# Stability of a system that has (Jacobian-like) matrix with eigenvalue of less than 1 that has $x$ as non-eigenvector

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.)

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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.
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