# Meaning of a zero in an eigenvector for the solution of a system of difference equations

In a system of first order difference equations, I get the following solution

$\begin{bmatrix} x_{1,t} \\ x_{2,t}\end{bmatrix} = \begin{bmatrix} \mathcal{A} \\ 1 \end{bmatrix} \lambda_1^t + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \lambda_{2}^t$

For some $\mathcal{A} \neq 0$.

I know I need one eigenvalue to be less than one and the to be other greater than one in absolute terms for the system to be saddle path stable, which is what I want.

My question is about the fact that I get a zero in the second eigenvector. Does this have any particular meaning?

For example (please correct me if I am wrong), if $|\lambda_1| < 1$ and $|\lambda_2| > 1$, $x_{2,t}$ will stabilise and $x_{1,t}$ will go to zero as $t\rightarrow 0$.

But if the converse is true, i.e. $|\lambda_1| > 1$ and $|\lambda_2| < 1$, then although $x_{2,t}$ continues to be stable, $x_{1,t}$ will now explode.

This would not happen if that zero wasn't there, correct? I would always have both $x_{1,t}$ and $x_{2,t}$ saddle path stable as long as one of the eigenvalues was greater than one and the other less than one in absolute terms. Is that correct?

PS: how do I call that format of solution I wrote on top? Is this the Jordan form?

Your equation implies $x_{2,t}=\lambda_1^t+\lambda_2^t$; if $\lambda_1,\lambda_2$ are on opposite sides of $1$ but both distinct from it, then as $t\to\infty$ the value of $x_{2,t}$ will rapidly tend to $\lambda_i^t$ where $\lambda_i>1$. This means it does not stabilise, but goes to infinity. If you are talking about stabilising, then maybe you are implicitly rescaling vectors? For the first variable you get $x_{1,t}=\mathcal A\lambda_1^t$ which does what it suggests.