Imagine to have a covariance matrix $2\times 2$ called $\Sigma^*$.
\begin{bmatrix}1+\sigma^2&\rho_{12}\\\rho_{21}&1+\sigma^2\end{bmatrix}
I know that $\rho_{12} = \rho_{21}$ because it's symmetric.
I proceed in calculating the eigenvalues from
$$\det(\Sigma^*-\lambda I) = \lambda^2 – 2\lambda(1+\sigma^2) + (1+\sigma^4+2\sigma^2-\rho_{12}^2).$$
and finally the eigenvalues are:
$\lambda_1 = 1+\sigma^2 + \rho_{12}$
$\lambda_2 = 1+\sigma^2 – \rho_{12}.$
Now I've tried to do the same with a $3 \times 3$ matrix
$$\Sigma^* = \begin{bmatrix}1+\sigma^2&\rho_{12}&\rho_{13}\\\rho_{21}&1+\sigma^2&\rho_{23}\\\rho_{31}&\rho_{32}&1+\sigma^2\end{bmatrix}.$$
I know that $\rho_{12} = \rho_{21}, \rho_{13} = \rho_{31}, \rho_{23} = \rho_{32}$ because it's symmetric.
Now I'm kinda stuck. I'll leave below what I did:
$$\det(Σ^*-\lambda I) = (1+\sigma^2 -\lambda)^3 – (1+\sigma^2 -\lambda)(\rho_{12}^2 + \rho_{13}^2 + \rho_{23}^2) + 2 \rho_{12}\rho_{13}\rho_{23}$$
and if $z = 1+\sigma^2 -\lambda$
$$\det(Σ^*-\lambda I) = z^3 – z(\rho_{12}^2 + \rho_{13}^2 + \rho_{23}^2) + 2 \rho_{12}\rho_{13}\rho_{23}.$$
How can I go further? Is it better to either find the values of z or to use another approach like SVD if applicable?
My main goal is to find those eigenvalues.
****** UPDATE ******
Following the answer below from @GNUSupporter 8964民主女神 地下教會, I see that since the discriminant is less or equal to zero, I have that
$u^3, v^3 = -\dfrac{q}{2} ± i\sqrt{-\Delta}$
their module is $R = \sqrt{(-\frac{q}{2})^2+(\sqrt{-\Delta})^2}$
Do I need $R$? If yes, I need to calculate $\theta = arctg (-\frac{\sqrt{-\Delta}}{-\frac{q}{2}})$
If not, is it possible to just say that
$$x = u + v = \sqrt[3]{-\dfrac{q}{2} + i\sqrt{-\Delta}} + \sqrt[3]{-\dfrac{q}{2} – i\sqrt{-\Delta}}?$$
Best Answer
To find $\Sigma^*$'s eigenvalues, apply Cardano's method with $p = -(\rho_{12}^2 + \rho_{23}^2 + \rho_{31}^2)$ and $q = 2 \rho_{12}\rho_{23}\rho_{31}$ to your last equality
We introduce variables $u$ and $v$ so that $z = u + v$
\begin{align} u^{3} &= -{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}} = -\rho_{12}\rho_{23}\rho_{31} + \sqrt{(\rho_{12}\rho_{23}\rho_{31})^2 - \frac{(\rho_{12}^2 + \rho_{23}^2 + \rho_{31}^2)^3}{27}} \tag2 \label2 \\ v^{3} &= -{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}} = -\rho_{12}\rho_{23}\rho_{31} - \sqrt{(\rho_{12}\rho_{23}\rho_{31})^2 - \frac{(\rho_{12}^2 + \rho_{23}^2 + \rho_{31}^2)^3}{27}}. \tag3 \label3 \end{align}
Note that by the a.m.-g.m.-inequality,
$$\sqrt[3]{\rho_{12}^2 \rho_{23}^2 \rho_{31}^2} \le \frac{\rho_{12}^2 + \rho_{23}^2 + \rho_{31}^2}{3},$$
so the discriminant $$\Delta = {q^{2} \over 4}+{p^{3} \over 27} \le 0.$$
$\Delta = 0$ iff equality holds in the AM-GM inequality iff $\rho_{12}^2 = \rho_{23}^2 = \rho_{31}^2$.
Observe from \eqref{2} and \eqref{3} that $u^3 = \overline{v^3}$. Denote $\omega$ the third root of unity not equal to $1$. (i.e. $\omega^3 = 1$ but $\omega \ne 1$.)
Since all eigenvalues of a real symmetric matrix are real, you just take $u + \overline{u}$, $\omega u + \overline{\omega u}$ and $\omega^2 u + \overline{\omega^2 u}$ as roots for \eqref{1}, where $u$ is fixed as any one of the three roots of \eqref{2}.
