# Additional material - trajectory

## The trajectory around the steady state

For the moment, we have established that the model converges towards a unique steady state following a saddle path.
**Note:** This means that there is only one combination of initial capital and consumption, $(k_{0}, c_{0})$, such that the economy converges towards it.
Any other initial value consumption at $t=0$ (capital is pre-determined and thus we *cannot* change it) has a diverging trajectory.

We now compute the exact behaviour of $(k_{t}, c_{t})$ around the steady state.

The model is highly non-linear, so we study a simpler linearised version around the steady state. First, let’s approximate the dynamics of capital and consumption around the steady state.

### The approximation around the steady state

The behaviour of the dynamical system around the steady state can be approximated using a first-order Taylor expansion around it. In that sense:

$$\begin{pmatrix} c_{t+1} - \bar{c} \\\ k_{t+1} - \bar{k} \end{pmatrix} \approx \bar{A} \begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix}.$$

Next, we solve this equation of difference equations, this is, we obtain $c_t = \mathcal{C}(c_0, k_0, t), k_t = \mathcal{K}(c_0, k_0, t).$

### Diagonalising the matrix: eigenvalues and eigenvectors

**Note:** based on the notes by Benjamin Moll.

To simplify the notation, let $y_{t} = \begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} -\bar{k} \end{pmatrix}.$ Hence, our problem becomes: $$y_{t+1} = \bar{A} y_{t}.$$

It $\bar{A}$ were a diagonal matrix, the solution to the system would be straightforward. In fact, if that were the case it would have been (I change variables to avoid confusion):

$$\begin{pmatrix} \phi_{t+1} \\ \omega_{t+1} \end{pmatrix} = \begin{pmatrix} a_{1,1} & 0 \\\ 0 & a_{2,2} \end{pmatrix} \begin{pmatrix} \phi_{t} \\\ \omega_{t} \end{pmatrix}.$$ So, it is clear that $\phi_{t+1} = a_{1,1} \phi_{t} \implies \phi_{t} = a_{1,1}^{t} \phi_{0}$ and we would find a similar expression for $\omega.$

$\bar{A}$ is not diagonal, but we can apply a spectral decomposition and obtain an equivalent system governed by a diagonal matrix. In particular, we are looking for an $2 \times 2$ invertible matrix $X$ such that $ X^{-1} \bar{A} X = \Lambda$, where $\Lambda$ is diagonal. We will then apply the following transformation to our system (pre-multiplying by $\color{red}{X^{-1}}$ on both sides and multiplying by $\color{green}{XX^{-1}}$ on the right-hand side). $$ \color{red}{X^{-1}} y_{t+1} = \color{red}{X^{-1}} \bar{A} \color{green}{(X X^{-1})} y_{t} = X^{-1} \bar{A} X (X^{-1} y_{t}) =$$ $$ = \Lambda X^{-1} y_{t}.$$

Denote $z \equiv X^{-1} y$. Thus, the system becomes:

$$z_{t+1} = \Lambda z_{t}.$$
Since $\Lambda$ is diagonal, the solutions are of the form:
$$z_{t} = \Lambda^{t} z_{0}.$$
**Note:** in general, the power of a matrix is complex.
However, for a *diagonal matrix* it is simply the power of each component.

Once we have the solution for the transformed system we must undo the transformation: in fact, we do not care about the evolution of $z$. This step involves using that $X^{-1} y = z \implies y = X z$.

#### The matrix $\Lambda$

In a spectral decomposition, the diagonal matrix $\Lambda$ consists of the matrix whose entries are its eigenvalues. Thus (we have a $2 \times 2$ system), $$\Lambda = \begin{pmatrix} \lambda_{1} & 0 \\\ 0 & \lambda_{2} \end{pmatrix}.$$

#### The matrix $X$

The matrix $X$ has as columns the eigenvectors of $\bar{A}.$
There is one eigenvector associated with each eigenvalue.
**Note:** it is important to correctly relate each eigenvector to its eigenvalue.

