The SIR epidemic model presents three differential equations for three time-dependent variables — $S(t)$, $I(t)$, $R(t)$.
\begin{aligned} \frac{dS}{dt} &= -\beta SI\\ \frac{dI}{dt} &= \beta SI - \gamma I\\ \frac{dR}{dt} &= \gamma I\\ \end{aligned}
For the SIR model, I calculate the Jacobian matrix $J$, as follows
\begin{equation*} J = \begin{pmatrix} \frac{\partial S}{\partial S} & \frac{\partial S}{\partial I} & \frac{\partial S}{\partial R} \\ \frac{\partial I}{\partial S} & \frac{\partial I}{\partial I} & \frac{\partial I}{\partial R} \\ \frac{\partial R}{\partial S} & \frac{\partial R}{\partial I} & \frac{\partial R}{\partial R} \end{pmatrix} = \begin{pmatrix} - \beta I & - \beta S & 0 \\ \beta I & \beta S -\gamma & 0 \\ 0 & \gamma & 0 \end{pmatrix} \end{equation*}
In this case, the characteristic polynomial is given by
$$(-\lambda) \left( \lambda^2+(\beta I-\beta S+\gamma)\lambda+\beta I\gamma \right)$$
To determine the stability of the disease-free equilibrium we substitute $S = 1$ and $I = 0$ for $S$ and $I$. Hence, the characteristic polynomial becomes
$$-\lambda^2(\lambda-\beta+\gamma)$$
This has three roots: $\lambda_1$, $\lambda_2 =0$ and $\lambda_3=\beta-\gamma$. The last eigenvalue to be negative is required that $\beta<\gamma$. How can I determine the stability if two of the three eigenvalues are $0$? I have seen some solutions for $2 \times 2$ matrix, but I don't understand them. Is there any easy reason for being stable or unstable?
You do not need any Jacobians here. First note that the third equation is separate from the other two, hence it can be ignored (mathematically, you have $S+R+I=1$, so the variables are not independent). Second, look at the second equation: $$ \dot I=\beta SI-\gamma I=I(\beta S-\gamma). $$ We clearly have the growth for $I$ if $\beta S>\gamma$, hence at the initial time moment, when $S=1$, we get the condition that we have epidemics (the disease free equilibrium is unstable) if $\beta>\gamma$, which is equivalent to what you have found.