Consider a general $su(3)$ matrix in usual Gell-Mann basis $\lambda = l_a \lambda_a$, with $a=1,...,8$. Are there any resources studying the eigenvalues /eigenvectors of $$\lambda^2 = \frac{2}{3} l^2 I + d^{abc}l_a l_b \lambda_c \quad ?$$ For one, I have computed the eigenvalues (which should be real, given $\lambda$ is hermitian) with the help of Mathematica and certain $su(3)$ structure constant identities to be $$ \lambda_1 = \frac{2}{3} l^2 - \frac{1}{3}R^{1/3}- \frac{1}{3}l^4R^{-1/3}\\ \lambda_2 = \frac{2}{3} l^2 + \frac{1}{6}(1-i \sqrt{3})R^{1/3}+ \frac{1}{6\sqrt{3}}(3 i + \sqrt{3})l^4R^{-1/3}\\ \lambda_3 = \frac{2}{3} l^2 + \frac{1}{6}(1+i \sqrt{3})R^{1/3}+ \frac{1}{6\sqrt{3}}(-3 i + \sqrt{3})l^4 R^{-1/3}, $$ where $$R=-6(d^{abc}l_al_bl_c)^2 +l^6+2|d^{abc}l_al_bl_c|\sqrt{3(d^{abc}l_al_bl_c)^2-l^6}$$ But I'm not entirely sure I can trust these or if there are simpler, less cryptic ways of writing them. I have learned from eq. 3.14 from doi:10.1007/BF01654302 that the argument of the square root is always negative but Mathematica complains about taking cubed roots when I try to give it numerical values respecting that (which I imagine comes from secret assumptions it made while solving the cubic characteristic equation).
finding eigenvectors seems rather difficult..
It might not be cryptic.
For a generic 3×3 matrix, the Cayley-Hamilton theorem yields a direct expression which, for your traceless case, reduces/collapses to $$ \lambda^3= \lambda {1\over 2}\operatorname{tr}\lambda^2 + 1\!\! 1 {1\over 3}\operatorname{tr}\lambda^3 , $$ so that $$ \operatorname{tr}\lambda^3 = 2 d^{abc} l_a l_b l_c \equiv m , $$ supplementing your $$ \operatorname{tr}\lambda=0, \qquad \operatorname{tr}\lambda^2 = 2l^2. $$
But these traces suffice to determine the eigenvalues n of λ, whose squares you are seeking, $\lambda_i=n_i^2$, $$ n_1+n_2+n_3=0, \qquad n_1^2+n_2^2+n_3^2=2 l^2, \qquad n_1^3+n_2^3+n_3^3=m= 3n_1 n_2 n_3 ~, $$ without fancy use of specific basis identities... The characteristic equation is already in depressed form with non-negative discriminant $(2l^2)^3 /2- 3m^2$.
For the irredeemably lazy amongst us, consider $l=(0,0,0,0,0,0,0,-\sqrt 3)$, so $2l^2=6$, m = 6, hence vanishing discriminant. Hence your $n_i$ are -1,-1, 2, and your $\lambda_i$ are 1,1,4.