<s>[INST] 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.
 [/INST] 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.
</s>