To finish this question, just do the substitution $\lambda = 1+\sigma^2 -z$.
Edit in response to OP's comment
Despite OP's edit to include some trigonometric function, in this section, I shall make no use of them since this is essentially an algebra problem for finding roots of cubic-equations, and that trigonometric functions are transcendental. I'll keep my existing working algebraic at the first place. By algebra, I don't mean to be arithmetic, that is, I avoid as much calculations as possible.
Each of equations \eqref{2} and \eqref{3} have three distinct roots, so there are at most $3 \times 3 = 9$ candidates for $z = u + v$.
The reason for \eqref{2} and \eqref{3} to have three distinct roots is that the cubic equation $\omega^3 = 1$ has three distinct roots: $1$ and the two conjugate complex roots $\omega$ and $\bar{\omega} = \omega ^2$, so that $\omega^2 + \omega + 1 = 0$. If you fix a root $u$, then $\omega u$ and $\omega^2 u$ are also roots to \eqref{2} since $(\omega u)^3 = \omega^3 u^3 = u^3$ and $(\omega^2 u)^3 = (\omega^3)^2 u^3 = u^3$. Since we want to represent all three roots of \eqref{2} as $u, \omega u, \omega^2 u$, we don't want $\omega = 1$.
To visualize $u$, you may consider it as a nonzero arrow on the complex plane, and $\omega$ as a rotation of 120°.
Third roots of unity
Image source: Wikimedia commons
Given that \eqref{2} is solved and that $u^3 = \overline{v^3}$, the easiest root for \eqref{3} would be $v = \bar{u}$. Apply the same argument above to see that roots of \eqref{3} are $\bar{u}, \omega \bar{u}, \omega^2 \bar{u}$.
The remaining work is too select the three real values from $\omega^i u + \omega^j \bar{u}$, $i,j \in \{0,1,2\}$. The easiest choice is the case when $i = j = 0$, so that $z = u + \bar{u} \in \Bbb{R}$. It should be quite obvious that we should choose $\omega u + \overline{\omega u}$ and $\omega^2 u + \overline{\omega^2 u}$. These three roots are distinct since rotation $\omega \ne 1$. Since \eqref{1} is at most three roots, we're done.
To conclude, to solve for $z$, fix any root $u$ in \eqref{2}, then the first root is $2\mathrm{Re}(u)$. Rotate $u$ by $120°$ (anti-)clockwisely to get another root $2\mathrm{Re}(\omega u)$, repeat this to get the last root $2\mathrm{Re}(\omega^2 u)$.
Remark: I used "third root of unity" in the first version of this answer since it's easier to type in words than its explicit expression.
Edit in response to updated question body
As shown by the previous section, we can solve the problem without introducing the modulus $$R = \sqrt{(-\frac{q}{2})^2+(\sqrt{-\Delta})^2}$$ and $$\theta = \arctan\frac{-\sqrt{-\Delta}}{-q/2}$$ of $u^3$ (or $v^3$).
Nonetheless, $R$ and $\theta$ provide a more compact expression for $u^3$ and $v^3$: $v^3 = Re^{i\theta}$ and $u^3 = Re^{-i\theta}$. Therefore, $u = \sqrt[3]{R} e^{-i(\theta + 2k\pi)/3}$ and $v = \sqrt[3]{R} e^{i(\theta + 2m\pi)/3}$, $k,m \in \{0,1,2\}$. The checking for $u + v \in \Bbb{R}$ is similar to the previous section, and I left this as exercise.
To litterally respond to your question, it depends on your personal choice. Nobody can answer that for you.
First, I suppose you mean $z = u + v$, since $x$ has never been used before. Second, this notation for cubic root $\sqrt[3]{\cdots}$ is ambiguous unless $\Delta = 0$. The notation $\sqrt[3]{\cdots}$ can mean any one of the tree cubic root unless the number inside the radical sign is real.
This ambiguity would give rise to at most $3 \times 3 = 9$ possible candidates of $z$, but a cubic equation admits exactly three roots in $\Bbb{C}$. Therefore, an additional condition has to be added.
This condition comes from Cardano's trick $z = u + v$, which transforms the depressed cubic equation $z^3 + pz + q = 0$ into $$\begin{aligned} (u+v)^3 + p(u+v) + q &= u^3 + v^3 + 3uv(u+v) + p(u+v) + q \\ &= u^3 + v^3 + (3uv + p)(u+v) + q = 0 \end{aligned}$$
Set $3uv + p = 0$. Since a new variable is introduced, we can add one more constraint. This is where the $uv = -\frac{p}{3}$ from. Then the above equation becomes $u^3 + v^3 = -q$. To solve for $u$ and $v$, we make use of the well-known quadratic formula on $u^3$ and $v^3$ to get \eqref{2} and \eqref{3}.