# 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:

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

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