To find an eigenvector associated with $\lambda_{1}$ (*there are infinitely many possible eigenvectors, we only want one*) we solve the following system:

$$\bar{A} \begin{pmatrix} v_{1} \\\ v_{2} \end{pmatrix} = \lambda_{1} \begin{pmatrix} v_{1} \\\ v_{2} \end{pmatrix} \quad \mathrm{or} \quad (\bar{A} - \lambda_{1} \mathbb{I}) \begin{pmatrix} v_{1} \\\ v_{2} \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \end{pmatrix}.$$

The matrix $X$ then becomes:

$$X = \begin{pmatrix} v_{1,1} & v_{2,1} \\\ 1 & 1 \end{pmatrix}.$$

### Solving the system

We begin with the solution for the transformed variable $z.$ We already know it takes the form:

$$z_{t} = \Lambda^{t} z_{0} \quad \mathrm{or} \quad \begin{pmatrix} z_{t}^{1} \\\ z_{t}^{2} \end{pmatrix} = \begin{pmatrix} \lambda_{1}^{t} & 0 \\\ 0 & \lambda_{2}^{t} \end{pmatrix} \begin{pmatrix} v_{0}^{1} \\\ v_{0}^{2} \end{pmatrix}.$$

Next, we reverse the transformation, this is, we obtain the dynamics of $y: y_{t} = X z_{t}.$ Therefore, we obtain:

$$y_{t} = X z_{t} = X \Lambda^{t} z_{0} = \begin{pmatrix} v_{1,1} & v_{2,1} \\\ v_{1,2} & v_{2,2} \end{pmatrix} \begin{pmatrix} \lambda_{1}^{t} & 0 \\\ 0 & \lambda_{2}^{t} \end{pmatrix} \begin{pmatrix} z_{0}^{1} \\\ z_{0}^{2} \end{pmatrix}.$$

Alternatively, multiplying the matrices and recalling that $y_{t}=\begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix}$ :

$$\begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix} = z_{0}^{1} \lambda_{1}^{t} \begin{pmatrix} v_{1,1} \\\ v_{1,2} \end{pmatrix} + z_{0}^{2} \lambda_{2}^{t} \begin{pmatrix} v_{2,1} \\\ v_{2,2} \end{pmatrix}.$$

The eigenvalues appear clearly in the solution.
**Remember** that we know that one eigenvalue is bigger than one, while the second lies within the unit circle.
Without loss of generality, assume that $\lambda_{1} \in (-1,1)$ and $\lambda_{2} > 1.$
According to the solution before, this implies an explosive behaviour: $\lambda_{2} > 1 \implies \lim_{t \rightarrow +\infty} \lambda_{2}^{t} = +\infty.$
To have well-behaved dynamics we must impose $z_{0}^{2} = 0$, which eliminates explosive behaviour.

After imposing $z_{0}^{2} = 0$ the system becomes:

$$\begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix} = z_{0}^{1} \lambda_{1}^{t} \begin{pmatrix} v_{1,1} \\\ v_{1,2} \end{pmatrix}.$$

#### Closing the model with initial values

We know that the economy begins with a level of capital $k_{t=0} = k_{0} > 0.$ Substituting $t=0$ in the equation above, and focusing on capital, we obtain:

$$k_{0} - \bar{k} = z_{0}^{1} v_{1,2} \implies \color{red}{z_{0}^{1} = \frac{k_{0} - \bar{k}}{v_{1,2}}}.$$ Therefore, we can substitute the value for $z_{0}^{1}$ to obtain the dynamics of capital:

$$k_{t} - \bar{k} = \color{red}{z_{0}^{1}} \lambda_{1}^{t} v_{1,2} = \left(k_{0} - \bar{k} \right) \lambda^{t}.$$

Similarly,for consumption we have:

$$c_{t} - \bar{c} = \color{red}{z_{0}^{1}} \lambda_{1}^{t} v_{1,1} = \frac{v_{1,1}}{v_{1,2}} \lambda_{1}^{t} (k_{0} - \bar{k}).$$ Finally, the initial value of consumption $c_{0}$ that puts the economy in the saddle path is obtained by setting $t=0$ in the previous equation:

$$c_{0} - \bar{c} = \frac{v_{1,1}}{v_{1,2}} (k_{0} - \bar{k}).$$