The trace is the sum of the eigenvalues (counted by algebraic multiplicity). If the trace is $0$, at least one eigenvalue must have nonnegative real part. If some eigenvalue has positive real part, the critical point is unstable. However, it is possible that all eigenvalues have real part $0$, e.g. the matrix could be $$\pmatrix{0 & 1 & 0 & 0\cr -1 & 0 & 0 & 0\cr 0 & 0 & 0 & -1\cr 0 & 0 & 1 & 0\cr}$$ in which case the linearized system is stable but not asymptotically stable, but the nonlinear system could be either stable or unstable, depending on the nonlinear terms.