# The system of $x(t+1) = Ax$ growing and retaining stability possible?

This is about general equilibrium:

Suppose that $x(t)$ represents outputs of all sectors and parts of the whole economy - represented as matrix. How outputs evolve to $x(t+1)$ is determined by the matrix $A$ - $A$ representes how previous outputs are used as inputs to produce new ouputs - so $A$ can be said as table of outputs produced from inputs. $$x(t+1) = A \cdot x(t)$$

and it continues on to say that if the matrix $A$ has (absolute value of) all eigenvalues less than 1, the system is stable, while if not, is unstable.

The question is, I do get that when when $x$ is the eigenvector of $A$, it becomes unstable, but what if it's not - then can we still say that the system is unstable? (so what I am saying is, let's assume that the system will never have $x$ as eigenvector of $A$. Then what happens?) If so, can anyone show the proof of it?

(The linked question seems to assume that $x$ is an eigenvector of $A$ - I do not assume that at here.)

and the response goes like:

Let's take the simplest case, where $A$ is $n\times n$ and ${\bf R}^n$ has a basis consisting of eigenvectors of $A$ (in other words, the case where $A$ is diagonalizable). Let a basis of eigenvectors be $v_1,\dots,v_n$ with corresponding eigenvalues $b_1,\dots,b_n$, respectively. You can write $x(0)$ as a linear combination of the basis vectors, $$x(0)=c_1v_1+\cdots+c_nv_n$$ Then $$x(m)=c_1b_1^mv_1+\cdots+c_nb_n^mv_n$$ Now if, say, $|b_1|\gt1$, then $x(m)$ blows up as $m$ increases. $x(0)$ doesn't have to be the eigenvector $v_1$, it just has to have a nonzero component in the $v_1$ direction in order for there to be instability.

The question is then, how can the system of $x(t+1) = A \cdot x(t)$ grow in a stable fashion? As all eigenvalues of $A$ must be smaller than 1, the system seems to approach zero - not growing! Am I confused?

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anyone..........? – La Ventana Dec 9 '12 at 4:18

It says, "this is about general equilibrium". A system is in equilibrium if it doesn't change at all. In the current case, that means $x(t+1)=x(t)$, which means $Ax(t)=x(t)$, which means $x(t)$ is an eigenvector of $A$ with eigenvalue $b_1=1$.
Now one can ask whether the equilibrium is stable. What that means is, if the system is pushed a little bit away from equilibrium, does it return to equilibrium, or does it move farther away from equilibrium? If there's an eigenvalue exceeding $1$ in absolute value, then (using the formula for $x(m)$ in terms of eigenvalues and eigenvectors, you can see that) the equilibrium is unstable. If all the eigenvalues other than $b_1$ are less than $1$ in absolute value, then the equilibrium is stable. Things are a bit trickier if there are no eigenvalues bigger than $1$ in absolute value, but there are one or more eigenvalues other than $b_1$ that equal $1$ in absolute value; the system might orbit around the equilibrium point, never going far from it but not converging to it, or, if there is a repeated eigenvalue, the equilibrium will be unstable.
If one or more eigenvalues are equal to $1$, then your system stabilizes as $m$ increases.