 Solution

Write $R^2={\sum }_i{x}_i^2$ and impose tracelessness by a Lagrange multiplier $\zeta$. Differentiating $V-\zeta {\sum }_i{x}_i$ givesFor $b\ne 0$, each eigenvalue is a real root of the same nonzero cubic polynomial, so at most three distinct values occur at any stationary point, in particular at any existing minimum. If the multiplicities are $n_1,\dots ,n_k$, the unbroken subgroup is the block centralizerIts connected Lie algebra is ${⨁}_{\alpha }\mathfrak{su}(n_{\alpha })\oplus \mathfrak{u}(1{)}^{k-1}$. The partitions allowed by this upper bound are $5$, $4+1$, $3+2$, $3+1+1$, and $2+2+1$. These are candidate patterns, not a proof that every one is a stable minimum for given couplings. For three distinct roots, the absent quadratic coefficient also implies that their unweighted sum is zero; tracelessness imposes an additional multiplicity-weighted relation.

The unrestricted printed claim has an exception. If $b=0$, $a>0$ and $m>0$, every traceless matrix with $R^2=m^2/(2a)$ is a global minimum. For examplehas five distinct eigenvalues and attains that minimum. Its unbroken subgroup is the maximal torus $U(1{)}^4$. The cubic argument therefore requires a nonzero $b$, or another condition excluding the identically vanishing stationarity polynomial. Couplings for which the potential is unbounded below have no global minimum to classify.