 Solution

For real $\lambda \ge 0$, draw $y=\lambda +\beta$ and $y=-\gamma e^{-\lambda }$. The first is nonnegative and increasing, while the second is negative and approaches zero from below. They cannot intersect on the nonnegative real axis. At zero the characteristic function equals $\beta +\gamma >0$, so a stationary zero-root crossing is excluded as well.

Consequently a loss of stability from a stable regime cannot occur by a real eigenvalue crossing through zero. It must involve a conjugate pair crossing the imaginary axis at nonzero frequency, yielding oscillatory instability. This does not mean that every characteristic root is nonreal: negative real roots may exist and are harmless for stability. The graph and imaginary-axis conditions are shown together below.