Can saddle path not go through the origin in Ramsey model? In my case, the utility function is CEIS and discrete, the production fuction is $f(k_{t})=k_{t}^\alpha$, the budget constraint is $f(k_{t})+(1-\delta)k_{t}=c_{t} + k_{t+1}$. I use Jacobian matrix and Schur factorization to get the linearized policy function for consumption, therefore i can plot saddle path and unstable arms. In the end they look like below. However I read that saddle path must go through the origin, which is not right in my plot. 

So my question is: does saddle path always go through the origin?
I guess you already went trough the algebra below, but just for context, the problem you're trying to solve is 

$$
\max_{c}\sum_{t=0}^{+\infty}\beta^t u(c_t) \\
\text{s.t.}~~ f(k_t) + (1- \delta)k_t = c_t + k_{t+1} \tag{1}
$$

where $f(k_t) = k_t^\alpha$ and

$$
u(c_t) = \frac{c_t^{1-\gamma}}{1-\gamma} - 1 \tag{2}
$$

The problem in (1) can be cast into the two coupled equations

\begin{eqnarray}
u'(c_t) &amp;=&amp; \beta[1 + f'(k_{t+1}) - \delta]u'(c_{t+1}) \\
k_{t+1} &amp;=&amp; f(k_t) + (1-\delta)k_t - c_t \tag{3}
\end{eqnarray}

where $u'(x) = x^{-\gamma}$, and $f'(x) = \alpha x^{\alpha-1}$. These first of Eqns. (3) can be inverted to obtain an expression for $c_{t+1}$ in terms of $(k_t,c_t)$, leading to

\begin{eqnarray}
c_{t+1} &amp;=&amp; \beta^{1/\gamma}c_t [1 + \alpha[k_t^\alpha + (1-\delta)k_t - c_t]^{\alpha-1} - \delta]^{1/\gamma} \\
k_{t+1} &amp;=&amp; f(k_t) + (1-\delta)k_t - c_t \tag{4}
\end{eqnarray}

Which can be expressed as

$$
{\bf x}_{t+1} = {\bf F}({\bf x}_{t})~~~\mbox{with}~~~ {\bf x}_t = \left(\begin{array}{c}c_{t}\\k_{t}\end{array}\right) \tag{5}
$$

A fixed point ${\bf x}^*$ of the map ${\bf F}$ is such that

$$
{\bf x}^* = {\bf F}({\bf x}^*) \tag{6}
$$

that is, a point for which the system does not evolve. If you use $\gamma=2$, $\beta=0.9964$, $\alpha=0.36$, $\delta=0.025$ this point is (found by solving Eq. (6)),

$$
{\bf x}^* = \left(\begin{array}{c}c^*\\k^*\end{array}\right) = \left(\begin{array}{c}2.84829\\52.2808\end{array}\right)  \tag{7}
$$

which is clearly different from zero! You can linearize ${\bf F}$ around ${\bf x}^*$ and write the result as 

$$
{\bf y}_{t+1} = {\bf J}{\bf y}_t ~~~\mbox{where}~~~ {\bf y}_t = {\bf x}_{t} - {\bf x}^*, ~~~ {\bf J} = \left.\frac{\partial{\bf J}}{\partial {\bf x}}\right|_{{\bf x} = {\bf x}^*} \tag{8}
$$

Is this last system the one that has a saddle point at ${\bf y} = 0$