eigenvalue $1$ and its $1$-dimensional orthogonal complement is an eigenspace of $Q$ with eigenvalue $0.$ But that same $2$-dimensional column space of $Q$ is an eigenspace of $P$ with eigenvalue $0$ and its orthogonal complement is an eigenspace of $P$ with eigenvalue $1.$
So we have an orthonormal basis with respect to which we have
$$
P = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \text{ and } Q = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]
$$
and therefore the matrix we started with becomes
$$
\left[ \begin{array}{ccc} 1+2\rho & 0 & 0 \\ 0 & 1-\rho & 0 \\ 0 & 0 & 1-\rho \end{array} \right].
$$
So there are the eigenvalues.
A: There's another way that is in keeping with your method, but it'll be a lot easier to find the eigenvalues of this matrix in particular.
I'm taking it up from here:

$$\operatorname{chp}(x) = \det\left(\mathbf{\rho} - xI\right) = \det\begin{bmatrix} 1-x & \rho & \rho \\ \rho & 1-x & \rho \\ \rho & \rho & 1-x \end{bmatrix}$$

Now, to find the determinant of the above matrix, you can make use of some elementary row and column transformations before expanding the matrix along a row(or a column) to make the calculations a lot more simpler.
$$\begin{align}
\det\begin{bmatrix}
1-x & \rho & \rho\\
\rho & 1-x & \rho\\
\rho & \rho & 1-x
\end{bmatrix}&=
\det\begin{bmatrix}
1+2\rho-x & \rho & \rho\\
1+2\rho-x & 1-x & \rho\\
1+2\rho-x & \rho & 1-x
\end{bmatrix}&C_1\rightarrow C_1+C_2+C_3\\
&=
\left(1+2\rho-x\right)\det\begin{bmatrix}
1 & \rho & \rho\\
1 & 1-x & \rho\\
1 & \rho & 1-x
\end{bmatrix}&\text{Taking $1+2\rho-x$ out from $C_1$}\\
&=
\left(1+2\rho-x\right)\det\begin{bmatrix}
1 & \rho & \rho\\
0 & 1-x-\rho & 0\\
0 & 0 & 1-x-\rho
\end{bmatrix}& \begin{array}{1}
R_2\rightarrow R_2-R_1\\
R_3\rightarrow R_3-R_1
\end{array}
\end{align}$$
Now expanding by $C_1$, we get,
$$\det\begin{bmatrix}
1-x & \rho & \rho\\
\rho & 1-x & \rho\\
\rho & \rho & 1-x
\end{bmatrix}=\left(1+2\rho-x\right)\left(1-x-\rho\right)^2$$
I hope you can take it up from here and find the eigenvalues easily now.
